Split (1)

train · 300 rows

id stringlengths 12 12	source stringlengths 76 76	problem stringlengths 59 2.04k	solutions listlengths 0 11
USAMO-1972-1	https://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_1	The symbols $(a,b,\ldots,g)$ and $[a,b,\ldots, g]$ denote the greatest common divisor and least common multiple, respectively, of the positive integers $a,b,\ldots, g$. For example, $(3,6,18)=3$ and $[6,15]=30$. Prove that \[ \frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)} \] .	[ "Consider an arbitrary prime \$p\$. Let \$p^\\alpha\$, \$p^\\beta\$, and \$p^\\gamma\$ be the greatest powers of \$p\$ that divide \$a\$, \$b\$, and \$c\$. WLOG let \$\\alpha \\leq \\beta \\leq \\gamma\$.\n\nExamining each factor in the equation, we see that the largest power of \$p\$ that divides the left hand side is \$2\\gamma -(\\beta+\\gamma+\\gamma) = -\\beta\$, and the largest power of \$p\$ that divides the right hand side is 2\\alpha -(\$\\alpha + \\beta + \\alpha) = -\\beta\$. Since every prime has the same power in both expressions, the expressions are equal. \$\\blacksquare\$", "Let \$p = (a, b, c)\$, \$pq = (a, b)\$, \$pr = (b, c)\$, and \$ps = (c, a)\$. Then it follows that \$q, r, s\$ are pairwise coprime and \$a = pqsa'\$, \$b = pqrb'\$, and \$c = prsc'\$, with \$a', b', c'\$ pairwise coprime as well. Then, we wish to show\n\n\\[\n\\frac{(pqrsa'b'c')^2}{(pqrsa'b')(pqrsb'c')(pqrsc'a')} = \\frac{p^2}{(pq)(pr)(ps)},\n\\]\n\nwhich can be checked fairly easily.", "Note: This solution is more of a \"bashy\" solution, and is easier to think of than the first 2 solutions.\n\nDividing both sides of the equation by \$(a,b,c)^2\$, and then multiplying both sides by \$[a,b][b,c][c,a]\$ gives us\n\n\\[\n\\left(\\frac{[a,b,c]}{(a,b,c)}\\right)^2 = \\frac{[a,b][b,c][c,a]}{(a,b)(b,c)(c,a)}.\n\\]\n\nNow, we look at the prime factorisations of \$a,b,c\$. Let\n\n\\[\na=2^{x_1}\\cdot 3^{x_2}\\cdot 5^{x_3}\\cdots,\n\\]\n\n\\[\nb=2^{y_1}\\cdot 3^{y_2}\\cdot 5^{y_3}\\cdots,\n\\]\n\n\\[\nc=2^{z_1}\\cdot 3^{z_2}\\cdot 5^{z_3}\\cdots,\n\\]\n\nwhere all of the \$x_1, y_1, z_1\$ are nonnegative integers (they could be 0). Thus, we have\n\n\\[\n[a,b,c]=2^{\\max(x_1,y_1,z_1)} \\cdot 3^{\\max(x_2,y_2,z_2)} \\cdot 5^{\\max(x_3,y_3,z_3)} \\cdots\n\\]\n\nand\n\n\\[\n(a,b,c)=2^{\\min(x_1,y_1,z_1)} \\cdot 3^{\\min(x_2,y_2,z_2)} \\cdot 5^{\\min(x_3,y_3,z_3)} \\cdots.\n\\]\n\nNow, we see that \$\\left(\\frac{[a,b,c]}{(a,b,c)}\\right) = 2^{\\max(x_1,y_1,z_1) - \\min(x_1,y_1,z_1)} \\cdot 3^{\\max(x_2,y_2,z_2) - \\min(x_2,y_2,z_2)} \\cdot 5^{\\max(x_3,y_3,z_3) - \\min(x_3,y_3,z_3)} \\cdots.\$ Squaring the LHS will just double all the exponents on the RHS.\n\nAlso, we have \$[a,b]=2^{\\max(x_1,y_1)} \\cdot 3^{\\max(x_2,y_2)} \\cdot 5^{\\max(x_3,y_3)} \\cdots\$. We can also build similar equations for \$[b,c],[c,a],(a,b),(b,c)(c,a)\$, using 2 of \$x,y,z\$ and either the minimum of the exponents or the maximum. We see that when we multiply \$[a,b]\$, \$[b,c]\$, and \$[a,c]\$ together, the exponents of the prime factorization will be \$\\max(x_i,y_i)+ \\max(y_i,z_i)+ \\max(z_i,x_1)\$, where \$i\$ is chosen for the \$i\$'th prime. When we multiply \$(a,b)\$, \$(b,c)\$, and \$(a,c)\$ together, the exponents of the prime factorization will be \$\\min(x_i,y_i)+ \\min(y_i,z_i)+ \\min(z_i,x_1)\$, where \$i\$ is chosen for the \$i\$'th prime.\n\nThus, when we divide the former by the latter, we have \$\\max(x_i,y_i)+ \\max(y_i,z_i)+ \\max(x_i,z_i) - (\\min(x_i,y_i)+ \\min(y_i,z_i)+ \\min(x_i,z_i))\$. We wish to prove that this is equal to the exponents we got from \$([a,b,c]/(a,b,c))^2\$.\n\nIn other words, this problem is now down to proving that\n\n\\[\n2(\\max(x_i,y_i,z_i) - \\min(x_i,y_i,z_i)) = \\max(x_i,y_i)+ \\max(y_i,z_i)+ \\max(x_i,z_i) - (\\min(x_i,y_i)+ \\min(y_i,z_i)+ \\min(x_i,z_i))\n\\]\n\nfor any nonnegative integers \$x_i, y_i, z_i\$. WLOG, let \$x_i \\le y_i \\le z_i\$. Thus, computing maximums and minimums, this equation turns into\n\n\\[\n2(z_i - x_i) = y_i + z_i + z_i - x_i - y_i - x_i.\n\\]\n\nFinally, canceling out the \$y_i\$'s and collecting like terms, we see the RHS is \$2z_i - 2x_i\$, which is equal to the LHS. Thus, this equation is true, and we are done. \$\\square\$", "Let\n\n\\[\na=dzxA\n\\]\n\n\\[\nb=dxyB\n\\]\n\n\\[\nc=dyzC\n\\]\n\nwhere \$\\gcd(A,B,C)=\\gcd(x,y,z)=1\$. (Existence proof missing.) Then, \$\\operatorname{lcm}(a,b,c)^2=(dxyzABC)^2\$ and \$\\operatorname{lcm}(a,b)=dxyzAB\$. Using this cyclically, we have\n\n\\[\n\\text{LHS}=\\frac{d^2x^2y^2z^2A^2B^2C^2}{(dxyzAB)(dxyzBC)(dxyzCA)}=\\frac{1}{dxyz}.\n\\]\n\nNow, note that \$\\gcd(a,b,c)^2=d^2\$ and \$\\gcd(a,b)=dx\$. Using this cyclically, we have\n\n\\[\n\\text{RHS}=\\frac{d^2}{(dx)(dy)(dz)}=\\frac{1}{dxyz}.\n\\]\n\n-brainiacmaniac31\n\n## Note\n\nIf one thinks of the problem from a set theory vantage point, it is quite intuitive to recognize from PIE:\n\n\\[\n\\text{lcm}(a,b,c) = \\frac{a \\cdot b \\cdot c \\cdot \\text{gcd}(a,b,c)}{\\text{gcd}(a,b) \\cdot \\text{gcd}(b,c) \\cdot \\text{gcd}(c,a)}\n\\]\n\nwhere \$\\text{lcm}(a,b,c)\$ can be thought of as the union, \$a \\cdot b \\cdot c\$ can be thought of as the raw addition of sets, and the \$\\text{gcd}\$ function can be thought of as intersections." ]
USAMO-1972-2	https://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_2	A given tetrahedron $ABCD$ is isosceles, that is, $AB=CD, AC=BD, AD=BC$. Show that the faces of the tetrahedron are acute-angled triangles.	[ "Suppose \$\\triangle ABD\$ is fixed. By the equality conditions, it follows that the maximal possible value of \$BC\$ occurs when the four vertices are coplanar, with \$C\$ on the opposite side of \$\\overline{AD}\$ as \$B\$. In this case, the tetrahedron is not actually a tetrahedron, so this maximum isn't actually attainable.\n\nFor the sake of contradiction, suppose \$\\angle ABD\$ is non-acute. Then, \$(AD)^2\\geq (AB)^2+(BD)^2\$. In our optimal case noted above, \$ACDB\$ is a parallelogram, so\n\n\\[\n\\begin{align} 2(BD)^2 + 2(AB)^2 &= (AD)^2 + (CB)^2 \\\\ &= 2(AD)^2 \\\\ &\\geq 2(BD)^2+2(AB)^2. \\end{align}\n\\]\n\nHowever, as stated, equality cannot be attained, so we get our desired contradiction.", "It's not hard to see that the four faces are congruent from SSS Congruence. Without loss of generality, assume that \$AB\\leq BC \\leq CA\$. Now assume, for the sake of contradiction, that each face is non-acute; that is, right or obtuse. Consider triangles \$\\triangle ABC\$ and \$\\triangle ABD\$. They share side \$AB\$. Let \$k\$ and \$l\$ be the planes passing through \$A\$ and \$B\$, respectively, that are perpendicular to side \$AB\$. We have that triangles \$ABC\$ and \$ABD\$ are non-acute, so \$C\$ and \$D\$ are not strictly between planes \$k\$ and \$l\$. Therefore the length of \$CD\$ is at least the distance between the planes, which is \$AB\$. However, if \$CD=AB\$, then the four points \$A\$, \$B\$, \$C\$, and \$D\$ are coplanar, and the volume of \$ABCD\$ would be zero. Therefore \$CD>AB\$. However, we were given that \$CD=AB\$ in the problem, which leads to a contradiction. Therefore the faces of the tetrahedron must all be acute.", "Let \$\\vec{a} = \\overrightarrow{DA}\$, \$\\vec{b} = \\overrightarrow{DB}\$, and \$\\vec{c} = \\overrightarrow{DC}\$. The conditions given translate to\n\n\\[\n\\begin{align} \\vec{a}\\cdot\\vec{a} &= \\vec{b}\\cdot\\vec{b} + \\vec{c}\\cdot\\vec{c} - 2(\\vec{b}\\cdot\\vec{c}) \\\\ \\vec{b}\\cdot\\vec{b} &= \\vec{c}\\cdot\\vec{c} + \\vec{a}\\cdot\\vec{a} - 2(\\vec{c}\\cdot\\vec{a}) \\\\ \\vec{c}\\cdot\\vec{c} &= \\vec{a}\\cdot\\vec{a} + \\vec{b}\\cdot\\vec{b} - 2(\\vec{a}\\cdot\\vec{b}) \\end{align}\n\\]\n\nWe wish to show that \$\\vec{a}\\cdot\\vec{b}\$, \$\\vec{b}\\cdot\\vec{c}\$, and \$\\vec{c}\\cdot\\vec{a}\$ are all positive. WLOG, \$\\vec{a}\\cdot\\vec{a}\\geq \\vec{b}\\cdot\\vec{b}, \\vec{c}\\cdot\\vec{c} > 0\$, so it immediately follows that \$\\vec{a}\\cdot\\vec{b}\$ and \$\\vec{a}\\cdot\\vec{c}\$ are positive. Adding all three equations,\n\n\\[\n\\vec{a}\\cdot\\vec{a} + \\vec{b}\\cdot\\vec{b} + \\vec{c}\\cdot\\vec{c} = 2(\\vec{a}\\cdot\\vec{b} + \\vec{a}\\cdot\\vec{c} + \\vec{b}\\cdot\\vec{c})\n\\]\n\nIn addition,\n\n\\[\n\\begin{align} (\\vec{a} - \\vec{b} - \\vec{c})\\cdot(\\vec{a} - \\vec{b} - \\vec{c})&\\geq 0 \\\\ \\vec{a}\\cdot\\vec{a} + \\vec{b}\\cdot\\vec{b} + \\vec{c}\\cdot\\vec{c}&\\geq 2(\\vec{a}\\cdot\\vec{b} + \\vec{a}\\cdot\\vec{c} - \\vec{b}\\cdot\\vec{c}) \\\\ 2(\\vec{a}\\cdot\\vec{b} + \\vec{a}\\cdot\\vec{c} + \\vec{b}\\cdot\\vec{c})&\\geq 2(\\vec{a}\\cdot\\vec{b} + \\vec{a}\\cdot\\vec{c} - \\vec{b}\\cdot\\vec{c}) \\\\ \\vec{b}\\cdot\\vec{c}&\\geq 0 \\end{align}\n\\]\n\nEquality could only occur if \$\\vec{a} = \\vec{b} + \\vec{c}\$, which requires the vectors to be coplanar and the original tetrahedron to be degenerate.", "Suppose for the sake of contradiction that \$\\angle BAC\$ is not acute. Since all three sides of triangles \$BAC\$ and \$CDB\$ are congruent, those two triangles are congruent, meaning \$\\angle BDC=\\angle BAC>90^{\\circ}\$. Construct a sphere with diameter \$BC\$. Since angles \$BAC\$ and \$BDC\$ are both not acute, \$A\$ and \$D\$ both lie on or inside the sphere. We seek to make \$AD=BC\$ to satisfy the conditions of the problem. This can only occur when \$AD\$ is a diameter of the sphere, since both points lie on or inside the sphere. However, for \$AD\$ to be a diameter, all four points must be coplanar, as all diameters intersect at the center of the sphere. This would make tetrahedron \$ABCD\$ degenerate, creating a contradiction. Thus, all angles on a face of an isosceles tetrahedron are acute.", "Proof by contradiction: Assume at least one of the tetrahedron's faces are obtuse. WLOG, assume \$\\angle BAC\$ is an obtuse angle. Using SSS congruence to prove that all four faces of the tetrahedron are congruent also shows that the angles surrounding point \$A\$ are congruent to angles in triangle \$ABC\$; namely, \$\\angle BAD\$ is congruent to \$\\angle ABC\$, and \$\\angle CAD\$ is congruent to \$\\angle ACB\$. Since the internal angles of triangle \$ABC\$ must add to 180 degrees, so do the angles surrounding point A. Now lay triangle \$ABC\$ on a flat surface. A diagram would make it clear that from the perspective of an aerial view, the \"apparent\" measures of \$\\angle BAD\$ and \$\\angle CAD\$ (which are most likely distorted visually, assuming these angles stick up vertically from the flat surface on which triangle \$ABC\$ lies) can never exceed the true measures of those angles (equality happens when these angles also lie flat on top of triangle \$ABC\$). This means that these two angles can never join to form side AD (because \$\\angle BAC\$ is more than the sum of \$\\angle BAD\$ and \$\\angle CAD\$ - a direct consequence of the facts that \$\\angle BAC\$ is obtuse and all three angles add up to 180 degrees), so the tetrahedron with obtuse triangle faces is impossible.", "Lemma: given triangle \$ABC\$ and the midpoint of \$BC\$, which we will call \$M\$, we can say that if \$AM > \\frac{BC}{2}\$, then \$\\angle A < 90\$. Proof: Since \$M\$ is the midpoint of \$BC\$, \$BM = MC = \\frac{BC}{2}\$. Since it is given that \$AM > \\frac{BC}{2}\$, we can substitute \$\\frac{BC}{2}\$ to get two inequalities:\n\n\\[\nAM > CM, \\quad AM > BM.\n\\]\n\nThe above inequalities imply that \$\\angle C > \\angle CAM\$ and \$\\angle B > \\angle BAM\$. Adding these inequalities and simplifying the RHS, we have that \$\\angle C + \\angle B > \\angle A\$. Adding \$\\angle A\$ to both sides, replacing the LHS with \$180\$ and dividing by \$2\$ gets us that \$\\angle A < 90\$. This is our desired inequality, so we are done. Note that all faces of this tetrahedron are congruent, by SSS. In particular, we will use that \$\\triangle ABD \\cong \\triangle BAC\$. WLOG, assume that \$\\angle ADB\$ is the largest angle in triangle \$ABD\$. Because \$\\triangle ABD \\cong \\triangle BAC\$, the median from \$D\$ to \$AB\$ is equal length to the median from \$C\$ to \$AB\$. These points meet at \$E\$, the midpoint of \$AB\$. By the triangle inequality,\n\n\\[\nDE + CE > CD.\n\\]\n\nBy substituting \$CD\$ with \$AB\$ (this is a given in the problem) and \$CE\$ with \$DE\$, and then dividing by \$2\$, we get that\n\n\\[\nDE > \\frac{AB}{2}.\n\\]\n\nBy the lemma we showed at the start, this implies that \$\\angle ADB < 90\$, and since we said that \$\\angle ADB\$ was the largest angle, triangle \$ADB\$ must be acute. Since all of the faces of this tetrahedron are congruent, then, all of the faces must be acute." ]
USAMO-1972-3	https://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_3	A random number selector can only select one of the nine integers 1, 2, ..., 9, and it makes these selections with equal probability. Determine the probability that after $n$ selections ($n>1$), the product of the $n$ numbers selected will be divisible by 10.	[ "For the product to be divisible by 10, there must be a factor of 2 and a factor of 5 in there.\n\nThe probability that there is no 5 is \$\\left( \\frac{8}{9}\\right)^n\$.\n\nThe probability that there is no 2 is \$\\left( \\frac{5}{9}\\right)^n\$.\n\nThe probability that there is neither a 2 nor 5 is \$\\left( \\frac{4}{9}\\right)^n\$, which is included in both previous cases.\n\nThe only possibility left is getting a 2 and a 5, making the product divisible by 10. By complementarity and principle of inclusion-exclusion, the probability of that is \$1- \\left( \\left( \\frac{8}{9}\\right)^n + \\left( \\frac{5}{9}\\right)^n - \\left( \\frac{4}{9}\\right)^n\\right)=\\boxed{1-(8/9)^n-(5/9)^n+(4/9)^n}\$.", "Define \$a_n\$ as the probability that the product is divisible by \$10\$ after selection \$n\$. Likewise, define \$b_n\$ and \$c_n\$ with divisibility by \$2\$ and \$5\$, respectively. Define \$d_n\$ to be the chance of dividing neither \$2\$ nor \$5\$ (and thus not \$10\$ either) similarly.\n\nIt is clear that \$d_n=\\left(\\frac{4}{9}\\right)^n\$. Now we can define other recursive formulas:\n\nWe are able to reach a \$b\$ state via selecting a non-\$5\$ from a \$b\$ state and selecting an even number from a \$d\$ state. Thus its formula is \$b_n=\\frac{8}{9}b_{n-1}+\\frac{4}{9}d_{n-1}\$.\n\nWe are able to reach a \$c\$ state via selecting a non-even number from a \$c\$ state and selecting a \$5\$ from a \$d\$ state. Thus its formula is \$c_n=\\frac{5}{9}c_{n-1}+\\frac{1}{9}d_{n-1}\$.\n\nFinally, to reach an \$a\$ state, we can select a \$5\$ from a \$b\$ state and select an even number from a \$c\$ state. We can also reach \$a_n\$ from \$a_{n-1}\$ because of the fact that once the product is divisible by \$10\$, it will always be divisible by \$10\$ regardless of the following selections. Thus its formula is \$a_n=a_{n-1}+\\frac{1}{9}b_{n-1}+\\frac{4}{9}c_{n-1}\$.\n\nFor our formula for \$b_n\$, we can substitute to find that \$b_n=\\frac{8}{9}b_{n-1}+\\left(\\frac{4}{9}\\right)^n\$. Solving this recursion yields \$b_n=\\left(\\frac{8}{9}\\right)^n-\\left(\\frac{4}{9}\\right)^n\$.\n\nFor our formula for \$c_n\$, we can substitute to find that \$c_n=\\frac{5}{9}c_{n-1}+\\frac{1}{9}\\cdot\\left(\\frac{4}{9}\\right)^{n-1}\$. Solving this recursion yields \$c_n=\\left(\\frac{5}{9}\\right)^n-\\left(\\frac{4}{9}\\right)^n\$.\n\nFinally, substituting the values into the \$a_n\$ formula yields \$a_n=a_{n-1}+\\frac{1}{9}\\left(\\left(\\frac{8}{9}\\right)^{n-1}+4\\cdot\\left(\\frac{5}{9}\\right)^{n-1}-5\\cdot\\left(\\frac{4}{9}\\right)^{n-1}\\right)\$. Solving this recursion yields the final answer, \$\\boxed{1-\\left(\\frac{8}{9}\\right)^n-\\left(\\frac{5}{9}\\right)^n+\\left(\\frac{4}{9}\\right)^n}\$.\n\n~eevee9406" ]
USAMO-1972-4	https://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_4	Let $R$ denote a non-negative rational number. Determine a fixed set of integers $a,b,c,d,e,f$, such that for every choice of $R$, \[ \left\|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}\right\|<\|R-\sqrt[3]{2}\| \]	[ "Note that when \$R\$ approaches \$\\sqrt[3]{2}\$, \$\\frac{aR^2+bR+c}{dR^2+eR+f}\$ must also approach \$\\sqrt[3]{2}\$ for the given inequality to hold. Therefore\n\n\\[\n\\lim_{R\\rightarrow \\sqrt[3]{2}} \\frac{aR^2+bR+c}{dR^2+eR+f}=\\sqrt[3]{2}\n\\]\n\nwhich happens if and only if\n\n\\[\n\\frac{a\\sqrt[3]{4}+b\\sqrt[3]{2}+c}{d\\sqrt[3]{4}+e\\sqrt[3]{2}+f}=\\sqrt[3]{2}\n\\]\n\nWe cross multiply to get \$a\\sqrt[3]{4}+b\\sqrt[3]{2}+c=2d+e\\sqrt[3]{4}+f\\sqrt[3]{2}\$. It's not hard to show that, since \$a\$, \$b\$, \$c\$, \$d\$, \$e\$, and \$f\$ are integers, then \$a=e\$, \$b=f\$, and \$c=2d\$.\n\nNote, however, that this is a necessary but insufficient condition. For example, we must also have \$a^2<2bc\$ to ensure the function does not have any vertical asymptotes (which would violate the desired property). A simple search shows that \$a=0\$, \$b=2\$, and \$c=2\$ works." ]
USAMO-1972-5	https://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_5	A given convex pentagon $ABCDE$ has the property that the area of each of the five triangles $ABC$, $BCD$, $CDE$, $DEA$, and $EAB$ is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the above area property. \[ [asy] size(120); defaultpen(fontsize(10)); pair A=dir(90), B=dir(90-72), C=dir(90-272), D=dir(90-372), E=dir(90-4*72); draw(A--B--C--D--E--cycle); draw(A--C--E--B--D--cycle); label("A",A,A);label("B",B,B);label("C",C,C);label("D",D,D);label("E",E,E); [/asy] \]	[ "Lemma: Convex pentagon \$A_0A_1A_2A_3A_4\$ has the property that \$[A_0A_1A_2] = [A_1A_2A_3] = [A_2A_3A_4] = [A_3A_4A_0] = [A_4A_0A_1]\$ if and only if \$\\overline{A_{n - 1}A_{n + 1}}\\parallel\\overline{A_{n - 2}A_{n + 2}}\$ for \$n = 0, 1, 2, 3, 4\$ (indices taken mod 5).\n\nProof: For the \"only if\" direction, since \$[A_0A_1A_2] = [A_1A_2A_3]\$, \$A_0\$ and \$A_3\$ are equidistant from \$\\overline{A_1A_2}\$, and since the pentagon is convex, \$\\overline{A_0A_3}\\parallel\\overline{A_1A_2}\$. The other four pairs of parallel lines are established by a symmetrical argument, and the proof for the other direction is just this, but reversed.\n\n\\[\n[asy] size(120); defaultpen(fontsize(10)); pathpen = black; pair A=MP(\"A\",dir(90),dir(90)), B=MP(\"B\",dir(90-72),dir(90-72)), C=MP(\"C\",dir(90-272),dir(90-272)), D=MP(\"D\",dir(90-372),dir(90-372)), E=MP(\"E\",dir(90-472),dir(90-472)); D(A--B--C--D--E--cycle); D(A--C--E--B--D--cycle); pair Ap = MP(\"A'\",IP(B--D,C--E),dir(270)), Bp = MP(\"B'\",IP(A--D,C--E),dir(270-72)), Cp = MP(\"C'\",IP(A--D,B--E),dir(270-722)), Dp = MP(\"D'\",IP(A--C,B--E),dir(270-723)), Ep = MP(\"E'\",IP(A--C,B--D),dir(270-724)); [/asy]\n\\]\n\nLet \$A'B'C'D'E'\$ be the inner pentagon, labeled so that \$A\$ and \$A'\$ are opposite each other, and let \$a, b, c, d, e\$ be the side lengths of the inner pentagon labeled opposite their corresponding vertices. From the lemma, pentagon \$A'B'C'D'E'\$ is similar to pentagon \$ABCDE\$ with \$AB = m(A'B')\$ and parallelogram \$ABCB'\$ (and cyclic) has area 2. Supposing\n\n\\[\n\\begin{align} P &= [ABCDE] \\\\ Q &= [A'B'C'D'E'] \\\\ R &= \\sum_{\\text{cyc}}[AC'D'] \\\\ S &= \\sum_{\\text{cyc}}[ABD'], \\end{align}\n\\]\n\nwe have\n\n\\[\n\\begin{align} \\sum_{\\text{cyc}}[ABCB'] &= 5Q + 3R + 2S \\\\ 10 &= 3Q + R + 2P. \\end{align}\n\\]\n\nSince \$BCDC'\$ and \$CDED'\$ are parallelograms, \$BD' = EC' = (m - 1)a\$. Triangle \$EB'C'\$ is similar to triangle \$ECB\$, so \$me = BC = \\left(\\frac{2m - 1}{m - 1}\\right)e\$, and with the requirement that \$m > 1\$,\n\n\\[\nm = \\frac{2m - 1}{m - 1}\\Rightarrow m = \\frac{3 + \\sqrt{5}}{2}.\n\\]\n\nNow, we compute that \$[AC'D'] = \\frac{1}{2m - 1} = \\sqrt{5} - 2\$, and similar computation for the other four triangles gives \$R = 5\\sqrt{5} - 10\$. From the aforementioned pentagon similarity, \$Q = P/m^2 = \\left(\\frac{7 - 3\\sqrt{5}}{2}\\right)P\$. Solving for \$P\$, we have\n\n\\[\n\\begin{align} 10 &= 3\\left(\\frac{7 - 3\\sqrt{5}}{2}\\right)P + (5\\sqrt{5} - 10) + 2P \\\\ 20 - 5\\sqrt{5} &= \\left(\\frac{25 - 9\\sqrt{5}}{2}\\right)P \\\\ P &= \\frac{40 - 10\\sqrt{5}}{25 - 9\\sqrt{5}} \\\\ &= \\frac{5 + \\sqrt{5}}{2}. \\end{align*}\n\\]\n\nTo show that there are infinitely many pentagons with the given property, we start with triangle \$A'B'C'\$ and construct the original pentagon using the required length ratios. This triangle has three continuous degrees of freedom, and the condition that the equal areas be specifically unity means that there are two continuous degrees of freedom in which the configuration can vary, which implies the desired result.\n\nAlternatively, one can shear a regular pentagon, a transformation which preserves areas and the property of a pair of lines being parallel." ]
USAMO-1973-1	https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_1	Two points $P$ and $Q$ lie in the interior of a regular tetrahedron $ABCD$. Prove that angle $PAQ < 60^\circ$.	[ "Let the side length of the regular tetrahedron be \$a\$. Link and extend \$AP\$ to meet the plane containing triangle \$BCD\$ at \$E\$; link \$AQ\$ and extend it to meet the same plane at \$F\$. We know that \$E\$ and \$F\$ are inside triangle \$BCD\$ and that \$\\angle PAQ = \\angle EAF\$\n\nNow let's look at the plane containing triangle \$BCD\$ with points \$E\$ and \$F\$ inside the triangle. Link and extend \$EF\$ on both sides to meet the sides of the triangle \$BCD\$ at \$I\$ and \$J\$, \$I\$ on \$BC\$ and \$J\$ on \$DC\$. We have \$\\angle EAF < \\angle IAJ\$\n\nBut since \$E\$ and \$F\$ are interior of the tetrahedron, points \$I\$ and \$J\$ cannot be both at the vertices and \$IJ < a\$, \$\\angle IAJ < \\angle BAD = 60\$. Therefore, \$\\angle PAQ < 60\$.\n\nSolution with graphs posted at\n\nhttp://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.USA1973Problem1", "Let the side length of the regular tetrahedron \$ABCD\$ be \$s\$.\n\nExtend \$AP\$ and \$AQ\$ to intersect the plane \$BCD\$ at points \$P'\$ and \$Q'\$ inside \$\\Delta BCD\$, respectively. We can observe that \$\\angle PAQ=\\angle P'AQ'\$.\n\nExtend \$P'Q'\$ to intersect the sides of \$\\Delta BCD\$ at \$P''\$ and \$Q''\$. We can observe that \$\\angle P'AQ'<\\angle P''AQ''\$. We assume that, WLOG, \$P''\$ is on \$BD\$ and \$Q''\$ is on \$BC\$.\n\nLet the ratios \$\\frac{BP''}{BD}=\\mu\$ and \$\\frac{BQ''}{BC}=\\lambda\$. We can see that \$\\mu,\\lambda\\in (0,1)\$. It follows from Stewart's theorem that \$AP''^2=AD^2\\mu+AB^2(1-\\mu)-\\mu(1-\\mu)BD^2=(\\mu^2-\\mu+1)s\\\\ AQ''^2=AC^2\\lambda+AB^2(1-\\lambda)-\\lambda(1-\\lambda)BC^2=(\\lambda^2-\\lambda^2+1)s\$\n\nFurthermore, we have \$P''C^2=CD^2\\mu+CB^2(1-\\mu)-\\mu(1-\\mu)BD^2=(\\mu^2-\\mu+1)s\$\n\nUsing this, we can conclude that \$P''Q''^2=P''C^2\\lambda+(BD\\mu)^2(1-\\lambda)-\\lambda(1-\\lambda)BC^2\\\\ =(\\mu\\lambda-\\mu-\\lambda+2)s^2\$\n\nBy cosine law, we have \$\\cos \\angle P''AQ''=\\frac{P''A^2+Q''A^2-P''Q''^2}{2P''A\\cdot Q''A}\\\\ =\\frac{\\mu\\lambda-\\mu-\\lambda+2}{2\\sqrt{(\\mu^2-\\mu+1)(\\lambda^2-\\lambda+1)}}\$ for \$\\mu,\\lambda\\in(0,1)\$.\n\nWe first find a lower bound of the numerator. We write \$\\mu\\lambda-\\mu-\\lambda+2=(\\lambda-1)\\mu-\\lambda+1\$ as a linear function of \$\\mu\$ by letting \$\\lambda\$ be constant. Since \$\\lambda\\in (0,1)\$, the coefficient \$\\lambda-1\$ of \$\\mu\$ in this linear function is negative, meaning this expression is least when \$\\mu\$ is least \"at\" \$1\$. Hence, \$\\mu\\lambda-\\mu-\\lambda+2<1\$.\n\nNow we find an upper bound of the denominator. We complete the square in the expression \$2\\sqrt{(\\mu^2-\\mu+1)(\\lambda^2-\\lambda+1)}=2\\sqrt{((\\mu-\\frac{1}{2})^2+\\frac{3}{4})((\\lambda-\\frac{1}{2})^2+\\frac{3}{4})}\$ which is least when \$\\lambda\$ and \$\\mu\$ \"are\" both 1. Hence, \$2\\sqrt{(\\mu^2-\\mu+1)(\\lambda^2-\\lambda+1)}>2\$.\n\nHence, \$\\cos \\angle P''AQ''<\\frac{1}{2}\$, implying that \$\\angle P''AQ''<60^\\circ\$.\n\nTherefore, \$\\angle PAQ=\\angle P'AQ'<\\angle P''AQ''<60^\\circ\$.\n\nhurdler: Remark on solution 1: This proof is not rigorous, in the very last step. The last step needs more justification." ]
USAMO-1973-2	https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_2	Let $\{X_n\}$ and $\{Y_n\}$ denote two sequences of integers defined as follows: \[ X_0=1, \] \[ X_1=1, \] \[ X_{n+1}=X_n+2X_{n-1} \] \[ (n=1,2,3,\dots), \] \[ Y_0=1, \] \[ Y_1=7, \] \[ Y_{n+1}=2Y_n+3Y_{n-1} \] \[ (n=1,2,3,\dots). \] Thus, the first few terms of the sequences are: \[ X:1, 1, 3, 5, 11, 21, \dots, \] \[ Y:1, 7, 17, 55, 161, 487, \dots. \] Prove that, except for the "1", there is no term which occurs in both sequences.	[ "We can look at each sequence \$\\bmod{8}\$:\n\n\\[\nX\n\\]\n\n:\n\n\\[\n1\n\\]\n\n,\n\n\\[\n1\n\\]\n\n,\n\n\\[\n3\n\\]\n\n,\n\n\\[\n5\n\\]\n\n,\n\n\\[\n3\n\\]\n\n,\n\n\\[\n5\n\\]\n\n,\n\n\\[\n\\dots\n\\]\n\n,\n\n\\[\nY\n\\]\n\n:\n\n\\[\n1\n\\]\n\n,\n\n\\[\n7\n\\]\n\n,\n\n\\[\n1\n\\]\n\n,\n\n\\[\n7\n\\]\n\n,\n\n\\[\n1\n\\]\n\n,\n\n\\[\n7\n\\]\n\n,\n\n\\[\n\\dots\n\\]\n\n.\n\n- Proof that \$X\$ repeats \$\\bmod{8}\$:\n\nThe third and fourth terms are \$3\$ and \$5\$ \$\\bmod{8}\$. Plugging into the formula, we see that the next term is \$11\\equiv 3\\bmod{8}\$, and plugging \$5\$ and \$3\$, we get that the next term is \$13\\equiv 5\\bmod{8}\$. Thus the sequence \$X\$ repeats, and the pattern is \$3,5,3,5,\\dots\$.\n\n- Proof that \$Y\$ repeats \$\\bmod{8}\$:\n\nThe first and second terms are \$1\$ and \$7\$ \$\\bmod{8}\$. Plugging into the formula, we see that the next term is \$17\\equiv 1\\bmod{8}\$, and plugging \$7\$ and \$1\$, we get that the next term is \$23\\equiv 7\\bmod{8}\$. Thus the sequence \$Y\$ repeats, and the pattern is \$1,7,1,7,\\dots\$.\n\nCombining both results, we see that \$X_i\$ and \$Y_j\$ are not congruent \$\\bmod{8}\$ when \$i\\geq 3\$ and \$j\\geq 2\$. Thus after the \"1\", the terms of each sequence are not equal.", "We can solve this problem by finding a particular solution for each linear recurrence.\n\n\$X_{n+1} - X_n - 2X_{n-1} = 0\$ with characteristic polynomial\n\n\$X^2 - X - 2 = (X - 2)(X + 1)\$, so \$X = -1, 2\$\n\n\\[\nX_n = A(2)^n + B(-1)^n\n\\]\n\nAfter plugging in to find the particular solution:\n\n\\[\nX_0 = 1 = A + B\n\\]\n\n\\[\nX_1 = 1 = 2A - B\n\\]\n\n\$A = \\frac{1}{3}\$ and \$B = \\frac{2}{3}\$, so\n\n\\[\nX_n = \\frac{1}{3}(2)^n + \\frac{2}{3}(-1)^n\n\\]\n\nDoing the same for \$Y_n\$, we get\n\n\\[\nY_n = 2(3)^n - (-1)^n\n\\]\n\nWe know they're equal at \$n = 0\$, so let's set them equal and compare.\n\n\\[\n2(3)^n - (-1)^n = \\frac{1}{3}((2)^n + 2(-1)^n)\n\\]\n\n\\[\n6(3)^n - 3(-1)^n = (2)^n + 2(-1)^n\n\\]\n\n\\[\n6(3)^n - 2^n = 5(-1)^n\n\\]\n\nBy induction, we know in general that \$(\\alpha + 1)^n > \\alpha^n\$ for all \$\\alpha, n > 0\$, so \$6(3)^n > 2^n\$ for all \$n > 1\$.\n\nTherefore, the left hand side is always increasing and will never equal 5 again, so equality between \$X_n\$ and \$Y_n\$ only holds when \$n = 0\$, so they only share 1 term." ]
USAMO-1973-3	https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_3	Three distinct vertices are chosen at random from the vertices of a given regular polygon of $(2n+1)$ sides. If all such choices are equally likely, what is the probability that the center of the given polygon lies in the interior of the triangle determined by the three chosen random points?	[ "There are \$\\binom{2n+1}{3}\$ ways how to pick the three vertices. We will now count the ways where the interior does NOT contain the center. These are obviously exactly the ways where all three picked vertices lie among some \$n+1\$ consecutive vertices of the polygon. We will count these as follows: We will go clockwise around the polygon. We can pick the first vertex arbitrarily (\$2n+1\$ possibilities). Once we pick it, we have to pick \$2\$ out of the next \$n\$ vertices (\$\\binom{n}{2}\$ possibilities).\n\nThen the probability that our triangle does NOT contain the center is\n\n\\[\np = \\frac{ (2n+1){\\binom{n}{2}} }{ {\\binom{2n+1}{3} } } = \\frac{ (1/2)(2n+1)(n)(n-1) }{ (1/6)(2n+1)(2n)(2n-1) } = \\frac{ 3n-3 }{ 4n-2 }\n\\]\n\nAnd then the probability we seek is\n\n\\[\n1-p = \\frac{ (4n-2) - (3n-3) }{ (4n-2) } = \\boxed{\\frac{n+1}{4n-2}}\n\\]\n\n## Perplexing Edge Case\n\nHowever, for \$n = 1\$, (a rare edge case), the circumcenter of the triangle must lie in the center of the triangle, which is notably difficult to prove if not assumed, resulting in a probability of \$1\$.\n\nObserve that the probability tends to \$\\frac{1}{4}\$ as \$n\$ approaches infinity. Why does this occur?" ]
USAMO-1973-4	https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_4	Determine all the roots, real or complex, of the system of simultaneous equations \[ x+y+z=3, \] \[ x^2+y^2+z^2=3, \] \[ x^3+y^3+z^3=3 \] .	[ "Let \$x\$, \$y\$, and \$z\$ be the roots of the cubic polynomial \$t^3+at^2+bt+c\$. Let \$S_1=x+y+z=3\$, \$S_2=x^2+y^2+z^2=3\$, and \$S_3=x^3+y^3+z^3=3\$. From this, \$S_1+a=0\$, \$S_2+aS_1+2b=0\$, and \$S_3+aS_2+bS_1+3c=0\$. Solving each of these, \$a=-3\$, \$b=3\$, and \$c=-1\$. Thus \$x\$, \$y\$, and \$z\$ are the roots of the polynomial \$t^3-3t^2+3t-1=(t-1)^3\$. Thus \$x=y=z=1\$, and there are no other solutions.", "Let \$P(t)=t^3-at^2+bt-c\$ have roots x, y, and z. Then\n\n\\[\n0=P(x)+P(y)+P(z)=3-3a+3b-3c\n\\]\n\nusing our system of equations, so \$P(1)=0\$. Thus, at least one of x, y, and z is equal to 1; without loss of generality, let \$x=1\$. Then we can use the system of equations to find that \$y=z=1\$ as well, and so \$\\boxed{(1,1,1)}\$ is the only solution to the system of equations.", "Let \$a=x-1,\$ \$b=y-1\$ and \$c=z-1.\$ Then\n\n\\[\na+b+c=0,\n\\]\n\n\\[\na^2+b^2+c^2=0,\n\\]\n\n\\[\na^3+b^3+c^3=0.\n\\]\n\nWe have\n\n\\[\n\\begin{align} 0&=(a+b+c)^3\\\\ &=(a^3+b^3+c^3)+3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+6abc\\\\ &=0-3a^3-3b^3-3c^3+6abc\\\\ &=6abc. \\end{align}\n\\]\n\nThen one of \$a, b\$ and \$c\$ has to be 0, and easy to prove the other two are also 0. So \$\\boxed{(1,1,1)}\$ is the only solution to the system of equations.\n\nJ.Z.", "We are going to use Intermediate Algebra Techniques to solve this equation.\n\nLet's start with the first one: \$x+y+z=3\$. This will be referred as the FIRST equation.\n\nWe are going to use the first equation to relate to the SECOND one (\$x^2+y^2+z^2=3\$) and the THIRD one (\$x^3+y^3+z^3=3)\$.\n\nSquaring this equation: \$x^2+y^2+z^2+2xy+2yz+2xz=9\$\n\nSubtracting this equation from the 2nd equation in the problem, we have \$2xy+2yz+2xz=6\$, so \$xy+xz+yz=3\$.\n\nNow we try the same idea with the cubed terms. Cube the first equation:\n\n\$x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3yz^2+3zx^2+3zy^2=27\$. Plug in \$x^3+y^3+z^3=3\$ and factor partially:\n\n\$3+3(x^2(y+z)+y^2(x+z)+z^2(x+y))+6xyz=27\$.\n\nNow here is the key step. Note that \$z=3-x-y, y=3-x-z, x=3-y-z\$. So we are going to substitute \$y+z, x+z, x+y\$ for each of the expressions and we get: \$-x^3+3x^2-y^3+3y^2-z^3+3z^2=8-2xyz\$ (I rearranged it a bit).\n\nResubstituting in the second and third equation: \$-3+3(3)=8-2xyz\$. So \$xyz=1\$.\n\nSo now we have three equations for the elementary symmetric sums of \$x,y,z\$:\n\nEquation 4: \$x+y+z=3\$ (this is also equation 1)\n\nEquation 5: \$xy+yz+xz=3\$\n\nEquation 6: \$xyz=1\$.\n\nIf we call the solutions of \$t^3-3t^2+3t-1=0\$ (Equation 7) \$a,b,c\$, then \$x,y,z\$ are the three roots \$a,b,c\$ but in some order. Notice that Equation 7 can be factored as \$(t-1)^3=0\$, which means that \$t=1\$. Therefore \$(x,y,z)\$ are permutations of \$(1,1,1)\$ in some order, which can be only \$(1,1,1)\$. (This step uses Vieta's formulas)\n\nTherefore, the only solution is \$\\boxed{(x,y,z)=(1,1,1)}\$.\n\n~hastapasta", "From the question\n\nEquation 1 : \$x+y+z=3\$\n\nEquation 2 : \$x^2+y^2+z^2=3\$\n\nEquation 3 : \$x^3+y^3+z^3=3\$\n\nFrom Eq(1) and Eq(2),\n\nEquation 4 : \$(x+y+z)^2=9 \\implies x^2+y^2+z^2+2\\sum xy = 9 \\implies \\sum xy = 3\$\n\nFrom Eq(1), Eq(3), and the fact that \$(x+y+z)^3=x^3+y^3+z^3+3\\prod(x+y)\$,\n\nEquation 5 : \$(x+y+z)^3=x^3+y^3+z^3+3\\prod(x+y)= 27 \\implies \\prod(x+y)=8\$\n\nAs from Eq(1), \$\\ x+y=3-z\$,\$\\ y+z=3-x\$, \$\\ z+x = 3- y\$\n\nEquation 6 : \$\\prod(3-x) =8\$\n\nNow let there be a cubic function \$f(t)\$ with the roots \$x,y,z\$ so from the relation between roots and coefficients,\n\nEquation 7 : \$f(t)= t^3-3t^2+3t-d\$ (As we don't know the product let the constant be \$d\$)\n\nAs \$f(t) = (t-x)(t-y)(t-z)\$ then at \$t=3\$,\n\nEquation 8 : \$f(3)=9-d=(3-x)(3-y)(3-z)\$\n\nAnd from Eq(6),\n\nEquation 9 : \$d= -1\$\n\nSo finally,\n\nEquation 10 : \$f(t)=t^3-3t^2+3t-1=(t-1)^{3}\$\n\nSo, \$\\ (x,y,z)=(1,1,1)\$\n\n~Sparrow_1827" ]
USAMO-1973-5	https://artofproblemsolving.com/wiki/index.php/1973_USAMO_Problems/Problem_5	Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.	[ "Assume that the cube roots of three distinct prime numbers can be three terms of an arithmetic progression. Let the three distinct prime numbers be \$p\$, \$q\$, and \$r.\$ WLOG, let \$p<q<r.\$\n\nThen,\n\n\\[\nq^{\\dfrac{1}{3}}=p^{\\dfrac{1}{3}}+md\n\\]\n\n\\[\nr^{\\dfrac{1}{3}}=p^{\\dfrac{1}{3}}+nd\n\\]\n\nwhere \$m\$, \$n\$ are distinct integers, and \$d\$ is the common difference in the progression. Then we have\n\n\\[\n\\begin{array}{rcl} nq^{\\dfrac{1}{3}}-mr^{\\dfrac{1}{3}}&=&(n-m)p^{\\dfrac{1}{3}} \\\\ \\\\n^{3}q-3n^{2}mq^{\\dfrac{2}{3}}r^{\\dfrac{1}{3}}+3nm^{2}q^{\\dfrac{1}{3}}r^{\\dfrac{2}{3}}-m^{3}r&=&(n-m)^{3}p\\\\ \\\\3nm^{2}q^{\\dfrac{1}{3}}r^{\\dfrac{2}{3}}-3n^{2}mq^{\\dfrac{2}{3}}r^{\\dfrac{1}{3}}&=&(n-m)^{3}p+m^{3}r-n^{3}q\\\\ \\\$3nmq^{\\dfrac{1}{3}}r^{\\dfrac{1}{3}})(mr^{\\dfrac{1}{3}}-nq^{\\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q\\\\ \\\\(3nmq^{\\dfrac{1}{3}}r^{\\dfrac{1}{3}})((m-n)p^{\\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q\\\\ \\\\q^{\\dfrac{1}{3}}r^{\\dfrac{1}{3}}p^{\\dfrac{1}{3}}&=&\\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)}\\\\ \\\\(pqr)^{\\dfrac{1}{3}}&=&\\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)} \\end{array}\n\\]\n\nBecause \\(p\$, \$q\$, \$r\$ are distinct primes, \$pqr\$ is not a perfect cube. Thus, the LHS is irrational but the RHS is rational, which is a contradiction. So, the cube roots of three distinct prime numbers cannot be three terms of an arithmetic progression.", "Assume the contrary and suppose that the cube roots of three distinct prime numbers, \$p_1^\\frac{1}{3}\$, \$p_2^\\frac{1}{3}\$, and \$p_3^\\frac{1}{3}\$, form an increasing arithmetic progression. Then we can write \$p_2^\\frac{1}{3} = p_1^\\frac{1}{3} + md\$ and \$p_3^\\frac{1}{3} = p_1^\\frac{1}{3} + nd\$ for some positive integral \$d, m, n\$ with \$m < n\$. Thus,\n\n\\[\n\\frac{p_3^\\frac{1}{3} - p_1^\\frac{1}{3}}{p_2^\\frac{1}{3} - p_1^\\frac{1}{3}} = \\frac{md}{nd} = \\frac{m}{n} = r,\n\\]\n\nfor some real \$r\$. Because \$m\$ and \$n\$ are integral, r is rational.\n\nThis equation rearranges to\n\n\\[\np_3^\\frac{1}{3} = rp_2^\\frac{1}{3} - (r-1)p_1^\\frac{1}{3}.\n\\]\n\nNote that if \$a-b=x\$, then\n\n\\[\n(a-b)^3 = a^3 - b^3 - 3ab(a-b) = a^3 - b^3 - 3abx\n\\]\n\nfrom the binomial theorem. As a result, cubing both sides of the equation gives\n\n\\[\np_3 = r^3p_2 - (r-1)^3p_1 - 3r(r-1)(p_1p_2p_3)^\\frac{1}{3}. ()\n\\]\n\nLemma: The sum, product, and difference of two rational numbers are all rational, and that of a rational and irrational number are all irrational.\n\nProof: The first part is trivial if we write the fractions as \$\\frac{m}{n}\$ and \$\\frac{p}{q}\$, and find the common denominators or simplify fractions as needed. The second part follows by contradiction: if the sum of a rational number and irrational number is rational, then the irrational number must be the difference of two rational numbers, which is rational. This is a contradiction; likewise for difference and product.\n\nThe LHS (left-hand side) of () is clearly rational (in fact, integral), and so the RHS must also be rational. This is the case for the first two terms \$r^3p_2\$ and \$-(r-1)^3p_1\$. But it is not the case for the third term. The number \$(p_1p_2p_3)^\\frac{1}{3}\$ is irrational because \$p_1p_2p_3\$, the product of three distinct primes, cannot be a perfect cube. Hence, \$-3r(r-1)(p_1p_2p_3)^\\frac{1}{3}\$, the product of a rational and an irrational number, is irrational, and the RHS is therefore the sum of two rational terms and one irrational term. This is irrational by the lemma, and hence we have a rational equalling an irrational, contradiction. It follows that the cube roots of three distinct prime numbers cannot form an arithmetic sequence." ]
USAMO-1974-1	https://artofproblemsolving.com/wiki/index.php/1974_USAMO_Problems/Problem_1	Let $a$, $b$, and $c$ denote three distinct integers, and let $P$ denote a polynomial having all integral coefficients. Show that it is impossible that $P(a)=b$, $P(b)=c$, and $P(c)=a$.	[ "It suffices to show that if \$a,b,c\$ are integers such that \$P(a) = b\$, \$P(b)=c\$, and \$P(c)= a\$, then \$a=b=c\$.\n\nWe note that\n\n\\[\na-b \\mid P(a) - P(b) = b-c \\mid P(b)-P(c) = c-a \\mid P(c) - P(a) = a-b ,\n\\]\n\nso the quanitities \$(a-b), (b-c), (c-a)\$ must be equal in absolute value. In fact, two of them, say \$(a-b)\$ and \$(b-c)\$, must be equal. Then\n\n\\[\n0 = \\lvert (a-b) + (b-c) + (c-a) \\rvert = \\lvert 2(a-b) + (c-a) \\rvert \\ge 2 \\lvert a-b \\rvert - \\lvert c-a \\rvert = \\lvert a-b \\rvert ,\n\\]\n\nso \$a=b= P(a)\$, and \$c= P(b) = P(a) = b\$, so \$a\$, \$b\$, and \$c\$ are equal, as desired. \$\\blacksquare\$", "Let \$k\$ be the value \$(a-b), (b-c), (c-a)\$ are equal to in absolute value. Assume \$k\$ is nonzero. Then each of \$(a-b), (b-c), (c-a)\$ is equal to \$k\$ or \$-k\$, so\n\n\\[\n0 = (a-b) + (b-c) + (c-a) = mk,\n\\]\n\nwhere \$m\$ is one of -3, -1, 1, or 3. In particular, neither \$m\$ nor \$k\$ is zero, contradiction. Hence, \$k = 0\$, and \$a, b, c\$ are equal, yielding a final contradiction of the existence of these variables. \$\\blacksquare\$\n\nIn fact, this approach generalizes readily. Suppose that \$n\$ is an odd integer, and that there exist \$n\$ distinct integers \$a_1, a_2, ..., a_n\$ such that \$P(a_i) = a_{i+1}\$ and \$P(a_n) = 1\$, for \$1 \\le i \\le n-1\$. Then we have\n\n\\[\na_1 - a_2 \| a_2 - a_3 \| ... \| a_n - a_1 \| a_1 - a_2,\n\\]\n\nso the differences are all equal in absolute value and thus equal to \$k\$ or \$-k\$ for some integer \$k\$. Adding the differences (in the order they are written) gives \$0 = mk\$ for some odd integer \$m\$, so \$m\$ is nonzero and hence \$k = 0\$. Thus, all the \$a_i\$ are equal, a contradiction of their distinctness, so the sequence \$a_i\$ cannot exist.", "Consider the polynomial \$Q(x) = (b-a)P(x) - (c-b)x - b^2 + ac.\$ By using the facts that \$P(a) = b\$ and \$P(b) = c\$, we find that \$Q(a) = Q(b) = 0.\$ Thus, the polynomial \$Q(x)\$ has a and b as roots, and we can write \$Q(x) = (x-a)(x-b)R(x)\$ for some polynomial \$R(x)\$. Because \$Q(x)\$ and \$(x-a)(x-b)\$ are monic polynomials with integral coefficients, their quotient, \$R(x)\$, must also have integral coefficients, as can be demonstrated by simulating the long division. Thus, \$Q(c)\$, and hence \$Q(c) - (x-a)(x-b)\$, must be divisible by \$(c-a)(c-b)\$. But if \$x=c-a\$ and \$y=c-b\$, then we must have, after rearranging terms and substitution, that \$(x-y)^2\$ is divisible by \$xy\$. Equivalently, \$S = x^2 + y^2\$ is divisible by \$xy\$ (after canceling the \$2xy\$ which is clearly divisble by \$xy\$). Because S must be divisible by both x and y, we quickly deduce that x is divisible by y and y is divisible by x. Thus, x and y are equal in absolute value. A similar statement holds for a cyclic pemutation of the variables, and so x, y, and z are all equal in absolute value. The conclusion follows as in the first solution.", "Assume for the sake of contradiction that is possible for unique integers \$a,b,c\$. Let \$P(x)=d_1x^n+d_2x^{n-1}+\\cdots+d_n.\$ Note that\n\n\\[\nb=P(a)=d_1a^n+d_2a^{n-1}+\\cdots+d_n\n\\]\n\n\\[\nc=P(b)=d_1b^n+d_2b^{n-1}+\\cdots+d_n\n\\]\n\n\\[\na=P(c)=d_1c^n+d_2c^{n-1}+\\cdots+d_n\n\\]\n\nSubtracting the second from the first, third from second, and first from third gives\n\n\\[\nb-c=P(a)-P(b)=d_1(a^n-b^n)+d_2(a^{n-1}-b^{n-1})+\\cdots+ d_{n-1}(a-b)\n\\]\n\n\\[\nc-a=P(b)-P(c)=d_1(b^n-c^n)+d_2(b^{n-1}-c^{n-1})+\\cdots +d_{n-1}(b-c)\n\\]\n\n\\[\na-b=P(c)-P(a)=d_1(c^n-a^n)+d_2(c^{n-1}-a^{n-1})+\\cdots +d_{n-1}(c-a)\n\\]\n\nBy the RHS, note that \$a-b\\mid P(a)-P(b)\$ so \$\\lvert{a-b}\\rvert\\leq\\lvert{b-c}\\rvert.\$ Similarly, \$\\lvert{b-c}\\rvert\\leq\\lvert{c-a}\\rvert\$ and \$\\lvert{c-a}\\rvert\\leq\\lvert{a-b}\\rvert.\$ Hence, \$\\lvert{a-b}\\rvert\\leq\\lvert{b-c}\\rvert\\leq\\lvert{c-a}\\rvert\\leq\\lvert{a-b}\\rvert\$ so \$\\lvert{a-b}\\rvert=\\lvert{b-c}\\rvert=\\lvert{c-a}\\rvert.\$ Assume WLOG that \$a>b>c\$ so \$a-b=a-c\$ and \$b-c=c-a.\$ From the first equation, we get \$b=c\$ and substituting this in the second gives \$c=a.\$ Hence, \$a=b=c\$, contradicting the uniqueness of \$a,b,c.\$ \$\\blacksquare\$", "Assume that \$a, b\$ and \$c\$ are distinct integers which satisfy the given conditions. Then we have \$P(c) = P(P(b)) = P(P(P(a))) = a\$. Similarly \$P(P(P(b))) = b\$ and \$P(P(P(c))) = c\$. Since the polynomial has integer coefficients, this implies that the polynomial \$P\$ is \$P(x) = x\$. But this is a contradiction since then \$P(a) = b\$ can only be true if \$a = b\$ which contradicts the fact that \$a, b\$ and \$c\$ are distinct integers.", "We get that \$a-b\|b-c, b-c\|c-a, c-a\|a-b\$. Thus, we can say that \$(b-c) = (a-b)k, (c-a) = (b-c)j, (a-b) = (c-a)m\$, for non zero integers, not necesarily positive, k, j, and m. Multiplying these equations and canceling, we get that \$kjm\$=1. This means that either k=1, m=1, j=1, or two of the variables are equal to 1 and the other is equal to -1. If all of the variables are equal to zero, plugging in gives us that \$a=b=c\$, contradictory to the conditions given in the problem statement. If two of the variables are equal to -1, and the other is equal to 1, then WLOG let \$k=m=-1, j=1\$. But this implies that the distinct integer condition is not satisfied, and hence, this case is invalid. Thus, it is impossible. \$\\blacksquare\$" ]
USAMO-1974-2	https://artofproblemsolving.com/wiki/index.php/1974_USAMO_Problems/Problem_2	Prove that if $a$, $b$, and $c$ are positive real numbers, then \[ a^ab^bc^c\ge (abc)^{(a+b+c)/3} \]	[ "Consider the function \$f(x)=x\\ln{x}\$. \$f''(x)=\\frac{1}{x}>0\$ for \$x>0\$; therefore, it is a convex function and we can apply Jensen's Inequality:\n\n\\[\n\\frac{a\\ln{a}+b\\ln{b}+c\\ln{c}}{3}\\ge \\left(\\frac{a+b+c}{3}\\right)\\ln\\left(\\frac{a+b+c}{3}\\right)\n\\]\n\nApply AM-GM to get\n\n\\[\n\\frac{a+b+c}{3}\\ge \\sqrt[3]{abc}\n\\]\n\nwhich implies\n\n\\[\n\\frac{a\\ln{a}+b\\ln{b}+c\\ln{c}}{3}\\ge \\left(\\frac{a+b+c}{3}\\right)\\ln\\left(\\sqrt[3]{abc}\\right)\n\\]\n\nRearranging,\n\n\\[\na\\ln{a}+b\\ln{b}+c\\ln{c}\\ge\\left(\\frac{a+b+c}{3}\\right)\\ln\\left(abc\\right)\n\\]\n\nBecause \$f(x) = e^x\$ is an increasing function, we can conclude that:\n\n\\[\ne^{a\\ln{a}+b\\ln{b}+c\\ln{c}}\\ge{e}^{\\ln\\left(abc\\right)(a+b+c)/3}\n\\]\n\nwhich simplifies to the desired inequality.", "Note that \$(a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\\left((abc)^{(a+b+c)/3}\\right)^3\$.\n\nSo if we can prove that \$a^ab^bc^c\\ge a^bb^cc^a\$ and \$a^ab^bc^c\\ge a^cb^ac^b\$, then we are done.\n\nWLOG let \$a\\ge b\\ge c\$.\n\nNote that \$(a^ab^bc^c)\\cdot \\left(\\dfrac{c}{a}\\right)^{a-b}\\cdot \\left(\\dfrac{c}{b}\\right)^{b-c}=a^bb^cc^a\$. Since \$\\dfrac{c}{a} \\le 1\$, \$\\dfrac{c}{b} \\le 1\$, \$a-b \\ge 0\$, and \$b-c \\ge 0\$, it follows that \$a^ab^bc^c \\ge a^bb^cc^a\$.\n\nNote that \$(a^ab^bc^c)\\cdot \\left(\\dfrac{b}{a}\\right)^{a-b}\\cdot \\left(\\dfrac{c}{a}\\right)^{b-c}=a^cb^ac^b\$. Since \$\\dfrac{b}{a} \\le 1\$, \$\\dfrac{c}{a} \\le 1\$, \$a-b \\ge 0\$, and \$b-c \\ge 0\$, it follows that \$a^ab^bc^c \\ge a^cb^ac^b\$.\n\nThus, \$(a^ab^bc^c)^3\\ge (a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\\left((abc)^{(a+b+c)/3}\\right)^3\$, and cube-rooting both sides gives \$a^ab^bc^c\\ge (abc)^{(a+b+c)/3}\$ as desired.", "WLOG let \$a\\ge b\\ge c\$. Let \$b = ax\$ and \$c = ay\$, where \$x \\ge 1\$ and \$y \\ge 1\$.\n\nWe want to prove that \$(a)^{a}(ax)^{ax}(ay)^{ay} \\ge (a \\cdot ax \\cdot ay)^{\\frac{a + ax + ay}{3}}\$.\n\nSimplifying and combining terms on each side, we get \$a^{a + ax + ay}x^{ax}y^{ay} \\ge a^{a + ax + ay}(xy)^{\\frac{a + ax + ay}{3}}\$.\n\nSince \$a > 0\$, we can divide out \$a^{a + ax + ay}\$ to get \$x^{ax}y^{ay} \\ge (xy)^{\\frac{a + ax + ay}{3}}\$.\n\nTake the \$a\$th root of each side and then cube both sides to get \$x^{3x}y^{3y} \\ge (xy)^{1 + x + y}\$.\n\nThis simplifies to \$x^{2x-1}y^{2y-1} \\ge x^{y}y^{x}\$.\n\nSince \$2x - 1 \\ge x\$ and \$2y - 1 \\ge y\$, we only need to prove \$x^{x}y^{y} \\ge x^{y}y^{x}\$ for our given \$x, y\$.\n\nWLOG, let \$y \\ge x\$ and \$y =kx\$ for \$k \\ge 1\$. Then our expression becomes\n\n\\[\nx^{x}(xk)^{xk} \\ge x^{xk}(xk)^{x}\n\\]\n\n\\[\nx^{x+xk}k^{xk} \\ge x^{x+xk}k^{x}\n\\]\n\n\\[\nk^{xk} \\ge k^{x}\n\\]\n\n\\[\nk^k \\ge k\n\\]\n\nThis is clearly true for \$k \\ge 1\$.", "WLOG let \$a\\ge b\\ge c\$. Then sequence \$(a,b,c)\$ majorizes \$(\\frac{a+b+c}{3},\\frac{a+b+c}{3},\\frac{a+b+c}{3})\$. Thus by Muirhead's Inequality, we have \$\\sum_{sym} a^ab^bc^c \\ge \\sum_{sym} a^{\\frac{a+b+c}{3}}b^{\\frac{a+b+c}{3}}c^{\\frac{a+b+c}{3}}\$, so \$a^ab^bc^c \\ge (abc)^{\\frac{a+b+c}{3}}\$.", "Let \$x=\\frac{a}{\\sqrt[3]{abc}},\$ \$y=\\frac{b}{\\sqrt[3]{abc}}\$ and \$z=\\frac{c}{\\sqrt[3]{abc}}.\$ Then \$xyz=1\$ and a straightforward calculation reduces the problem to\n\n\\[\nx^xy^yz^z \\ge 1.\n\\]\n\nWLOG, assume \$x\\ge y\\ge z.\$ Then \$x\\ge 1,\$ \$z\\le 1\$ and \$xy=\\frac{1}{z} \\ge 1.\$ Therefore,\n\n\\[\nx^xy^yz^z=x^{x-y}(xy)^{y-z}(xyz)^z \\ge 1.\n\\]\n\nJ.Z.", "Cubing both sides of the given inequality gives\n\n\\[\na^{3a}b^{3b}c^{3c}\\ge(abc)^{a+b+c}\n\\]\n\nIf we take \$a^{3a}b^{3b}c^{3c}\$ as the product of \$3a\$ \$a\$'s, \$3b\$ \$b\$'s, and \$3c\$ \$c's\$, we get that\n\n\\[\n\\sqrt[3(a+b+c)]{a^{3a}b^{3b}c^{3c}}\\ge\\frac{3(a+b+c)}{\\frac{3a}{a}+\\frac{3b}{b}+\\frac{3c}{c}}=\\frac{a+b+c}{3}\\ge\\sqrt[3]{abc}\n\\]\n\nby GM-HM, as desired.", "Replacing \$a\$ with \$ak\$, \$b\$ with \$bk\$, and \$c\$ with \$ck\$, for some positive real \$k\$ we get: \$a^{ak}k^{ak}b^{bk}k^{bk}c^{ck}k^{ck} = a^{ak}b^{bk}c^{ck}k^{ak+bk+ck} \\ge {k^3abc}^{{ak+bk+ck}/3} = k^{ak+bk+ck}{abc}^{{ak+bk+ck}/3}\$ This means that this inequality is homogeneous since both sides have the same power of \$k\$ as a factor. Since the inequality is homogeneous, we can scale \$abc\$ so that their product is \$1\$, i.e. \$abc = 1\$. This makes the inequality turn into something much more nicer to deal with. Now we have to prove: \$a^ab^bc^c \\ge 1\$ given that \$abc = 1\$. Note that \$a^a \\ge a\$ even if \$a \\le 1\$. Therefore \$a^a \\ge a\$, \$b^b \\ge b\$, and \$c^c \\ge c\$. Multiplying these together we get: \$a^ab^bc^c \\ge abc = 1\$. This proves the desired result. Equality holds when \$a = b = c = 1\$.", "Let \$\\log_2a=x\$, \$\\log_2b=y\$ and \$\\log_2c=z\$, and WLOG \$a\\ge b\\ge c >0\$. Then we have both \$a\\ge b\\ge c\$ and \$x\\ge y \\ge z\$. By the rearrangement inequality, \$ax+by+cz\\ge ay + bz +cx\$ and \$ax+by +cz\\ge az+bx+cy\$. Summing, \$2(ax+by+cz)\\ge a(y+z)+b(x+z)+c(x+y)\$ Adding \$ax+by+cz\$,we get \$ax+by+cz\\ge \\frac{(a+b+c)(x+y+z)}{3}\$. Now we substitute back for \$x,y,z\$ to get: \$\\log_2a^ab^bc^c\\ge \\log_2(abc)^{\\frac{a+b+c}{3}}\$. Raising \$2\$ to the power of each side, we get \$a^ab^bc^c\\ge (abc)^{\\frac{a+b+c}{3}}\$", "Assume without loss of generality that \$a\\le b\\le c\$. Then, we have\n\n\\[\nb/a\\ge 1, c/a\\ge 1, c/b\\ge 1, b-a\\ge 0, c-a\\ge 0, c-b\\ge 0.\n\\]\n\nIt follows that\n\n\\[\n(b/a)^{(b-a)/3}\\cdot (c/a)^{(c-a)/3}\\cdot (c/b)^{(c-b)/3}\\ge 1.\n\\]\n\nWe can \"distribute\" the exponents, which gives\n\n\\[\n\\dfrac{b^{(b-a)/3}}{a^{(b-a)/3}}\\cdot \\dfrac{c^{(c-a)/3}}{a^{(c-a)/3}}\\cdot \\dfrac{c^{(c-b)/3}}{b^{(c-b)/3}}\\ge 1.\n\\]\n\nNotice that in the \$b^{(c-b)/3}\$ in the denominator of the third fraction can be brought out of the denominator by negating its exponent. This gives\n\n\\[\n\\dfrac{b^{(b-a)/3}}{a^{(b-a)/3}}\\cdot \\dfrac{c^{(c-a)/3}}{a^{(c-a)/3}}\\cdot c^{(c-b)/3} \\cdot b^{(b-c)/3}\\ge 1.\n\\]\n\nWe can move around terms on the left to get\n\n\\[\n\\left(\\dfrac{b^{(b-a)/3}\\cdot b^{(b-c)/3}}{a^{(b-a)/3}}\\right)\\left(\\dfrac{c^{(c-a)/3}\\cdot c^{(c-b)/3}}{a^{(c-a)/3}}\\right)\\ge 1.\n\\]\n\nNow we combine the exponents in the numerators. This gives\n\n\\[\n\\left(\\dfrac{b^{(2b-c-a)/3}}{a^{(b-a)/3}}\\right)\\left(\\dfrac{c^{(2c-b-a)/3}}{a^{(c-a)/3}}\\right)\\ge 1.\n\\]\n\nWe multiply both sides by \$\\left(a^{(b-a)/3}\\right)\\left(a^{(c-a)/3}\\right),\$ which gives\n\n\\[\n\\left(b^{(2b-c-a)/3}\\right)\\left(c^{(2c-b-a)/3}\\right)\\ge \\left(a^{(b-a)/3}\\right)\\left(a^{(c-a)/3}\\right).\n\\]\n\nWe then multiply both sides by \$\\left(b^{(c-a)/3}\\right)\\left(c^{(b-a)/3}\\right):\$\n\n\\[\n\\left(b^{(2b-2a)/3}\\right)\\left(c^{(2c-2a)/3}\\right)\\ge \\left(a^{(b-a)/3}\\right)\\left(c^{(b-a)/3}\\right)\\left(a^{(c-a)/3}\\right)\\left(b^{(c-a)/3}\\right).\n\\]\n\nMultiplying both sides by \$\\left(b^{(b-a)/3}\\right)\\left(c^{(c-a)/3}\\right),\$ then simplifying both sides with exponent laws, gives\n\n\\[\nb^{b-a}\\cdot c^{c-a}\\ge (abc)^{(b+c-2a)/3}.\n\\]\n\nFinally, we multiply both sides by \$(abc)^a.\$ Combining exponents one last time gives the desired\n\n\\[\na^a b^b c^c\\ge (abc)^{(a+b+c)/3}.\n\\]\n\n--MenuThreeOne" ]
USAMO-1974-3	https://artofproblemsolving.com/wiki/index.php/1974_USAMO_Problems/Problem_3	Two boundary points of a ball of radius 1 are joined by a curve contained in the ball and having length less than 2. Prove that the curve is contained entirely within some hemisphere of the given ball.	[ "Question: Why is the curve necessarily on a great circle? The curve is arbitrary–in space. Also, there is only one great circle through the two points, as three points determine a plane–a counterexample to this claim would then be readily found. Draw a Great Circle containing the two points and the curve. Then connect the two points with a chord in the circle. The circle has radius 1, so the circle has diameter 2, so the two points are all but diametrically opposite. Therefore we can draw a diameter \$EF\$ parallel to the chord but not on it.\n\nNow take the plane through EF perpendicular to AG. This creates two hemispheres, and the curve is in one and only one of them.", "Let the curve intersect the sphere at points A and B. Let B' be the point diametrically opposite A, and let \$Q\$ be the plane through the center O of the sphere perpendicular to \$BB'\$ and passing through the midpoint of \$BB'\$. We claim that the curve must be in the hemisphere divided by \$Q\$ containing points A and B.\n\nAssume the contrary and suppose that the curve intersects \$Q\$ at point C. Reflect the portion of the curve from C to B to obtain a same-length curve from C to B'. Then the length of the curve equals the length of the new curve from A to B', which is at least the distance from A to B', or 2. This is a contradiction to the claim that the length of the curve is less than 2, so the curve must be contained in the hemisphere divided by \$Q\$ containing points A and B." ]
USAMO-1974-4	https://artofproblemsolving.com/wiki/index.php/1974_USAMO_Problems/Problem_4	A father, mother and son hold a family tournament, playing a two person board game with no ties. The tournament rules are: (i) The weakest player chooses the first two contestants. (ii) The winner of any game plays the next game against the person left out. (iii) The first person to win two games wins the tournament. The father is the weakest player, the son the strongest, and it is assumed that any player's probability of winning an individual game from another player does not change during the tournament. Prove that the father's optimal strategy for winning the tournament is to play the first game with his wife.	[ "There are three possible strategies for the father:\n\n1) Sit out the first game\n\n2) Play the first game against the weaker of the other two players (the mother). This person is referred to throughout the solution as \"the weaker player\".\n\n3) Play the first game against the stronger of the other two players (the son). This person is referred to throughout the solution as \"the stronger player\".\n\nStrategy 1 is dominated by both strategy 2 and strategy 3. If strategy 1 is used and the weaker player wins the first game, it becomes strategy 2, but with the father starting behind 0-1. Likewise, if strategy 1 is used and the stronger player wins the first game, it becomes strategy 3, but with the father again starting behind 0-1.\n\nThus, the father must play the first game.\n\nBy the pigeonhole principle, there may be no more than 4 games; once the 4th game is finished, someone will have won two games. Let \$W\$ denote a game the father wins, let \$L\$ denote a game the father loses, and let \$x\$ denote a game the father does not participate in. Then, with the knowledge that the father plays the first game, the following ways for the father to win two games are possible:\n\n(A) \$WW\$\n\n(B) \$WLxW\$\n\n(C) \$LxWW\$\n\nIf the father wins the first game, then he must either win the second game (A), or sit out the third game to win the fourth game (B). If the father loses the first game, then he must sit out the second game and win the third and fourth games (C).\n\nNow we must determine which gives the largest probability of the father winning: playing the first game against the weaker player or the stronger player. We define the following probabilities:\n\n\$x\$ is the chance of the father defeating the weaker player. \$(1-x)\$ is the chance of the weaker player defeating the father.\n\n\$y\$ is the chance of the father defeating the stronger player. \$(1-y)\$ is the chance of the stronger player defeating the father.\n\n\$z\$ is the chance of the weaker player defeating the stronger player. \$(1-z)\$ is the chance of the stronger player defeating the weaker player.\n\nNote that \$x > y\$, and all of \$x, y\$ and \$z\$ are less than \$0.5\$.\n\nIf the father plays the weaker player first, we split the probability of winning the tournament in the three mutually exclusive cases above:\n\n\$P(A) = xy\$ because the father must defeat the weaker player, then the father must defeat the stronger player.\n\n\$P(B) = x(1-y)zx\$ because the father must defeat the weaker player, then stronger player must defeat the father. Third, the weaker player must defeat the stronger player (because if the stronger player wins a second game, the tournament is over). Fourth, the father plays the weaker player a second time and must win.\n\n\$P(C) = (1-x)(1-z)yx\$ because the weaker player must defeat the father, followed by the stronger player defeating the weaker player (because if the weaker player wins two in a row, he wins the tournament). The father must then defeat the stronger player and the weaker player in the third and fourth game, respectively.\n\nFrom this, the total probability of the father winning by playing the weaker player first is:\n\n\\[\nP(TOTAL) = P(A) + P(B) + P(C) = xy + x(1-y)zx + (1-x)(1-z)yx\n\\]\n\nLikewise, we compute the probability of the father winning if he plays the stronger player first. We do this by symmetry:\n\nAll \$x\$ will turn into \$y\$, because whenever the father defeated the weaker player before, he must defeat the stronger player now.\n\nAll \$y\$ will turn into \$x\$, because whenever the father defeated the stronger player before, he must defeat the weaker player now.\n\nAll \$z\$ will turn into \$(1-z)\$, because whenever the weaker player defeated the stronger player, now the stronger player must defeat the weaker player.\n\nAll \$(1-z)\$ will turn into \$z\$, because whenever the stronger player defeated the weaker player, now the weaker player must defeat the stronger player.\n\nApplying these transformations and calling the new probabilities \$P'\$, we find that:\n\n\\[\nP'(TOTAL) = P'(A) + P'(B) + P'(C) = yx + y(1-x)(1-z)y + (1-y)zxy\n\\]\n\nWe compare these probabilities.\n\n\$P(A) = P'(A)\$. This makes sense, as the probabilitiy of the father winning the first two games is the same, regardless of the order in which they are played.\n\n\$P(B) > P'(C)\$: Calculating \$P(B) / P'(C)\$, we get \$x^2(1-y)z / xy(1-y)z\$, which simplifies to \$x/y\$.\n\nSince \$x > y\$\n\n\\[\nx/y > 1\n\\]\n\n\\[\nP(B) / P'(C) > 1\n\\]\n\nso \$P(B) > P'(C)\$.\n\n\$P(C) > P'(B)\$: Calculating \$P(C) / P'(B)\$, we get \$(1-x)(1-z)yx / (1-x)(1-z)y^2\$, which simpllifies again to \$x/y\$. Thus, \$P(C) > P'(B)\$\n\nAdding the above equations and inequalities, we see that \$P(TOTAL) > P'(TOTAL)\$, and thus the father is better off playing the weaker of the two players (ie the mother) first.\n\n## TL;DR\n\nLet the probability of the father winning against the son be \$a\$, the father winning against the mother be \$b\$, and the son winning against the mother be \$c\$. Here \$a<b<\\frac{1}{2}\$ and \$c>\\frac{1}{2}\$.\n\nWe define three first game cases as above:\n\n1. Father vs Mother\n\n2. Father vs Son\n\n3. Mother vs Son\n\nThe probability that the father wins (1) (by casework or tree) is \$P(1)=ab+b^2(1-a)(1-c)+(1-b)abc\$.\n\nThe probability that the father wins (2) is \$P(2)=ab+a^2c(1-b)+ab(1-a)(1-c)\$.\n\nThe probability that the father wins (3) is \$P(3)=ab\$.\n\nEvidently \$P(1)>P(3)\$, so all that is left is to prove that \$P(1)>P(2)\$.\n\n\\[\nP(1)>P(2)\n\\]\n\n\\[\nab+b^2(1-a)(1-c)+(1-b)abc>ab+a^2c(1-b)+ab(1-a)(1-c)\n\\]\n\n\\[\nb^2(1-a)(1-c)+(1-b)abc>a^2c(1-b)+ab(1-a)(1-c)\n\\]\n\n\\[\nb(b-a)(1-a)(1-c)>ac(a-b)(1-b)\n\\]\n\nUsing the fact that \$a<b\$, we easily find that the LHS is positive while the RHS is negative, proving the inequality.\n\n~eevee9406" ]
USAMO-1974-5	https://artofproblemsolving.com/wiki/index.php/1974_USAMO_Problems/Problem_5	Consider the two triangles $\triangle ABC$ and $\triangle PQR$ shown in Figure 1. In $\triangle ABC$, $\angle ADB = \angle BDC = \angle CDA = 120^\circ$. Prove that $x=u+v+w$. \[ [asy] size(400); defaultpen(1); pair C=(0,1), A=(1,6), B=(4,1), D=(1.5,2); draw(D--A--B--C--D--B); draw(A--C); label("$A$",A,N); label("$B$",B,ESE); label("$C$",C,SW); label("$D$",D,NE); label("$a$",(B+C)/2,S); label("$b$",(C+A)/2,WNW); label("$c$",(A+B)/2,NE); label("$u$",(A+D)/2,E); label("$v$",(B+D)/2,N); label("$w$",(C+D)/2,N); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); draw(M--P--Q--R--M--Q); draw(P--R); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,W); label("$M$",M,NE); label("$x$",(P+Q)/2,S); label("$x$",(Q+R)/2,E); label("$x$",(R+P)/2,W); label("$a$",(P+M)/2,SE); label("$b$",(Q+M)/2,N); label("$c$",(R+M)/2,E); label("Figure 1",P-(0,1),S); [/asy] \]	[ "We rotate figure \$PRQM\$ by a clockwise angle of \$\\pi/3\$ about \$Q\$ to obtain figure \$RR'QM'\$:\n\n\\[\n[asy] size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label(\"$P$\",P,W); label(\"$Q$\",Q,E); label(\"$R$\",R,W); label(\"$M$\",M,NW); label(\"$R'$\",RR,NE); label(\"$M'$\",MM,ESE); [/asy]\n\\]\n\nEvidently, \$MM'Q\$ is an equilateral triangle, so triangles \$MRM'\$ and \$ABC\$ are congruent. Also, triangles \$PMQ\$ and \$RM'Q\$ are congruent, since they are images of each other under rotations. Then\n\n\\[\n[ABC] + \\frac{b^2 \\sqrt{3}}{4} = [MRM'] + [MM'Q] = [QMR] + [RM'Q] = [QMR] + [PMQ] .\n\\]\n\nThen by symmetry,\n\n\\[\n3 [ABC] + \\frac{(a^2+b^2+c^2)\\sqrt{3}}{4} = 2\\bigl( [PMQ] + [QMR] + [RMP] \\bigr) = 2 [PRQ] .\n\\]\n\nBut \$ABC\$ is composed of three smaller triangles. The one with sides \$w,v,a\$ has area \$\\tfrac{1}{2} wv \\sin 120^\\circ = \\frac{wv \\sqrt{3}}{4}\$. Therefore, the area of \$ABC\$ is\n\n\\[\n\\frac{(wv+vu+uw)\\sqrt{3}}{4} .\n\\]\n\nAlso, by the Law of Cosines on that small triangle of \$ABC\$, \$a^2 = w^2 + wv+ v^2\$, so by symmetry,\n\n\\[\n\\frac{(a^2 + b^2 + c^2)\\sqrt{3}}{4} = \\frac{\\bigl[2(u^2+w^2+v^2) + wv + vu + uw \\bigr] \\sqrt{3}}{4}.\n\\]\n\nTherefore\n\n\\[\n\\begin{align} \\frac{(u+v+w)^2 \\sqrt{3}}{2} &= 3 \\frac{(wv+vu+uw)\\sqrt{3}}{4} + \\frac{\\bigl[2(u^2+w^2+v^2) + wv + vu + uw \\bigr] \\sqrt{3}}{4} \\\\ &= 3 [ABC] + \\frac{(a^2+b^2+c^2)\\sqrt{3}}{4} = 2[PQR] . \\end{align}\n\\]\n\nBut the area of triangle \$PQR\$ is \$x^2 \\sqrt{3}/4\$. It follows that \$u+v+w=x\$, as desired. \$\\blacksquare\$", "Rotate \$\\triangle ABC\$ \$60\$ degrees clockwise about \$A\$ to get \$\\triangle AB'C'\$. Observe that \$\\triangle ADD'\$ is equilateral, which means \$D'D=AD=u\$. Also, \$B',D',D,C\$ are collinear because \$\\angle B'D'A + \\angle AD'D = 120+60=180\$ and \$\\angle CDA + \\angle DD'A = 120+60=180\$. The resulting \$\\triangle B'AC\$ has side lengths \$b,c,u+v+w\$ and the angle opposite side \$u+v+w\$ has magnitude \$A+60\$.\n\n\\[\n[asy] size(200); defaultpen(fontsize(8)); pair A=6expi(5/9pi), B=3expi(5/9pi+2pi/3), C=4expi(5/9pi+4pi/3), D=(0,0); path triabc = D--A--B--C--D--B..A--C; draw(triabc); label(\"$A$\",A,( 0, 1)); label(\"$B$\",B,(-1,-1));label(\"$b$\",(C+A)/2,( 1, 1)); label(\"$C$\",C,( 1,-1));label(\"$w$\",(C+D)/2,( 0, 1)); label(\"$D$\",D,( 1, 1)); transform rot60a = rotate(-60,A); pair A1 = rot60aA, B1 = rot60aB, C1 = rot60aC, D1 = rot60aD; path triabc1 = D1--A1--B1--C1--D1--B1..A1--C1; draw(triabc1, linetype(\"8 8\")); label(\"$B'$\",B1,(-1, 0));label(\"$v$\",(B1+D1)/2,( 0, 1)); label(\"$C'$\",C1,( 0,-1));label(\"$c$\",(A1+B1)/2,( 0, 1)); label(\"$D'$\",D1,(-1,-1));label(\"$u$\",(D1+D)/2,( 0, -1)); draw(A--B1--C--A, red+1.5); dot(A^^B1^^D1^^D^^C); [/asy]\n\\]\n\nIf we perform the rotation about points \$B\$ and \$C\$, we get two triangles. One has side lengths \$a,c,u+v+w\$ and the angle opposite side \$u+v+w\$ has magnitude \$B+60\$, and the other has side lengths \$b,c,u+v+w\$ and the angle opposite side \$u+v+w\$ has magnitude \$C+60\$.\n\n\\[\n[asy] size(400); defaultpen(fontsize(8)); picture transC; pair A=6expi(5/9pi), B=3expi(5/9pi+2pi/3), C=4expi(5/9pi+4pi/3), D=(0,0); path triabc = D--A--B--C--D--B..A--C; draw(triabc); label(\"$A$\",A,( 0, 1));label(\"$u$\",(A+D)/2,( 1, 0)); label(\"$B$\",B,(-1,-1)); label(\"$C$\",C,( 1,-1));label(\"$c$\",(A+B)/2,(-1, 1)); label(\"$D$\",D,( 1, 1)); transform rot60b = rotate(-60,B); pair A1 = rot60bA, B1 = rot60bB, C1 = rot60bC, D1 = rot60bD; path triabc1 = D1--A1--B1--C1--D1--B1..A1--C1; draw(triabc1, linetype(\"8 8\")); label(\"$A'$\",A1,( 1, 0));label(\"$a$\",(B1+C1)/2,(-1, 0)); label(\"$C'$\",C1,( 0,-1));label(\"$w$\",(C1+D1)/2,( 1, 0)); label(\"$D'$\",D1,(-1,-1));label(\"$v$\",(D1+D)/2,( 1, 1)); draw(A--B--C1--A, red+1.5); dot(A^^B^^D1^^D^^C1); draw(transC, triabc); label(transC, \"$A$\",A,( 0, 1));label(transC, \"$a$\",(B+C)/2,( 0,-1)); label(transC, \"$B$\",B,(-1,-1));label(transC, \"$v$\",(B+D)/2,( 0, 1)); label(transC, \"$C$\",C,( 1,-1)); label(transC, \"$D$\",D,( 0,-1)); transform rot60c = rotate(-60,C); pair A2 = rot60cA, B2 = rot60cB, C2 = rot60cC, D2 = rot60cD; path triabc2 = D2--A2--B2--C2--D2--B2..A2--C2; draw(transC, triabc2, linetype(\"8 8\")); label(transC, \"$A'$\",A2,( 1, 0));label(transC, \"$u$\",(A2+D2)/2,( 1,-1)); label(transC, \"$b$\",(C2+A2)/2,( 1,-1)); label(transC, \"$w$\",(D+D2)/2,( 1,-1)); label(transC, \"$D'$\",D2,( 0, 1)); draw(transC, A2--B--C--A2, red+1.5); dot(transC, C^^B^^D2^^D^^A2); add(shift(15right)transC); [/asy]\n\\]\n\nThese three triangles fit together because \$(A+60)+(B+60)+(C+60) = 360\$. The result is an equilateral triangle of side length \$u+v+w\$.", "\\[\n[asy] size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label(\"$P$\",P,W); label(\"$Q$\",Q,E); label(\"$R$\",R,W); label(\"$M$\",M,NW); label(\"$R'$\",RR,NE); label(\"$M'$\",MM,ESE); [/asy]\n\\]\n\nAs in the first solution, we rotate and establish that \$\\triangle MRM' \\cong \\triangle ABC\$.\n\nLet \$X\$ and \$Y\$ be points on \$\\overline{RQ}\$ and \$\\overline{PQ}\$, respectively, such that \$M\$ lies on \$\\overline{XY}\$ and \$\\overline{XY}\\parallel \\overline{PR}\$. We note that \$m\\angle RXM = 120^\\circ\$. The rotation then takes \$Y\$ to \$X\$, so \$m\\angle RXM'=m\\angle PYM = 120^\\circ\$. It follows that \$RX=BD=v\$, \$MX=AD=u\$, \$M'X=CD=w\$.\n\nSince \$m\\angle MXM' = 120^\\circ\$ and \$m\\angle MQM' = 60^\\circ\$, \$MXM'Q\$ is cyclic. By Ptolemy's theorem,\n\n\\[\n\\begin{align} (MX)(M'Q) + (XM')(QM) &= (MM')(XQ) \\\\ XQ &= MX + M'X \\\\ &= u+w. \\end{align}\n\\]\n\nFinally, \$RQ = RX+XQ = u+v+w\$, as desired." ]
USAMO-1975-1	https://artofproblemsolving.com/wiki/index.php/1975_USAMO_Problems/Problem_1	(a) Prove that \[ [5x]+[5y]\ge [3x+y]+[3y+x], \] where $x,y\ge 0$ and $[u]$ denotes the greatest integer $\le u$ (e.g., $[\sqrt{2}]=1$). (b) Using (a) or otherwise, prove that \[ \frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!} \] is integral for all positive integral $m$ and $n$.	[]
USAMO-1975-2	https://artofproblemsolving.com/wiki/index.php/1975_USAMO_Problems/Problem_2	Let $A,B,C,D$ denote four points in space and $AB$ the distance between $A$ and $B$, and so on. Show that \[ AC^2+BD^2+AD^2+BC^2\ge AB^2+CD^2. \]	[ "\\[\n[asy] defaultpen(fontsize(8)); pair A=(2,4), B=(0,0), C=(4,0), D=(4,2); label(\"A\",A,(0,1));label(\"D\",D,(1,0));label(\"B\",B,(-1,-1));label(\"C\",C,(1,-1)); axialshade(A--C--D--cycle, lightgray, A, gray, D); draw(A--B--C--A--D--C);draw(B--D, linetype(\"8 8\")); label(\"$m$\",(A+B)/2,(-1,1));label(\"$n$\",(C+D)/2,(1,0)); label(\"$c$\",(B+C)/2,(0,-1));label(\"$b$\",(A+C)/2,(-1,-1)); label(\"$a$\",(A+D)/2,(1,1));label(\"$d$\",(B+D)/2,(-1,1)); [/asy]\n\\]\n\nIf we project points \$A,B,C,D\$ onto the plane parallel to \$\\overline{AB}\$ and \$\\overline{CD}\$, \$AB\$ and \$CD\$ stay the same but \$BC, AC, AD, BD\$ all decrease, making the inequality sharper. Thus, it suffices to prove the inequality when \$A,B,C,D\$ are coplanar:\n\n\\[\n[asy] size(200); defaultpen(fontsize(8)); pair A=(8,3), B=(4,-5), C=(10,0), D=(0,0); draw(A--C--B--D--A--B);draw(C--D);draw(anglemark(A,B,D,40));draw(anglemark(C,B,A,55,60)); label(\"A\",A,(0,1));label(\"D\",D,(-1,0));label(\"B\",B,(0,-1));label(\"C\",C,(1,0)); label(\"$m$\",(A+B)/2,(1,0));label(\"$n$\",(C+D)/2,(0,1)); label(\"$c$\",(B+C)/2,(1,-1));label(\"$b$\",(A+C)/2,(-1,-1)); label(\"$a$\",(A+D)/2,(0,1));label(\"$d$\",(B+D)/2,(-1,-1)); label(\"$\\phi-\\theta$\",anglemark(A,B,D,40),(1,5));label(\"$\\theta$\",anglemark(C,B,A,55),(8,9)); [/asy]\n\\]\n\nLet \$AD=a, AC=b, BC=c, BD=d, AB=m, CD=n\$. We wish to prove that \$a^2+b^2+c^2+d^2\\ge m^2+n^2\$. Let us fix \$\\triangle BCD\$ and the length \$AB\$ and let \$A\$ vary on the circle centered at \$B\$ with radius \$m\$. If we find the minimum value of \$a^2+b^2\$, which is the only variable quantity, and prove that it is larger than \$m^2+n^2-c^2-d^2\$, we will be done.\n\nFirst, we express \$a^2+b^2\$ in terms of \$c,d,m,\\theta,\\phi\$, using the Law of Cosines:\n\n\\[\n\\begin{align} a^2+b^2 &= c^2+d^2+2m^2-2cm\\cos(\\theta)-2dm\\cos(\\phi-\\theta) \\\\ (a^2+b^2-c^2-d^2-2m^2)^2 &= 4m^2(c^2\\cos^2(\\theta)+d^2\\cos^2(\\phi-\\theta)+2cd\\cos(\\theta)\\cos(\\phi-\\theta)) \\end{align}\n\\]\n\n\$a^2+b^2\$ is a function of \$\\theta\$, so we take the derivative with respect to \$\\theta\$ and obtain that \$a^2+b^2\$ takes a minimum when\n\n\\[\n\\begin{align} c\\sin(\\theta)-d\\sin(\\phi-\\theta) &= 0 \\\\ c^2\\sin^2(\\theta)+d^2\\sin^2(\\phi-\\theta)-2cd\\sin(\\theta)\\sin(\\phi-\\theta) &= 0 \\\\ (a^2+b^2-c^2-d^2-2m^2)^2 &= 4m^2(c^2+d^2+2cd(\\cos(\\theta)\\cos(\\phi-\\theta)-\\sin(\\theta)\\sin(\\phi-\\theta))) \\\\ &= 4m^2(c^2+d^2+2cd\\cos{\\phi})\\\\ &= 4m^2(2c^2+2d^2-n^2) \\end{align}\n\\]\n\nDefine \$p=a^2+b^2\$ and \$q=c^2+d^2\$:\n\n\\[\n\\begin{align} (p-q-2m^2)^2 &= 4m^2(2q-n^2) \\\\ p^2+q^2+4m^4-4m^2p+4m^2q-2pq &= 8m^2q-4m^2n^2 \\\\ p^2+q^2+4m^4-4m^2p-4m^2q-2pq &= -4m^2n^2 \\\\ p^2-2pq+q^2-4m^2(p+q) &= -4m^2(m^2+n^2) \\\\ \\frac{(p-q)^2}{m^2} &= p+q-m^2-n^2\\geq 0 \\\\ a^2+b^2+c^2+d^2 &\\geq m^2+n^2 \\\\ \\end{align}\n\\]", "Let\n\n\\[\n\\begin{align} A &= (0,0,0) \\\\ B &= (1,0,0) \\\\ C &= (a,b,c) \\\\ D &= (x,y,z). \\end{align}\n\\]\n\nIt is clear that every other case can be reduced to this. Then, with the distance formula and expanding,\n\n\\[\n\\begin{align} AC^2 + BD^2 + AD^2 + BC^2 - AB^2 - CD^2 &= x^2-2x+1+y^2+z^2+a^2-2a+b^2+c^2+2ax+2by+2cz \\\\ &= (x+a-1)^2 + (y+b)^2 + (z+c)^2. \\\\ &\\geq 0, \\end{align}\n\\]\n\nwhich rearranges to the desired inequality.", "Because the distances are all squared, we must only prove the result in one dimension, and then we can just add up the three individual inequalities for the \$x\$, \$y\$, and \$z\$ dimension to get the desired result. Let \$x_a\$, \$x_b\$, \$x_c\$, and \$x_d\$ be the positions of \$A\$, \$B\$, \$C\$, and \$D\$ respectively. Then we must show that,\n\n\\[\n\\begin{align} (x_a - x_c)^2 + (x_b-x_d)^2 + (x_a - x_d)^2 + (x_b - x_c)^2 &\\geq (x_a - x_b)^2 + (x_c-x_d)^2 \\\\ x_a^2 + x_b^2 + x_c^2 + x_d^2 &\\geq 2x_a x_c + 2x_b x_d + 2x_a x_d + 2x_b x_c - 2x_a x_b - 2x_c x_d \\\\ (x_a + x_b)^2 + (x_c + x_d)^2 &\\geq 2(x_a +x_b)(x_c + x_d)\\\\ (x_a + x_b - x_c - x_d)^2 &\\geq 0. \\end{align}\n\\]\n\nSo we are done.", "Let \$a\$, \$b\$, \$c\$, \$d\$ correspond to the position vectors of points A, B, C, and D, respectively, with respect to an arbitrary origin O. Let us also for simplicity define \$a^2 = a \\cdot a = \|\|a\|\|^2\$, where \$\|\|a\|\|\$ is the magnitude of vector \$a\$. Because squares are non-negative, \$a^2\$ is non-negative for all vectors \$a\$. Thus,\n\n\\[\n(a + b - c - d)^2 \\ge 0\n\\]\n\nBecause dot product is linear, we expand to obtain\n\n\\[\na^2 + b^2 + c^2 + d^2 + 2a \\cdot b + 2 c \\cdot d - 2 a \\cdot c - 2 a \\cdot d - 2 b \\cdot c - 2 c \\cdot d \\ge 0,\n\\]\n\nfrom which we add \$a^2 + b^2 + c^2 + d^2\$ to both sides, rearrange, and complete the square to get\n\n\\[\n(a-c)^2 + (a-d)^2 + (b-c)^2 + (b-d)^2 \\ge (a-b)^2 + (c-d)^2.\n\\]\n\nAs \$(a-b)^2 = \|\|a-b\|\|^2 = \|\|AB\|\|^2 = AB^2\$ and likewise for the others,\n\n\\[\nAC^2 + AD^2 + BC^2 + BD^2 \\ge AB^2 + CD^2,\n\\]\n\nwhich is what we wanted to prove.\n\nNOTES:\n\n1. Equality holds when the vector equality \$a + b = c + d\$ holds, which occurs when A, B, C, and D are the vertices of a (planar) parallelogram, in that order.\n\n2. The algebra employed here is almost identical to the algebra used in Solution 3, which means that Solution 3 is just a simplification of this solution." ]
USAMO-1975-3	https://artofproblemsolving.com/wiki/index.php/1975_USAMO_Problems/Problem_3	If $P(x)$ denotes a polynomial of degree $n$ such that \[ P(k)=\frac{k}{k+1} \] for $k=0,1,2,\ldots,n$, determine $P(n+1)$.	[ "Let \$Q(x) = (x+1)P(x) - x\$, and clearly, \$Q(x)\$ has a degree of \$n+1\$.\n\nThen, for \$k=0,1,2,\\ldots,n\$, \$Q(k) = (k+1)P(k) - k = (k+1)\\cdot \\dfrac{k}{k+1} - k = 0\$.\n\nThus, \$k=0,1,2,\\ldots,n\$ are the roots of \$Q(x)\$.\n\nSince these are all \$n+1\$ of the roots of the \$n+1^{\\text{th}}\$ degree polynomial, by the Factor Theorem, we can write \$Q(x)\$ as\n\n\\[\nQ(x) = c(x)(x-1)(x-2) \\cdots (x-n)\n\\]\n\nwhere \$c\$ is a constant.\n\nThus,\n\n\\[\n(x+1)P(x) - x = c(x)(x-1)(x-2) \\cdots (x-n).\n\\]\n\nWe plug in \$x = -1\$ to cancel the \$(x+1)P(x)\$ and find \$c\$:\n\n\\[\n\\begin{align} -(-1) &= c(-1)(-1-1)(-1-2) \\cdots (-1-n) \\\\ 1 &= c(-1)^{n+1}(1)(2) \\cdots (n+1) \\\\ c &= (-1)^{n+1}\\dfrac{1}{(n+1)!} \\\\ \\end{align}\n\\]\n\nFinally, plugging in \$x = n+1\$ to find \$P(n+1)\$ gives:\n\n\\[\n\\begin{align} Q(n+1)&=(n+2)P(n+1)-(n+1)\\\\ (-1)^{n+1}\\dfrac{1}{(n+1)!}\\cdot(n+1)! &=(n+2)P(n+1)-(n+1)\\\\ (-1)^{n+1}&=(n+2)P(n+1)-(n+1)\\\\ (-1)^{n+1}+(n+1)&=(n+2)P(n+1)\\\\ P(n+1) &= \\dfrac{(-1)^{n+1} + (n+1)}{n+2}\\\\ \\end{align}\n\\]\n\nIf \$n\$ is even, this simplifies to \$P(n+1) = \\dfrac{n}{n+2}\$. If \$n\$ is odd, this simplifies to \$P(n+1) = 1\$. \$\\Box\$\n\n~Edits by BakedPotato66", "It is fairly natural to use Lagrange's Interpolation Formula on this problem:\n\n\\[\n\\begin{align} P(n+1) &= \\sum_{k=0}^n \\frac{k}{k+1} \\prod_{j \\ne k} \\frac{n+1-j}{k-j} \\\\ &= \\sum_{k=0}^n \\frac{k}{k+1} \\cdot \\frac{\\frac{(n+1)!}{n+1-k}}{k(k-1)(k-2) \\dots 1\\cdot (-1)(-2) \\dots (k-n)} \\\\ &= \\sum_{k=0}^n \\frac{k}{k+1} (-1)^{n-k}\\cdot \\frac{(n+1)!}{k!(n+1-k)!} \\\\ &= \\sum_{k=0}^n (-1)^{n-k} \\binom{n+1}{k} - \\sum_{k=0}^n \\frac{(n+1)!(-1)^{n-k}}{(k+1)!(n+1-k)!} \\\\ &= -\\left(\\sum_{k=0}^{n+1} (-1)^{n+1-k} \\binom{n+1}{k} - 1\\right) + \\frac{1}{n+2} \\cdot \\sum_{k=0}^n (-1)^{n+1-k} \\binom{n+2}{k+1} \\\\ &= 1 + \\frac{1}{n+2} \\left(\\sum_{k=-1}^{n+1} (-1)^{n+2 - (k+1)} \\binom{n+2}{k+1} - (-1)^{n+2} - 1\\right) \\\\ &= \\boxed{1 - \\frac{(-1)^n + 1}{n+2}} \\end{align}\n\\]\n\nthrough usage of the Binomial Theorem. \$\\square\$\n\n~lpieleanu (minor editing and reformatting)" ]
USAMO-1975-4	https://artofproblemsolving.com/wiki/index.php/1975_USAMO_Problems/Problem_4	Two given circles intersect in two points $P$ and $Q$. Show how to construct a segment $AB$ passing through $P$ and terminating on the two circles such that $AP\cdot PB$ is a maximum. \[ [asy] size(150); defaultpen(fontsize(7)); pair A=(0,0), B=(10,0), P=(4,0), Q=(3.7,-2.5); draw(A--B); draw(circumcircle(A,P,Q)); draw(circumcircle(B,P,Q)); label("A",A,(-1,1));label("P",P,(0,1.5));label("B",B,(1,1));label("Q",Q,(-0.5,-1.5)); [/asy] \]	[ "A maximum \$AP \\cdot PB\$ cannot be attained if \$AB\$ intersects segment \$O_1O_2\$ because a larger value can be attained by making one of \$A\$ or \$B\$ diametrically opposite \$P\$, which (as is easily checked) increases the value of both \$AP\$ and \$PB\$. Thus, assume \$AB\$ does not intersect \$O_1O_2\$.\n\nLet \$E\$ and \$F\$ be the centers of the small and big circles, respectively, and \$r\$ and \$R\$ be their respective radii.\n\nLet \$M\$ and \$N\$ be the feet of \$E\$ and \$F\$ to \$AB\$, and \$\\alpha = \\angle APE\$ and \$\\epsilon = \\angle BPF\$\n\nWe have:\n\n\\[\nAP \\times PB = 2r \\cos{\\alpha} \\times 2R \\cos{\\epsilon} = 4 rR \\cos{\\alpha} \\cos{\\epsilon}\n\\]\n\n\$AP\\times PB\$ is maximum when the product \$\\cos{\\alpha} \\cos{\\epsilon}\$ is a maximum.\n\nWe have \$\\cos{\\alpha} \\cos{\\epsilon}= \\frac{1}{2} [\\cos(\\alpha +\\epsilon) + \\cos(\\alpha -\\epsilon)]\$\n\nBut \$\\alpha +\\epsilon = 180^{\\circ} - \\angle EPF\$ and is fixed, so is \$\\cos(\\alpha +\\epsilon)\$.\n\nSo its maximum depends on \$cos(\\alpha -\\epsilon)\$ which occurs when \$\\alpha=\\epsilon\$. To draw the line \$AB\$:\n\nDraw a circle with center \$P\$ and radius \$PE\$ to cut the radius \$PF\$ at \$H\$. Draw the line parallel to \$EH\$ passing through \$P\$. This line meets the small and big circles at \$A\$ and \$B\$, respectively." ]
USAMO-1975-5	https://artofproblemsolving.com/wiki/index.php/1975_USAMO_Problems/Problem_5	A deck of $n$ playing cards, which contains three aces, is shuffled at random (it is assumed that all possible card distributions are equally likely). The cards are then turned up one by one from the top until the second ace appears. Prove that the expected (average) number of cards to be turned up is $(n+1)/2$.	[ "We begin by induction. Our base case is naturally when \$n = 3\$, as there can be no less than \$3\$ cards in the deck. The only way to turn up the second ace is to turn up the first, and then turn up the second, which requires \$2\$ moves. This indeed is equal to \$(3 + 1)/2\$. Next, we assume that the expected number of cards required to pull the second ace from a deck of n cards is in fact \$(n + 1)/2\$. Now, we see that to simulate the case with \$n + 1\$ cards, we simply must add a card to our deck. By our assumption, we can expect to turn up \$(n + 1)/2\$ cards to be turned up, including our second ace. Thus ahead of our second ace, we expect \$(n - 1)/2\$ cards , and after it \$(n - 1)/2\$ as well. When we add our card, it is ahead of the second ace half the time, and behind if half the time. Therefore we expect to add one card to the number of cards drawn half the time, and add nothing for half the time. Hence we expect to add half a card, and the expected value is now \$((n + 1)/2) + 1/2)\$, or \$(n + 2)/2\$, the desired outcome for the \$n + 1\$ case.", "Although we can induct as in the situation above, we can use a more direct approach.\n\nConsider a deck. The second ace is in the \$k\$th place. If we draw the deck backwards (which has equal probability as drawing it forwards), then we will draw it in the \$n+1-k\$th place, since the \$n\$th place will become the first, the \$n-1\$th the second, etc.\n\nAlso, note that when second ace is still the second ace when we draw the deck backwards. (we need to note this, else this argument holds for the first ace and third ace as well)\n\nNow, this implies that the expected location of the second ace is in the \$(k+n+1-k)/2=(n+1)/2\$ place.", "Like Solution 2, this solution is also a direct solution.\n\nThe three aces divide the deck into four piles with a non-negative number of cards. Let \$a\$, \$b\$, \$c\$, and \$d\$ be the number of cards in the first, second, third, and fourth piles, respectively, given a random deck \$D\$. Also, let \$E(a), E(b), E(c), E(d)\$ denote the expected value of \$a\$, \$b\$, \$c\$, and \$d\$ over all decks \$D\$. Then \$a + b + c + d = n - 3\$ (the deck minus the three aces).\n\nFor every deck \$D\$, there must exist other decks that permute the values of \$a\$, \$b\$, \$c\$, \$d\$. (i.e. if there is a deck with \$a = 2, b = 3, c = 1, d = 4,\$ there exists a deck with \$a = 1, b = 2, c = 3, d = 4\$.) Therefore, the value of \$a\$ over all decks \$D\$ is a permutation of the value of \$b\$ over all such decks, and as a result \$E(a) = E(b)\$. A similar result holds for \$b\$, \$c\$, and \$d\$, and so \$E(a) = E(b) = E(c) = E(d)\$. Because\n\n\\[\nE(a) + E(b) + E(c) + E(d) = E(a + b + c + d) = E(n-3) = n-3,\n\\]\n\n\$E(a) = E(b) = \\frac{n-3}{4}\$. We want to compute the number of cards up to and including the second ace, which is just\n\n\\[\nE(a) + 1 + E(b) + 1 = \\frac{n-3}{2} + 2 = \\frac{n+1}{2},\n\\]\n\nas desired." ]
USAMO-1976-1	https://artofproblemsolving.com/wiki/index.php/1976_USAMO_Problems/Problem_1	\[ [asy] void fillsq(int x, int y){ fill((x,y)--(x+1,y)--(x+1,y+1)--(x,y+1)--cycle, mediumgray); } int i; fillsq(1,0);fillsq(4,0);fillsq(6,0); fillsq(0,1);fillsq(1,1);fillsq(2,1);fillsq(4,1);fillsq(5,1); fillsq(0,2);fillsq(2,2);fillsq(4,2); fillsq(0,3);fillsq(1,3);fillsq(4,3);fillsq(5,3); for(i=0; i<=7; ++i){draw((i,0)--(i,4),black+0.5);} for(i=0; i<=4; ++i){draw((0,i)--(7,i),black+0.5);} draw((3,1)--(3,3)--(7,3)--(7,1)--cycle,black+1); [/asy] \] - (a) Suppose that each square of a $4\times 7$ chessboard, as shown above, is colored either black or white. Prove that with any such coloring, the board must contain a rectangle (formed by the horizontal and vertical lines of the board such as the one outlined in the figure) whose four distinct unit corner squares are all of the same color. - (b) Exhibit a black-white coloring of a $4\times 6$ board in which the four corner squares of every rectangle, as described above, are not all of the same color.	[ "There are many ways to prove the first part, we will show one.\n\nConsider the first row. It contains at least four cells of the same color, say white. Pick any four such cells. Let's now consider these four columns only. If any of the remaining three rows contains two white cells in some of these columns, we are done. Otherwise, we have the following situation: three rows times four columns, and in each row there is at most one white cell. We can now pick any two rows and then find two columns such that all four cells where they intersect are black, and we are done. (Note that this proof in fact shows that already in a \$3\\times 7\$ board there must be such a rectangle.)\n\nFor the second part, consider the board shown below. Its six columns contain the six different permutations of two white and two black squares. From this observation it is immediately obvious that no two columns agree on two cells of the same color.\n\n\\[\n[asy] void fillsq(int x, int y){ fill((x,y)--(x+1,y)--(x+1,y+1)--(x,y+1)--cycle, mediumgray); } int i; fillsq(0,0);fillsq(0,1); fillsq(1,0);fillsq(1,2); fillsq(2,0);fillsq(2,3); fillsq(3,1);fillsq(3,2); fillsq(4,1);fillsq(4,3); fillsq(5,2);fillsq(5,3); for(i=0; i<=6; ++i){draw((i,0)--(i,4),black+0.5);} for(i=0; i<=4; ++i){draw((0,i)--(6,i),black+0.5);} [/asy]\n\\]", "We will prove the first part by the pigeonhole principle. To form a rectangle, two squares have in one column have to be colored the same color as the corresponding squares in another column. In the example given, the two second and third squares of the fourth and seventh columns are all white. When coloring this rectangle, every column can be entirely one color, have 3 squares be of one color, or have 2 squares be of one color. Coloring each column with two white squares and two black squares will minimize the number of squares, since a smaller fraction of the column is one color (if more than 2 squares in the same column are the same color, then there is more than one coloring in another column that can create a rectangle). There are \${4}\\choose{2}\$, or \$6\$ different ways a column can be colored with two white and two black squares. We need the same coloring in two different columns to create a rectangle, and this is guaranteed with 7 columns by the pigeonhole principle.\n\nFor the second part, we can draw a rectangle with each column having a different coloring with 2 black and 2 white squares. One such rectangle is shown below:\n\n\\[\n[asy] void fillsq(int x, int y){ fill((x,y)--(x+1,y)--(x+1,y+1)--(x,y+1)--cycle, mediumgray); } int i; fillsq(0,3);fillsq(0,2); fillsq(1,3);fillsq(1,1); fillsq(2,3);fillsq(2,0); fillsq(3,2);fillsq(3,1); fillsq(4,2);fillsq(4,0); fillsq(5,1);fillsq(5,0); for(i=0; i<=6; ++i){draw((i,0)--(i,4),black+0.5);} for(i=0; i<=4; ++i){draw((0,i)--(6,i),black+0.5);} [/asy]\n\\]" ]
USAMO-1976-2	https://artofproblemsolving.com/wiki/index.php/1976_USAMO_Problems/Problem_2	If $A$ and $B$ are fixed points on a given circle and $XY$ is a variable diameter of the same circle, determine the locus of the point of intersection of lines $AX$ and $BY$. You may assume that $AB$ is not a diameter. \[ [asy] size(300); defaultpen(fontsize(8)); real r=10; picture pica, picb; pair A=rexpi(5pi/6), B=rexpi(pi/6), X=rexpi(pi/3), X1=rexpi(-pi/12), Y=rexpi(4pi/3), Y1=rexpi(11pi/12), O=(0,0), P, P1; P = extension(A,X,B,Y);P1 = extension(A,X1,B,Y1); path circ1 = Circle((0,0),r); draw(pica, circ1);draw(pica, B--A--P--Y--X);dot(pica,P^^O); label(pica,"$A$",A,(-1,1));label(pica,"$B$",B,(1,0));label(pica,"$X$",X,(0,1));label(pica,"$Y$",Y,(0,-1));label(pica,"$P$",P,(1,1));label(pica,"$O$",O,(-1,1));label(pica,"(a)",O+(0,-13),(0,0)); draw(picb, circ1);draw(picb, B--A--X1--Y1--B);dot(picb,P1^^O); label(picb,"$A$",A,(-1,1));label(picb,"$B$",B,(1,1));label(picb,"$X$",X1,(1,-1));label(picb,"$Y$",Y1,(-1,0));label(picb,"$P'$",P1,(-1,-1));label(picb,"$O$",O,(-1,-1)); label(picb,"(b)",O+(0,-13),(0,0)); add(pica); add(shift(30right)*picb); [/asy] \]	[ "WLOG, assume that the circle is the unit circle centered at the origin. Then the points \$A\$ and \$B\$ have coordinates \$(-a,b)\$ and \$(a,b)\$ respectively and \$X\$ and \$Y\$ have coordinates \$(r,s)\$ and \$(-r,-s)\$. Note that these coordinates satisfy \$a^2 + b^2 = 1\$ and \$r^2 + s^2 = 1\$ since these points are on a unit circle. Now we can find equations for the lines:\n\n\\[\n\\begin{align} AX \\longrightarrow y &= \\frac{(s-b)x+rb+sa}{r+a}\\\\ BY \\longrightarrow y &= \\frac{(s+b)x+rb-sa}{r+a}. \\end{align}\n\\]\n\nSolving these simultaneous equations gives coordinates for \$P\$ in terms of \$a, b, r,\$ and \$s\$: \$P = \\left(\\frac{as}{b},\\frac{1 - ar}{b}\\right)\$. These coordinates can be parametrized in Cartesian variables as follows:\n\n\\[\n\\begin{align} x &= \\frac{as}{b}\\\\ y &= \\frac{1 - ar}{b}. \\end{align}\n\\]\n\nNow solve for \$r\$ and \$s\$ to get \$r = \\frac{1-by}{a}\$ and \$s = \\frac{bx}{a}\$ . Then since \$r^2 + s^2 = 1, \\left(\\frac{bx}{a}\\right)^2 + \\left(\\frac{1-by}{a}\\right)^2 = 1\$ which reduces to \$x^2 + (y-1/b)^2 = \\frac{a^2}{b^2}.\$ This equation defines a circle and is the locus of all intersection points \$P\$. In order to define this locus more generally, find the slope of this circle function using implicit differentiation:\n\n\\[\n\\begin{align} 2x + 2(y-1/b)y' &= 0\\\\ (y-1/b)y' &= -x\\\\ y' &= \\frac{-x}{y-1/b}. \\end{align}\n\\]\n\nNow note that at points \$A\$ and \$B\$, this slope expression reduces to \$y' = \\frac{-b}{a}\$ and \$y' = \\frac{b}{a}\$ respectively, values which are identical to the slopes of lines \$AO\$ and \$BO\$. Thus we conclude that the complete locus of intersection points is the circle tangent to lines \$AO\$ and \$BO\$ at points \$A\$ and \$B\$ respectively.", "Notice that \$m\\angle AYP=\\frac{1}{2}m\\overarc{AB}\$ (Inscried angle theorem) and that \$m\\angle XAY=90\$ since \$XY\$ is a diameter, and thus subtends an arc of \$180\$. This will hold for all \$X\$ and all \$Y\$, and so by AA similarity, the angle \$APY\$ will be constant for all P, thus implying that the points A, B, and all P will be concyclic. If we assume that the center of this circle is \$H\$, we know that \$m\\angle AHB=2m\\angle APB=2(90-\\frac{1}{2}\\overarc{AB})=180-\\overarc{AB}\$. We can assume that \$\\overline{AH}\$ and \$\\overline{BH}\$ intersect the original circle at points \$M\$, and \$N\$ respectively. This will give us that \$m\\angle AHB=\\frac{1}{2}\|m\\overarc{MN}-m\\overarc{AB}\|\$ (Since \$H\$ lies on the perpendicular bisector of \$AB\$, we know that \$M\$, \$N\$ will be on the same side of \$AB\$.) Now we also know that \$m\\overarc{MN}\\leq m\\overarc{AB}\$ or \$m\\overarc{MN}\\leq 360-m\\overarc{AB}\$. The only case where \$m\\overarc{MN}\$ satisfies the measure of \$\\angle AHB\$, is when \$M=A\$ and \$N=B\$, implying that \$AH\$ and \$BH\$ are tangents, and so \$H\$ is the intersection of the tangents from \$A\$ and \$B\$ to the original circle." ]
USAMO-1976-3	https://artofproblemsolving.com/wiki/index.php/1976_USAMO_Problems/Problem_3	Determine all integral solutions of $a^2+b^2+c^2=a^2b^2$.	[ "Either \$a^2=0\$ or \$a^2>0\$. If \$a^2=0\$, then \$b^2=c^2=0\$. Symmetry applies for \$b\$ as well. If \$a^2,b^2\\neq 0\$, then \$c^2\\neq 0\$. Now we look at \$a^2\\bmod{4}\$:\n\n\$a^2\\equiv 0\\bmod{4}\$: Since a square is either 1 or 0 mod 4, then all the other squares are 0 mod 4. Let \$a=2a_1\$, \$b=2b_1\$, and \$c=2c_1\$. Thus \$a_1^2+b_1^2+c_1^2=4a_1^2b_1^2\$. Since the LHS is divisible by four, all the variables are divisible by 4, and we must do this over and over again, and from infinite descent, there are no non-zero solutions when \$a^2\\equiv 0\\bmod{4}\$.\n\n\$a^2\\equiv 1\\bmod{4}\$: Since \$b^2\\neq 0\\bmod{4}\$, \$b^2\\equiv 1\\bmod{4}\$, and \$2+c^2\\equiv 1\\bmod{4}\$. But for this to be true, \$c^2\\equiv 3\\bmod{4}\$, which is an impossibility. Thus there are no non-zero solutions when \$a^2\\equiv 1\\bmod{4}\$.\n\nThus the only solution is the solution above: \$(a,b,c)=0\$." ]
USAMO-1976-4	https://artofproblemsolving.com/wiki/index.php/1976_USAMO_Problems/Problem_4	If the sum of the lengths of the six edges of a trirectangular tetrahedron $PABC$ (i.e., $\angle APB=\angle BPC=\angle CPA=90^o$) is $S$, determine its maximum volume.	[ "Let the side lengths of \$AP\$, \$BP\$, and \$CP\$ be \$a\$, \$b\$, and \$c\$, respectively. Therefore \$S=a+b+c+\\sqrt{a^2+b^2}+\\sqrt{b^2+c^2}+\\sqrt{c^2+a^2}\$. Let the volume of the tetrahedron be \$V\$. Therefore \$V=\\frac{abc}{6}\$.\n\nNote that \$(a-b)^2\\geq 0\$ implies \$\\frac{a^2-2ab+b^2}{2}\\geq 0\$, which means \$\\frac{a^2+b^2}{2}\\geq ab\$, which implies \$a^2+b^2\\geq ab+\\frac{a^2+b^2}{2}\$, which means \$a^2+b^2\\geq \\frac{(a+b)^2}{2}\$, which implies \$\\sqrt{a^2+b^2}\\geq \\frac{1}{\\sqrt{2}} \\cdot (a+b)\$. Equality holds only when \$a=b\$. Therefore\n\n\\[\nS\\geq a+b+c+\\frac{1}{\\sqrt{2}} \\cdot (a+b)+\\frac{1}{\\sqrt{2}} \\cdot (c+b)+\\frac{1}{\\sqrt{2}} \\cdot (a+c)\n\\]\n\n\$=(a+b+c)(1+\\sqrt{2})\$.\n\n\$\\frac{a+b+c}{3}\\geq \\sqrt[3]{abc}\$ is true from AM-GM, with equality only when \$a=b=c\$. So \$S\\geq (a+b+c)(1+\\sqrt{2})\\geq 3(1+\\sqrt{2})\\sqrt[3]{abc}=3(1+\\sqrt{2})\\sqrt[3]{6V}\$. This means that \$\\frac{S}{3(1+\\sqrt{2})}=\\frac{S(\\sqrt{2}-1)}{3}\\geq \\sqrt[3]{6V}\$, or \$6V\\leq \\frac{S^3(\\sqrt{2}-1)^3}{27}\$, or \$V\\leq \\frac{S^3(\\sqrt{2}-1)^3}{162}\$, with equality only when \$a=b=c\$. Therefore the maximum volume is \$\\frac{S^3(\\sqrt{2}-1)^3}{162}\$.", "Note that by AM-GM\n\n\\[\nS = a + b + c + \\sqrt{a^2 + b^2} + \\sqrt{b^2 + c^2} + \\sqrt{a^2 + c^2} \\ge 3\\sqrt[3]{abc} + \\sqrt{2ab} + \\sqrt{2bc} + \\sqrt{2ac} \\ge 3\\sqrt[3]{abc} + 3\\sqrt{2}\\sqrt[3]{abc},\n\\]\n\nso\n\n\\[\n\\sqrt[3]{abc} \\le \\frac{S}{3 + 3\\sqrt{2}} = \\frac{S}{3} \\cdot (\\sqrt{2} - 1).\n\\]\n\nProceed as before." ]
USAMO-1976-5	https://artofproblemsolving.com/wiki/index.php/1976_USAMO_Problems/Problem_5	If $P(x)$, $Q(x)$, $R(x)$, and $S(x)$ are all polynomials such that \[ P(x^5) + xQ(x^5) + x^2 R(x^5) = (x^4 + x^3 + x^2 + x +1) S(x), \] prove that $x-1$ is a factor of $P(x)$.	[ "In general we will show that if \$m\$ is an integer less than \$n\$ and \$P_0, \\dotsc, P_{m-1}\$ and \$S\$ are polynomials satisfying\n\n\\[\nP_0(x^n) + x P_1(x^n) + \\dotsb + x^{m-1} P_{m-1}(x^n) = \\sum_{k=0}^{n-1} x^k S(x),\n\\]\n\nthen \$P_k(1) = 0\$, for all integers \$0 \\le k \\le m-1\$. For the problem, we may set \$n=5\$, \$m=3\$, and then note that since \$P(1)= 0\$, \$(x-1)\$ is a factor of \$P(x)\$.\n\nIndeed, let \$\\omega_1, \\dotsc, \\omega_{n-1}\$ be the \$n\$th roots of unity other than 1. Then for all integers \$1 \\le i \\le n-1\$,\n\n\\[\n\\sum_{k=0}^{m-1} \\omega_i^k P_k(\\omega_i^n) = \\sum_{k=0}^{m-1} \\omega_i^k P_k(1) = \\sum_{k=0}^{n-1} \\omega_i^k S(\\omega_i) = \\frac{\\omega_i^{n-1} - 1}{\\omega_i - 1} S(\\omega_i) = 0 ,\n\\]\n\nfor all integers \$1 \\le i \\le n\$. This means that the \$(m-1)\$th degree polynomial\n\n\\[\n\\sum_{k=0}^{m-1} x^k P_k(1)\n\\]\n\nhas \$n-1 > m-1\$ distinct roots. Therefore all its coefficients must be zero, so \$P_k = 0\$ for all integers \$0 \\le k \\le m-1\$, as desired. \$\\blacksquare\$", "Let \$\\zeta, \\xi, \\rho\$ be three distinct primitive fifth roots of unity. Setting \$x = \\zeta, \\xi\$, we have\n\n\\[\n\\begin{align} P(1) + \\zeta Q(1) + \\zeta^2 R(1) &= \\frac{\\zeta^5-1}{\\zeta-1} S(\\zeta) = 0, \\\\ P(1) + \\xi Q(1) + \\xi^2 R(1) &= \\frac{\\xi^5-1}{\\xi-1} S(\\xi) = 0 . \\end{align}\n\\]\n\nThese equations imply that\n\n\\[\n\\zeta Q(1) + \\zeta^2 R(1) = \\xi Q(1) + \\xi^2 R(1),\n\\]\n\nor\n\n\\[\nQ(1) = - ( \\zeta + \\xi)R(1).\n\\]\n\nBut by symmetry,\n\n\\[\nQ(1) = - (\\zeta + \\rho)R(1) .\n\\]\n\nSince \$\\xi \\neq \\rho\$, it follows that \$Q(1) = R(1) = 0\$. Then, as noted above,\n\n\\[\nP(1) = P(1) + \\zeta Q(1) + \\zeta^2 R(1) = 0,\n\\]\n\nso \$(x-1)\$ is a factor of \$P(x)\$, as desired. \$\\blacksquare\$", "Let \$z, z^2, z^3\$ be three of the 5th roots of unity not equal to one that satisfy \$1 + z + z^2 + z^3 + z^4 = 0\$ as a result. Plugging them into the equation gives the linear system of equations in \$(A(1), B(1), C(1))\$:\n\n\\[\nA(1) + zB(1) + z^2C(1) = 0\n\\]\n\n\\[\nA(1) + z^2B(1) + z^4C(1) = 0\n\\]\n\n\\[\nA(1) + z^3B(1) + z^6C(1) = 0\n\\]\n\nDirect observation gives \$(A(1), B(1), C(1)) = (0,0,0)\$ as a solution, and there are no others because the system of equations is linear. Hence, \$A(1) = 0\$, and so \$(x-1)\$ is a factor of \$A(x)\$, as desired. \$\\blacksquare\$\n\nNote: We can generalize this approach to prove the claim in Solution 1 in an equally fast, concise, and readily understandable fashion." ]
USAMO-1977-1	https://artofproblemsolving.com/wiki/index.php/1977_USAMO_Problems/Problem_1	Determine all pairs of positive integers $(m,n)$ such that $(1+x^n+x^{2n}+\cdots+x^{mn})$ is divisible by $(1+x+x^2+\cdots+x^{m})$.	[ "Denote the first and larger polynomial to be \$f(x)\$ and the second one to be \$g(x)\$. In order for \$f(x)\$ to be divisible by \$g(x)\$ they must have the same roots. The roots of \$g(x)\$ are the (m+1)th roots of unity, except for 1. When plugging into \$f(x)\$, the root of unity is a root of \$f(x)\$ if and only if the terms \$x^n, x^{2n}, x^{3n}, \\cdots x^{mn}\$ all represent a different (m+1)th root of unity not equal to 1.\n\nNote that if \$\\\\gcd(m+1,n)=1\$, the numbers \$n, 2n, 3n, \\cdots, mn\$ represent a complete set of residues minus 0 modulo \$m+1\$. However, if \$gcd(m+1,n)=a\$ not equal to 1, then \$\\frac{(m+1)(n)}{a}\$ is congruent to \$0 \\pmod {m+1}\$ and thus a complete set is not formed. Therefore, \$f(x)\$ divides \$g(x)\$ if and only if \$\\boxed{\\\\gcd(m+1,n)=1}.\$ \$\\blacksquare\$", "We could instead consider \$f(x)\$ modulo \$g(x)\$. Notice that \$x^{m+1} = 1 \\pmod {g(x)}\$, and thus we can reduce the exponents of \$f(x)\$ to their equivalent modulo \$m+1\$. We want the resulting \$h(x)\$ with degree less than \$m+1\$ to be equal to \$g(x)\$ (of degree \$m\$), which implies that the exponents of \$f(x)\$ must be all different modulo \$m+1\$. This can only occur if and only if \$gcd(m+1, n) = 1\$, and this is our answer, as shown in Solution 1.", "Notice that \$1+x^n+x^{2n}+...+x^{mn}=\\frac{x^{mn+n}-1}{x^n-1}\$ and \$1+x+x^2+...+x^m=\\frac{x^{m+1}-1}{x-1}\$, so it remains to prove that \$(x^n-1)(x^{m+1}-1) \\mid (x^{mn+n}-1)(x-1)\$. It is clear that \$(x^n-1) \\mid (x^{mn+n}-1)\$ and \$(x^{m+1}-1) \\mid (x^{mn+n}-1)\$. For any \$k \\in \\mathbb{N}\$, we can use the fact that \$x^k-1=\\prod_{d \\mid k} \\Phi _d(x)\$, where \$\\Phi _d(x)\$ is the dth cyclotomic polynomial. If \$\\gcd(m+1,n) = d > 1\$, then \$x^{m+1}-1\$ and \$x^n-1\$ share a common cyclotomic polynomial; namely, \$\\Phi _d(x)\$. But since all the factors of \$x^{mn+n}-1\$ are distinct, \$(x^{mn+n}-1)(x-1)\$ cannot be divisible by \$(x^n-1)(x^{m+1}-1)\$. We find that \$\\gcd(m+1,n) = 1\$ must be the solution, since the only shared polynomial is \$\\Phi _1(x)=x-1\$, and we are done." ]
USAMO-1977-2	https://artofproblemsolving.com/wiki/index.php/1977_USAMO_Problems/Problem_2	$ABC$ and $A'B'C'$ are two triangles in the same plane such that the lines $AA',BB',CC'$ are mutually parallel. Let $[ABC]$ denote the area of triangle $ABC$ with an appropriate $\pm$ sign, etc.; prove that \[ 3([ABC]+ [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B']+ [A'BC]+[B'CA] + [C'AB]. \]	[]
USAMO-1977-3	https://artofproblemsolving.com/wiki/index.php/1977_USAMO_Problems/Problem_3	If $a$ and $b$ are two of the roots of $x^4+x^3-1=0$, prove that $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$.	[ "Given the roots \$a,b,c,d\$ of the equation \$x^{4}+x^{3}-1=0\$.\n\nFirst, Vieta's relations give \$a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abc+abd+acd+bcd=0, abcd = -1\$.\n\nThen \$cd=-\\frac{1}{ab}\$ and \$c+d=-1-(a+b)\$.\n\nThe other coefficients give \$ab+(a+b)(c+d)+cd = 0\$ or \$ab+(a+b)[-1-(a+b)]-\\frac{1}{ab}=0\$.\n\nLet \$a+b=s\$ and \$ab=p\$.\n\nThus, \$0=ab+ac+ad+bc+bd+cd=p+s(-1-s)-\\frac{1}{p}\$. (1)\n\nAlso, \$0=abc+abd+acd+bcd=p(-1-s)-s/p\$.\n\nSolving this equation for \$s\$, \$s= \\frac{-p^2}{p^2+1}\$.\n\nSubstituting into (1): \$\\frac{p^{6}+p^{4}+p^{3}-p^{2}-1}{p(p^2+1)^2}=0\$.\n\nConclusion: \$p =ab\$ is a root of \$x^{6}+x^{4}+x^{3}-x^{2}-1=0\$." ]
USAMO-1977-4	https://artofproblemsolving.com/wiki/index.php/1977_USAMO_Problems/Problem_4	Prove that if the opposite sides of a skew (non-planar) quadrilateral are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals, and conversely, if the line joining the midpoints of the two diagonals of a skew quadrilateral is perpendicular to these diagonals, then the opposite sides of the quadrilateral are congruent.	[ "We first prove that if the opposite sides are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals.\n\nLet the vertices of the quadrilateral be A, B, C, D, in that order. Thus, the opposite sides congruent condition translates to (after squaring):\n\n\\[\nAB^2 = CD^2, BC = AD^2\n\\]\n\nLet the position vectors of A, B, C, D be a, b, c, d with respect to an arbitrary origin \$O\$. Define the square of a vector \$a^2 = a \\cdot a\$. Thus, \$(a-b)^2 = (c-d)^2\$ and so\n\n\\[\na^2 - 2ab + b^2 = c^2 - 2cd + d^2.\n\\]\n\nSimilarly,\n\n\\[\nb^2 - 2bc + c^2 = a^2 - 2ad + d^2.\n\\]\n\nSubtracting the second equation from the first eventually results in\n\n\\[\na^2 - ab + cd + bc - ad - c^2 = 0,\n\\]\n\nwhich factors as\n\n\\[\n(a - b + c - d)(a - c) = 0.\n\\]\n\nLet M and N be the midpoints of \$AC\$ and \$BD\$, with position vectors \$m = \\frac{a+c}{2}\$ and \$n = \\frac{b+d}{2}\$, respectively. Then notice that the perpendicularity condition for vectors gives \$MN\$ perpendicular to \$AC\$. Similarly (by adding the equations), we find that \$MN\$ is perpendicular to \$BD\$, completing the first part of the proof.\n\nProving the converse is straightforward also: indeed, \$MN\$ perpendicular to both diagonals gives\n\n\\[\na^2 - ab + cd + bc - ad - c^2 = 0\n\\]\n\n\\[\nb^2 - ab + cd - bc + ad - d^2 = 0\n\\]\n\nAdding and subtracting the equations eventually simplifies to\n\n\\[\n(a-b)^2 = (c-d)^2\n\\]\n\n\\[\n(a-d)^2 = (b-c)^2,\n\\]\n\nor \$AB^2 = CD^2\$ and \$AD^2 = BC^2\$. This reduces to \$AB = CD\$ and \$AD = BC\$, as desired. \$\\blacksquare\$", "Let \$A, B, C, D\$ be position vectors rather than the specified points. We're given that \$\|A-B\| = \|C-D\| \\rightarrow \|A-B\|^2=\|C-D\|^2 \\rightarrow (A-B) \\cdot (A-B) = (C-D) \\cdot (C-D) \\rightarrow \\\\ A \\cdot A - 2A \\cdot B + B \\cdot B - C \\cdot C + 2C \\cdot D - D \\cdot D=0\$. We're also given that \$\|A-D\| = \|C-B\| \\rightarrow \|A-D\|^2=\|C-B\|^2 \\rightarrow (A-D) \\cdot (A-D) = (C-B) \\cdot (C-B) \\rightarrow \\\\ A \\cdot A - 2A \\cdot D + D \\cdot D - C \\cdot C + 2C \\cdot B - B \\cdot B=0\$. Adding the two equations gives \$2A \\cdot A - 2A \\cdot B - 2A \\cdot D - 2 C \\cdot C + 2C \\cdot D + 2C \\cdot B = 0\$. Dividing by 4 and rearranging gives \$\\frac{A+C-B-D}{2} \\cdot (A-C) = 0\$. On the left hand side of this equation, one of the two factors is the vector between the midpoints of the two diagonals, and the other factor is on of the two diagonals. Their dot product is \$0\$, so they are perpendicular. The same methods can be used to show that the vector between the two midpoints is perpendicular to the other diagonal. We can employ this process in reverse to get the converse (this isn't completely simple, but I'm lazy so it's left as an exercise i guess)." ]
USAMO-1977-5	https://artofproblemsolving.com/wiki/index.php/1977_USAMO_Problems/Problem_5	If $a,b,c,d,e$ are positive numbers bounded by $p$ and $q$, i.e, if they lie in $[p,q], 0 < p$, prove that \[ (a+b +c +d +e)\left(\frac{1}{a} +\frac {1}{b} +\frac{1}{c} + \frac{1}{d} +\frac{1}{e}\right) \le 25 + 6\left(\sqrt{\frac {p}{q}} - \sqrt {\frac{q}{p}}\right)^2 \] and determine when there is equality.	[ "Fix four of the variables and allow the other to vary. Suppose, for example, we fix all but \$x\$. Then the expression on the LHS has the form \$(r + x)(s + \\frac{1}{x}) = (rs + 1) + sx + \\frac{r}{x}\$, where \$r\$ and \$s\$ are fixed. But this is convex. That is to say, as \$x\$ increases if first decreases, then increases. So its maximum must occur at \$x = p\$ or \$x = q\$. This is true for each variable.\n\nSuppose all five are \$p\$ or all five are \$q\$, then the LHS is 25, so the inequality is true and strict unless \$p = q\$. If four are \$p\$ and one is \$q\$, then the LHS is \$17 + 4\\left(\\frac{p}{q} + \\frac{q}{p}\\right)\$. Similarly if four are \$q\$ and one is \$p\$. If three are \$p\$ and two are \$q\$, then the LHS is \$13 + 6\\left(\\frac{p}{q} + \\frac{q}{p}\\right)\$. Similarly if three are \$q\$ and two are \$p\$.\n\n\$\\frac{p}{q} + \\frac{q}{p} \\geq 2\$ with equality iff \$p = q\$, so if \$p < q\$, then three of one and two of the other gives a larger LHS than four of one and one of the other. Finally, we note that the RHS is in fact \$13 + 6\\left(\\frac{p}{q} + \\frac{q}{p}\\right)\$, so the inequality is true with equality iff either (1) \$p = q\$ or (2) three of \$v, w, x, y, z\$ are \$p\$ and two are \$q\$ or vice versa." ]
USAMO-1978-1	https://artofproblemsolving.com/wiki/index.php/1978_USAMO_Problems/Problem_1	Given that $a,b,c,d,e$ are real numbers such that \[ a+b+c+d+e=8, \] \[ a^2+b^2+c^2+d^2+e^2=16. \] Determine the maximum value of $e$.	[ "By Cauchy Schwarz, we can see that \$(1+1+1+1)(a^2+b^2+c^2+d^2)\\geq (a+b+c+d)^2\$ thus \$4(16-e^2)\\geq (8-e)^2\$ Finally, \$e(5e-16) \\geq 0\$ which means \$\\frac{16}{5} \\geq e \\geq 0\$ so the maximum value of \$e\$ is \$\\frac{16}{5}\$.\n\nfrom: Image from Gon Mathcenter.net", "Seeing as we have an inequality with constraints, we can use Lagrange multipliers to solve this problem. We get the following equations:\n\n\\[\n(1)\\hspace{0.5cm} a+b+c+d+e=8\\\\ (2)\\hspace{0.5cm} a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16\\\\ (3)\\hspace{0.5cm} 0=\\lambda+2a\\mu\\\\ (4)\\hspace{0.5cm} 0=\\lambda+2b\\mu\\\\ (5)\\hspace{0.5cm} 0=\\lambda+2c\\mu\\\\ (6)\\hspace{0.5cm} 0=\\lambda+2d\\mu\\\\ (7)\\hspace*{0.5cm} 1=\\lambda+2e\\mu\n\\]\n\nIf \$\\mu=0\$, then \$\\lambda=0\$ according to \$(6)\$ and \$\\lambda=1\$ according to \$(7)\$, so \$\\mu \\neq 0\$. Setting the right sides of \$(3)\$ and \$(4)\$ equal yields \$\\lambda+2a \\mu= \\lambda+2b \\mu \\implies 2a\\mu=2b \\mu \\implies a=b\$. Similar steps yield that \$a=b=c=d\$. Thus, \$(1)\$ becomes \$4d+e=8\$ and \$(2)\$ becomes \$4d^{2}+e^{2}=16\$. Solving the system yields \$e=0,\\frac{16}{5}\$, so the maximum possible value of \$e\$ is \$\\frac{16}{5}\$.", "A re-writing of Solution 1 to avoid the use of Cauchy Schwarz. We have\n\n\\[\n(a+b+c+d)^2=(8-e)^2,\n\\]\n\nand\n\n\\[\na^2+b^2+c^2+d^2=16-e^2.\n\\]\n\nThe second equation times 4, then minus the first equation,\n\n\\[\n(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2=4(16-e^2)-(8-e)^2.\n\\]\n\nThe rest follows.\n\nJ.Z.", "By the Principle of Insufficient Reasons, since \$a,b,c,d\$ are indistinguishable variables, the maximum of \$e\$ is acheived when \$a=b=c=d\$, so we have\n\n\\[\n4a+\\max e=8\n\\]\n\n\\[\n4a^2+(\\max e)^2=16\n\\]\n\n\\[\n\\implies \\max e=\\boxed{\\frac{16}{5}}\n\\]\n\n. \$\\square\$ ~Ddk001\n\n- Note: For some reason I think this solution is missing something." ]
USAMO-1978-2	https://artofproblemsolving.com/wiki/index.php/1978_USAMO_Problems/Problem_2	$ABCD$ and $A'B'C'D'$ are square maps of the same region, drawn to different scales and superimposed as shown in the figure. Prove that there is only one point $O$ on the small map that lies directly over point $O'$ of the large map such that $O$ and $O'$ each represent the same place of the country. Also, give a Euclidean construction (straight edge and compass) for $O$. \[ [asy] defaultpen(linewidth(0.7)+fontsize(10)); real theta = -100, r = 0.3; pair D2 = (0.3,0.76); string[] lbl = {'A', 'B', 'C', 'D'}; draw(unitsquare); draw(shift(D2)rotate(theta)scale(r)unitsquare); for(int i = 0; i < lbl.length; ++i) { pair Q = dir(135-90i), P = (.5,.5)+Q/2^.5; label("$"+lbl[i]+"'$", P, Q); label("$"+lbl[i]+"$",D2+rotate(theta)(rP), rotate(theta)*Q); }[/asy] \]	[ "The point is obviously unique, because the two maps have different scales (but if P and Q where two fixed points the distance between them would be the same on both maps).\n\nLet the small map square be A'B'C'D' and the large be ABCD, where X and X' are corresponding points. We deal first with the special case where A'B' is parallel to AB. In this case let AA' and BB' meet at O. Then triangles OAB and OA'B' are similar, so O must represent the same point. So assume A'B' is not parallel to AB.\n\nLet the lines A'B' and AB meet at W, the lines B'C' and BC meet at X, the lines C'D' and CD meet at Y, and the lines D'A' and DA meet at Z. We claim that the segments WY and XZ meet at a point O inside the smaller square. W cannot lie between A' and B' (or one of the vertices A', B' of the smaller square would lie outside the larger square). If it lies on the opposite side of A' to B', then Y must lie on the opposite side of C' to D'. Thus the segment WY must cut the side A'D' at some point Z' and the side B'C' at some point X'. The same conclusion holds if W lies on the opposite side of B' to A', because then Y must lie on the opposite side of D' to C'. Similarly, the segment XZ must cut the side A'B' at some point W' and the side C'D' at some point Y'. But now the segments X'Z' and W'Y' join pairs of points on opposite sides of the small square and so they must meet at some point O inside the small square.\n\nNow the triangles WOW' and YOY' are similar (WW' and YY' are parallel). Hence OW/OY = OW'/OY'. So if we set up coordinate systems with AB as the x-axis and AD as the y-axis (for the large square) and A'B' as the x'-axis and A'D' as the y'-axis (for the small square) so that corresponding points have the same coordinates, then the y coordinate of O equals the y' coordinate of O. Similarly, XOX' and ZOZ' are similar, so OX/OZ = OX'/OZ', so the x-coordinate of O equals its x'-coordinate. In other words, O represents the same point on both maps.", "The point is obviously unique because of the different scale and angle, which can also be seen as the point can and must be the unique center of spiral similarity taking ABCD to A'B'C'D', as under the spiral similarity the center does not move. (Recall that a spiral similarity is a dilation combined with a rotation, and that there is a unique spiral similarity taking any segment to another segment, provided the four points do not make a parallelogram.) To construct this center of spiral similarity, reference EGMO Lemma 10.1. Draw line DD' and AA', and let them intersect at point X. Draw circles through points X,A,D and X,A',D' (using the perpendicular bisector constructions to find the center of the circle using two chords). The intersection point of the two circles other than X is the center of spiral similarity, and hence is O, which represents the same point on both maps." ]
USAMO-1978-3	https://artofproblemsolving.com/wiki/index.php/1978_USAMO_Problems/Problem_3	An integer $n$ will be called good if we can write $n=a_1+a_2+\cdots+a_k$, where $a_1,a_2, \ldots, a_k$ are positive integers (not necessarily distinct) satisfying $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_k}=1$. Given the information that the integers 33 through 73 are good, prove that every integer $\ge 33$ is good.	[ "Lemma: If \$g\$ is a good integer, then so is \$2g + 2\$ and \$2g + 9\$.\n\nProof: Let \$g = a_1 + a_2 + \\cdots + a_k\$ such that\n\n\\[\n\\frac{1}{a_1}+\\frac{1}{a_2}+\\cdots+\\frac{1}{a_k}=1,\n\\]\n\nwhere the \$a_i\$ are positive integers. Then, notice that\n\n\\[\n1 = \\frac{1}{2} + \\frac{1}{2} = \\frac{1}{2} + \\frac{1}{2a_1}+\\frac{1}{2a_2}+\\cdots+\\frac{1}{2a_k},\n\\]\n\nso \$g_1 = 2 + 2a_1 + 2a_2 + ... + 2a_k = 2 + 2g\$ is also a good integer. Furthermore,\n\n\\[\n1 = \\frac{1}{3} + \\frac{1}{6} + \\frac{1}{2} = \\frac{1}{3} + \\frac{1}{6} + \\frac{1}{2a_1}+\\frac{1}{2a_2}+\\cdots+\\frac{1}{2a_k},\n\\]\n\nso \$g_2 = 3 + 6 + 2a_1 + 2a_2 + ... + a_k = 9 + 2g\$ is also good.\n\nThis completes the lemma. Now, we use induction to show that the integers from 33 to \$n\$ are all good, where \$n \\ge 33\$.\n\nBase case: Our given takes care of \$33 \\le n \\le 73\$. Inductive step: Assume the integers between 33 and \$k > 73\$, inclusive, are all good. Now, we casework on parity of \$k+1\$.\n\nIf \$k+1\$ is odd, then write \$k+1 = 2g + 9\$ for some integer \$g\$.\n\nIt follows that because \$k \\ge 74\$, we must have\n\n\\[\ng = \\frac{k}{2} - 4 \\ge 37 - 4 = 33.\n\\]\n\nThus, \$33 \\le g < \\frac{k}{2}\$, and so \$g\$ is good by the inductive hypothesis. Applying the lemma gives \$2g + 9 = k+1\$ as a good integer.\n\nIf \$k+1\$ is even, write \$k+1 = 2g+2\$ for some integer \$g\$. Therefore,\n\n\\[\ng = \\frac{k-1}{2} > \\frac{73-1}{2} = 36 > 33,\n\\]\n\nso that \$33 < g < \\frac{k}{2}\$, meaning that \$g\$ is a good integer by the inductive hypothesis. From the lemma, \$k+1\$ must be good as well.\n\nThis completes the inductive step and thus the induction. Therefore, all integers \$n \\ge 33\$ are good integers." ]
USAMO-1978-4	https://artofproblemsolving.com/wiki/index.php/1978_USAMO_Problems/Problem_4	(a) Prove that if the six dihedral (i.e. angles between pairs of faces) of a given tetrahedron are congruent, then the tetrahedron is regular. (b) Is a tetrahedron necessarily regular if five dihedral angles are congruent?	[ "(a) Let \$ABCD\$ be the said tetrahedron, and let the inscribed sphere of \$ABCD\$ touch the faces at \$W, X, Y, Z\$. Then, \$OW, OX, OY, OZ\$ are normals to the respective faces. We know that the angle between any two normals is equal, so we have \$\|OW\|=\|OX\|=\|OY\|=\|OZ\|\$ at equal angles. Now, since\n\n\\[\nWX=2\\cdot OW\\sin\\frac{\\angle{WOX}}{2},\n\\]\n\nand similar for the other sides, we have that \$WXYZ\$ is a regular tetrahedron. Now, the faces of \$ABCD\$ are the tangent planes at \$W\$, \$X\$, \$Y\$, and \$Z\$. Then, consider a \$120^{\\circ}\$ rotation about \$OW\$. The rotation sends \$X\\mapsto Y\$, \$Y\\mapsto Z\$, and \$Z\\mapsto X\$. Thus we have \$AB=AC\$, \$AC=AD\$, \$AD=AB\$, and \$BC=CD=DB\$. Performing rotations about the other axes yields that \$ABCD\$ has equal edges, so it is regular.\n\n(b) Consider the normals \$OW, OX, OY, OZ\$. We can perform a transformation in which we slightly shift \$X\$, \$Y\$, and \$Z\$ closer such that \$\\triangle XYZ\$ is still equilateral, and such that all angles between every pair of normals is less that \$180^{\\circ}\$. Then, shift \$W\$ such that \$WX=WY=XY\$. Then five of the distances are equal but the sixth is not.\n\n~ tc1729" ]
USAMO-1978-5	https://artofproblemsolving.com/wiki/index.php/1978_USAMO_Problems/Problem_5	Nine mathematicians meet at an international conference and discover that among any three of them, at least two speak a common language. If each of the mathematicians speak at most three languages, prove that there are at least three of the mathematicians who can speak the same language.	[ "Suppose no three mathematicians speak the same language. Then person A can share some language with at most 3 other delegates, because if he shared some language with 4 delegates there would be 3 with the same language. So there are 5 delegates who do not share a language with A. Let one of them be B. Using the same logic, one the remaining 4, let it be C, does not share a language with B. Now A, B, C do not have any common languages. Contradiction." ]
USAMO-1979-1	https://artofproblemsolving.com/wiki/index.php/1979_USAMO_Problems/Problem_1	Determine all non-negative integral solutions $(n_1,n_2,\dots , n_{14})$ if any, apart from permutations, of the Diophantine Equation $n_1^4+n_2^4+\cdots +n_{14}^4=1599$.	[ "Recall that \$n_i^4\\equiv 0,1\\bmod{16}\$ for all integers \$n_i\$. Thus the sum we have is anything from 0 to 14 modulo 16. But \$1599\\equiv 15\\bmod{16}\$, and thus there are no integral solutions to the given Diophantine equation.", "In base \$16\$, this equation would look like:\n\n\\[\nn_1^4+n_2^4+\\cdots +n_{14}^4=63F_{16}\n\\]\n\nWe notice that the unit digits of the LHS of this equation should equal to \$F_{16}\$. In base \$16\$, the only unit digits of fourth powers are \$0\$ and \$1\$. Thus, the maximum of these \$14\$ terms is 14 \$1's\$ or \$E_{16}\$. Since \$E_{16}\$ is less than \$F_{16}\$, there are no integral solutions for this equation." ]
USAMO-1979-2	https://artofproblemsolving.com/wiki/index.php/1979_USAMO_Problems/Problem_2	$N$ is the north pole. $A$ and $B$ are points on a great circle through $N$ equidistant from $N$. $C$ is a point on the equator. Show that the great circle through $C$ and $N$ bisects the angle $ACB$ in the spherical triangle $ABC$ (a spherical triangle has great circle arcs as sides).	[ "\\[\nUSAMO 1979 P2a.png\n\\]\n\nSince \$N\$ is the north pole, we define the Earth with a sphere of radius one in space with \$N=(0,0,1)\$ and sphere center \$O=(0,0,0)\$ We then pick point \$N\$ on the sphere and define the \$xz\$-plane as the plane that contains great circle points \$A\$ , \$B\$, and \$N\$ with the \$x\$-axis perpendicular to the \$z\$-axis and in the direction of \$A\$.\n\nUsing this coordinate system and \$x\$, \$y\$, and \$z\$ axes \$A=(cos(\\phi),0,sin(\\phi))\$ where \$\\phi\$ is the angle from the \$xy\$-plane to \$A\$ or latitude on this sphere with \$\\frac{-\\pi}{2} < \\phi < \\frac{\\pi}{2}\$\n\nSince \$A\$ and \$B\$ are points on a great circle through \$N\$ equidistant from \$N\$, then \$B=(-cos(\\phi),0,sin(\\phi))\$\n\nSince \$C\$ is a point on the equator, then \$C=(cos(\\theta),sin(\\theta),0)\$ where \$\\theta\$ is the angle on the \$xy\$-plane from the origin to \$C\$ or longitude on this sphere with \$-\\pi < \\phi \\le \\pi\$\n\nWe note that vectors from the origin to points \$N\$, \$A\$, \$B\$, and \$C\$ are all unit vectors because all those points are on the unit sphere.\n\nSo, we're going to define points \$N\$, \$A\$, \$B\$, and \$C\$ as unit vectors with their coordinates.\n\nWe also define the following vectors as follows:\n\nVector \$\\overrightarrow{V_{CN}}\$ is the unit vector in the direction of arc \$CN\$ and tangent to the great circle of \$CN\$ at \$C\$\n\nVector \$\\overrightarrow{V_{CA}}\$ is the unit vector in the direction of arc \$CA\$ and tangent to the great circle of \$CA\$ at \$C\$\n\nVector \$\\overrightarrow{V_{CB}}\$ is the unit vector in the direction of arc \$CB\$ and tangent to the great circle of \$CB\$ at \$C\$\n\nTo calculate each of these vectors we shall use the cross product as follows:\n\n\\[\n\\overrightarrow{V_{CN}}=(\\overrightarrow{C}\\times\\overrightarrow{N})\\times\\overrightarrow{C}\n\\]\n\n\\[\n\\overrightarrow{V_{CN}}=(\\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\\times\\left\\langle 0,0,1 \\right\\rangle)\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\n\\]\n\n\\[\n\\overrightarrow{V_{CN}}=\\left\\langle sin(\\theta),-cos(\\theta),0 \\right\\rangle\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\n\\]\n\n\\[\n\\overrightarrow{V_{CN}}=\\left\\langle 0,0,sin^{2}(\\theta)+cos^{2}(\\theta) \\right\\rangle\n\\]\n\n\\[\n\\overrightarrow{V_{CN}}=\\left\\langle 0,0,1 \\right\\rangle\n\\]\n\nVector \$\\overrightarrow{V_{CA}}\$:\n\n\\[\n\\overrightarrow{V_{CA}}=(\\overrightarrow{C}\\times\\overrightarrow{A})\\times\\overrightarrow{C}\n\\]\n\n\\[\n\\overrightarrow{V_{CA}}=(\\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\\times\\left\\langle cos(\\phi),0,sin(\\phi) \\right\\rangle)\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\n\\]\n\n\\[\n\\overrightarrow{V_{CA}}=\\left\\langle sin(\\theta)sin(\\phi),-cos(\\theta)sin(\\phi),-sin(\\theta)cos(\\phi) \\right\\rangle\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\n\\]\n\nSince we're only interested in the \$z\$ component of the vector\n\n\\[\n\\overrightarrow{V_{CA}}=\\left\\langle V_{CA_{x}},V_{CA_{y}},sin^{2}(\\theta)sin(\\phi)+cos^{2}(\\theta)sin(\\phi) \\right\\rangle\n\\]\n\n\\[\n\\overrightarrow{V_{CA}}=\\left\\langle V_{CA_{x}},V_{CA_{y}},sin(\\phi) \\right\\rangle\n\\]\n\nVector \$\\overrightarrow{V_{CB}}\$:\n\n\\[\n\\overrightarrow{V_{CB}}=(\\overrightarrow{C}\\times\\overrightarrow{b})\\times\\overrightarrow{C}\n\\]\n\n\\[\n\\overrightarrow{V_{CB}}=(\\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\\times\\left\\langle -cos(\\phi),0,sin(\\phi) \\right\\rangle)\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\n\\]\n\n\\[\n\\overrightarrow{V_{CB}}=\\left\\langle sin(\\theta)sin(\\phi),-cos(\\theta)sin(\\phi),sin(\\theta)cos(\\phi) \\right\\rangle\\times \\left\\langle cos(\\theta),sin(\\theta),0 \\right\\rangle\n\\]\n\nSince we're only interested in the \$z\$ component of the vector\n\n\\[\n\\overrightarrow{V_{CB}}=\\left\\langle V_{CB_{x}},V_{CB_{y}},sin^{2}(\\theta)sin(\\phi)+cos^{2}(\\theta)sin(\\phi) \\right\\rangle\n\\]\n\n\\[\n\\overrightarrow{V_{CB}}=\\left\\langle V_{CB_{x}},V_{CB_{y}},sin(\\phi) \\right\\rangle\n\\]\n\nSince we're working with unit vectors, then we can use dot products on the vectors with their angles as follows:\n\n\\[\ncos(\\angle ACN) = \\overrightarrow{V_{CA}}\\cdot \\overrightarrow{V_{CN}}\n\\]\n\n\\[\ncos(\\angle ACN) = \\left\\langle V_{CA_{x}},V_{CA_{y}},sin(\\phi) \\right\\rangle \\cdot \\left\\langle 0,0,1 \\right\\rangle = 0V_{CA_{x}}+0V_{CA_{y}}+1sin(\\phi)=sin(\\phi)\n\\]\n\nLikewise,\n\n\\[\ncos(\\angle BCN) = \\overrightarrow{V_{CB}}\\cdot \\overrightarrow{V_{CN}}\n\\]\n\n\\[\ncos(\\angle BCN) = \\left\\langle V_{CB_{x}},V_{CB_{y}},sin(\\phi) \\right\\rangle \\cdot \\left\\langle 0,0,1 \\right\\rangle = 0V_{CB_{x}}+0V_{CB_{y}}+1sin(\\phi)=sin(\\phi)\n\\]\n\nTherefore,\n\n\$cos(\\angle ACN) = cos(\\angle BCN)\$ and thus \$\\angle ACN = \\angle BCN\$\n\nSince those angles are equal, it proves that the great circle through \$C\$ and \$N\$ bisects the \$\\angle ACB\$ in the spherical triangle \$ABC\$\n\n~Tomas Diaz. orders@tomasdiaz.com" ]
USAMO-1979-3	https://artofproblemsolving.com/wiki/index.php/1979_USAMO_Problems/Problem_3	$a_1, a_2, \ldots, a_n$ is an arbitrary sequence of positive integers. A member of the sequence is picked at random. Its value is $a$. Another member is picked at random, independently of the first. Its value is $b$. Then a third value, $c$. Show that the probability that $a + b +c$ is divisible by $3$ is at least $\frac14$.	[ "Let x equal the probability of picking an element of the sequence equivalent to 0 mod 3, y equal those 1 mod 3, and z equal those 2 mod 3. Then, considering that 0+0+0, 1+1+1, 2+2+2, and 0+1+2 are divisible by 3, we obtain the equivalent inequality in the First Hint. This simplifies to \$x^3 + y^3 + z^3 + 6xyz \\ge x^2(y+z) + y^2(x+z) + z^2(x+y).\$ Now, we attempt to prove that \$x^3 + y^3 + z^3 + 3xyz \\ge x^2(y+z) + y^2(x+z) + z^2(x+y).\$ for all nonnegative real numbers x, y, z; our inequality will follow from this. But this is just Schur's Inequality with r = 1! This completes the proof.\n\nNote: Schur's Inequality states that \$x^r(x-y)(x-z) + y^r(y-z)(y-x) + z^r(z-x)(z-y) \\ge 0\$. This can be proven by setting WLOG \$x \\ge y \\ge z\$ and bunching together terms that are nonnegative. Challenge: when does equality hold? When does equality hold in the problem?" ]
USAMO-1979-4	https://artofproblemsolving.com/wiki/index.php/1979_USAMO_Problems/Problem_4	$P$ lies between the rays $OA$ and $OB$. Find $Q$ on $OA$ and $R$ on $OB$ collinear with $P$ so that $\frac{1}{PQ} + \frac{1}{PR}$ is as large as possible.	[ "Perform the inversion with center \$P\$ and radius \$\\overline{PO}.\$ Lines \$OA,OB\$ go to the circles \$(O_1),(O_2)\$ passing through \$P,O\$ and the line \$QR\$ cuts \$(O_1),(O_2)\$ again at the inverses \$Q',R'\$ of \$Q,R.\$ Hence\n\n\\[\n\\frac{1}{PQ}+\\frac{1}{PR}=\\frac{PQ'+PR'}{PO^2}=\\frac{Q'R'}{PO^2}\n\\]\n\nThus, it suffices to find the line through \$P\$ that maximizes the length of the segment \$\\overline{Q'R'}.\$ If \$M,N\$ are the midpoints of \$PQ',PR',\$ i.e. the projections of \$O_1,O_2\$ onto \$QR,\$ then from the right trapezoid \$O_1O_2NM,\$ we deduce that \$O_1O_2 \\ge MN = \\frac{_1}{^2}Q'R'.\$ Consequently, \$2 \\cdot O_1O_2\$ is the greatest possible length of \$Q'R',\$ which obviously occurs when \$O_1O_2NM\$ is a rectangle. Hence, \$Q,R\$ are the intersections of \$OA,OB\$ with the perpendicular to \$PO\$ at \$P.\$", "\\[\n[asy] pair O = (0,0), A = (14,28), Q = (20,40), B = (16,0), R = (25,0), P = (23,16); dot(O); dot(A); dot(Q); dot(B); dot(R); dot(P); label(\"O\", O, S); label(\"A\", A, W); label(\"Q\", Q, W); label(\"B\", B, S); label(\"R\", R, S); label(\"P\", P, E); draw(O--R--Q--O); draw(O--P); label(\"r\", O--P, N); [/asy]\n\\]\n\nLet \$r = OP, x = \\angle OPR, a = \\angle POR,\$ and \$b = \\angle POQ.\$ Then \$\\angle ORP = \\pi - x - a\$ and \$\\angle OQP = x - b.\$ Using the Law of Sines on \$\\triangle OPR\$ gives\n\n\\[\nPR = \\sin a * \\frac{r}{\\sin(\\pi - x - a)} = \\sin a * \\frac{r}{\\sin(x + a)},\n\\]\n\nand using the Law of Sines on \$\\triangle OPQ\$ gives\n\n\\[\nPQ = \\sin b * \\frac{r}{\\sin(x - b)}.\n\\]\n\nNote that \$r, a,\$ and \$b\$ are given constants. Hence,\n\n\\[\n\\frac{1}{PR} + \\frac{1}{PQ} = \\frac{1}{r} (\\frac{\\sin(x + a)}{\\sin a} + \\frac{\\sin(x - b)}{\\sin b}) = \\frac{\\sin(x+a)\\sin b + \\sin(x-b)\\sin a}{r \\sin a \\sin b} = \\frac{\\sin x\\cos a\\sin b + \\sin a\\cos x\\sin b + \\sin x\\cos b\\sin a - \\cos x\\sin b\\sin a}{r \\sin a \\sin b} = \\frac{\\sin x(\\sin b \\cos a + \\sin a \\cos b)}{r \\sin a \\sin b}\n\\]\n\nClearly, this quantity is maximized when \$\\sin x = 1.\$ Because \$x\$ must be less than \$\\pi\$, \$\\frac{1}{PQ} + \\frac{1}{PR}\$ is as large as possible when \$x = \\frac{\\pi}{2},\$ or when line \$QR\$ is perpendicular to line \$PO\$." ]
USAMO-1979-5	https://artofproblemsolving.com/wiki/index.php/1979_USAMO_Problems/Problem_5	Let $A_1,A_2,...,A_{n+1}$ be distinct subsets of $[n]$ with $\|A_1\|=\|A_2\|=\cdots =\|A_{n+1}\|=3$. Prove that $\|A_i\cap A_j\|=1$ for some pair $\{i,j\}$.	[ "Suppose the problem statement does not hold. It is clear that \$n \\ge 4\$. Choose the smallest \$n\$ such that \$\|A_i \\cap A_j\| \\neq 1\$ for each \$\\{i, j\\}\$.\n\nFirst, the \$(n+1)\$ subsets have \$3(n+1)\$ elements (some repeated) in conjunction. Because there are \$n\$ elements of \$[n]\$ total, by the Pigeonhole Principle one element of \$[n]\$, say \$k\$, is in at least four subsets. Let these subsets be \$A_1, A_2, A_3, A_4\$, without loss of generality, and let \$A_1\$ have elements \$k, m, n\$. Then without loss of generality let \$A_2\$ have elements \$k, m, p\$. If set \$A_3\$ has elements \$k, n, p\$, then simple casework shows that it is impossible to create \$A_4\$ without having \$A_4\$ intersect one of \$A_1, A_2, A_3\$ at exactly one element. Thus assume \$A_3\$ has elements \$k, m, q\$. Then \$A_4\$ has elements \$k, m, r\$. Consider each remaining set \$A_i\$. Then \$A_i\$ either contains both \$k, m\$ or none of them. Because there are \$(n-2)\$ distinct elements of \$[n]\$ apart from \$k, m\$, at most \$(n-2)\$ subsets can contain \$k, m\$. Then at least 3 subsets do not contain \$k, m\$, and it is easy to see that they are disjoint from those subsets that do contain \$k, m\$. Thus, we can partition \$[n]\$ into two subsets, one of which is the union of the \$t\$ subsets that do contain \$k, m\$, and the other is the union of the \$(n+1)-t\$ subsets that do not contain \$k, m\$. Because the latter subset has \$(n-t-2) < (n+1)-t - 1\$ elements, we may use infinite descent to contradict the minimality of \$n\$. The proof is complete." ]
USAMO-1980-1	https://artofproblemsolving.com/wiki/index.php/1980_USAMO_Problems/Problem_1	A balance has unequal arms and pans of unequal weight. It is used to weigh three objects. The first object balances against a weight $A$, when placed in the left pan and against a weight $a$, when placed in the right pan. The corresponding weights for the second object are $B$ and $b$. The third object balances against a weight $C$, when placed in the left pan. What is its true weight?	[ "The effect of the unequal arms and pans is that if an object of weight \$x\$ in the left pan balances an object of weight \$y\$ in the right pan, then \$x = hy + k\$ for some constants \$h\$ and \$k\$. Thus if the first object has true weight x, then \$x = hA + k, a = hx + k\$.\n\nSo \$a = h^2A + (h+1)k\$.\n\nSimilarly, \$b = h^2B + (h+1)k\$. Subtracting gives \$h^2 = \\frac{a-b}{A-B}\$ and so\n\n\\[\n(h+1)k = a - h^2A = \\frac{bA - aB}{A - B}\n\\]\n\n.\n\nThe true weight of the third object is thus:\n\n\\[\nhC + k = \\\\ \\boxed{\\sqrt{ \\frac{a-b}{A-B}} C + \\frac{bA - aB}{(A - B)(\\sqrt{ \\frac {a-b}{A-B}}+ 1)}}\n\\]\n\n.\n\nMore readably:\n\n\\[\n\\boxed{ h=\\sqrt{\\frac{a-b}{A-B}} ; \\\\ \\text{weight} = hC + \\frac{bA - aB}{(A - B)(h + 1)}}\n\\]\n\nCredit: John Scholes https://prase.cz/kalva/usa/usoln/usol801.html" ]
USAMO-1980-2	https://artofproblemsolving.com/wiki/index.php/1980_USAMO_Problems/Problem_2	Find the maximum possible number of three term arithmetic progressions in a monotone sequence of $n$ distinct reals.	[ "Consider the first few cases for \$n\$ with the entire \$n\$ numbers forming an arithmetic sequence\n\n\\[\n(1, 2, 3, \\ldots, n)\n\\]\n\nIf \$n = 3\$, there will be one ascending triplet (123). Let's only consider the ascending order for now. If \$n = 4\$, the first 3 numbers give 1 triplet, the addition of the 4 gives one more, for 2 in total. If \$n = 5\$, the first 4 numbers give 2 triplets, and the 5th number gives 2 more triplets (135 and 345). Repeating a few more times, we can quickly see that if \$n\$ is even, the nth number will give\n\n\\[\n\\frac{n}{2} - 1\n\\]\n\nmore triplets in addition to all the prior triplets from the first \$n-1\$ numbers. If \$n\$ is odd, the \$n\$th number will give\n\n\\[\n\\frac{n-1}{2}\n\\]\n\nmore triplets. Let \$f(n)\$ denote the total number of triplets for \$n\$ numbers. The above two statements are summarized as follows: If \$n\$ is even,\n\n\\[\nf(n) = f(n-1) + \\frac{n}2 - 1\n\\]\n\nIf \$n\$ is odd,\n\n\\[\nf(n) = f(n-1) + \\frac{n-1}2\n\\]\n\nLet's obtain the closed form for when \$n\$ is even:\n\n\\[\n\\begin{align} f(n) &= f(n-2) + n-2\\\\ f(n) &= f(n-4) + (n-2) + (n-4)\\\\ f(n) &= \\sum_{i=1}^{n/2} n - 2i\\\\ \\Aboxed{f(n\\ \\text{even}) &= \\frac{n^2 - 2n}4} \\end{align}\n\\]\n\nNow obtain the closed form when \$n\$ is odd by using the previous result for when \$n\$ is even:\n\n\\[\n\\begin{align} f(n) &= f(n-1) + \\frac{n-1}2\\\\ f(n) &= \\frac{{(n-1)}^2 - 2(n-1)}4 + \\frac{n-1}2\\\\ \\Aboxed{f(n\\ \\text{odd}) &= \\frac{{(n-1)}^2}4} \\end{align}\n\\]\n\nNote the ambiguous wording in the question! If the \"arithmetic progression\" is allowed to be a disordered subsequence, then every progression counts twice, both as an ascending progression and as a descending progression. Double the expression to account for the descending versions of each triple, to obtain:\n\n\\[\n\\begin{align} f(n\\ \\text{even}) &= \\frac{n^2 - 2n}2\\\\ f(n\\ \\text{odd}) &= \\frac{{(n-1)}^2}2\\\\ \\Aboxed{f(n) &= \\biggl\\lfloor\\frac{(n-1)^2}2\\biggr\\rfloor} \\end{align}\n\\]\n\n~Lopkiloinm (corrected by integralarefun)" ]
USAMO-1980-3	https://artofproblemsolving.com/wiki/index.php/1980_USAMO_Problems/Problem_3	$A + B + C$ is an integral multiple of $\pi$. $x, y,$ and $z$ are real numbers. If $x\sin(A)+y\sin(B)+z\sin(C)=x^2\sin(2A)+y^2\sin(2B)+z^2\sin(2C)=0$, show that $x^n\sin(nA)+y^n \sin(nB) +z^n \sin(nC)=0$ for any positive integer $n$.	[ "Let \$a=xe^{iA}\$, \$b=ye^{iB}\$, \$c=ze^{iC}\$ be numbers in the complex plane.\n\nNote that \$A+B+C=k\\pi\$ implies \$abc=xyz(e^{ik\\pi})=\\pm xyz\$ which is real. Also note that \$x\\sin(A), y\\sin(B), z\\sin(C)\$ are the imaginary parts of \$a, b, c\$ and that \$x^2\\sin(2A), y^2\\sin(2B), z^2\\sin(2C)\$ are the imaginary parts of \$a^2, b^2, c^2\$ by de Moivre's Theorem. Therefore, \$a+b+c\$ and \$a^2+b^2+c^2\$ are real because their imaginary parts sum to zero.\n\nFinally, note that \$\\frac{1}{2}\\left((a+b+c)^2-(a^2+b^2+c^2)\\right)=ab+bc+ac\$ is real as well.\n\nIt suffices to show that \$P_n=a^n+b^n+c^n\$ is real for all positive integer \$n\$, which can be shown by induction.\n\nNewton Sums gives the following relationship between sums of the form \$P_k=a^k+b^k+c^k\$\n\n\\[\nP_k-S_1P_{k-1}+S_2P_{k-2}-S_3P_{k-3}=0\n\\]\n\nWhere \$S_1=a+b+c\$, \$S_2=ab+bc+ac\$, and \$S_3=abc\$. It is given that \$P_0, P_1, P_2\$ are real. Note that if \$P_{k-1}, P_{k-2}, P_{k-3}\$ are real, then clearly \$P_k\$ is real because all other parts of the above equation are real, completing the induction." ]
USAMO-1980-4	https://artofproblemsolving.com/wiki/index.php/1980_USAMO_Problems/Problem_4	The insphere of a tetrahedron touches each face at its centroid. Show that the tetrahedron is regular.	[ "Let \$ABCD\$ be the tetrahedron, and let \$P\$ and \$Q\$ be the points at which the insphere touches faces \$ABC\$ and \$ABD\$ respectively (and therefore the centroids of those faces). Looking at the plane containing \$P\$, \$A\$, and \$Q\$, we see that the intersection of the sphere and the plane is a circle, and that \$AP\$ and \$AQ\$ are both tangent to said circle. \$[AP]\$ and \$[AQ]\$ are tangents to the circle from the same point and thus have the same length. The same goes for \$[BP]\$ and \$[BQ]\$. Thus, triangles \$ABP\$ and \$ABQ\$ are congruent, and \$\\mathbf{\\angle PAB = \\angle QAB}\$.\n\nLet the intersection of \$AP\$ and \$BC\$ be \$M\$, and let the intersection of \$AQ\$ and \$BD\$ be \$N\$. Then \$AM\$ and \$AN\$ are medians of \$ABC\$ and \$ABD\$, and thus \$\|AM\| = \\frac{3}{2}\|AP\| = \\frac{3}{2}\|AQ\| = \|AN\|\$. We already know from the previous congruence that \$\\angle{MAB} = \\angle{PAB} = \\angle{QAB} = \\angle{NAB}\$, and \$\|AB\|\$ is equal to itself. Thus, \$MAB\$ and \$NAB\$ are also congruent to each other. Finally, \$\|BC\| = 2\|BM\| = 2\|BN\| = \|BD\|\$ (Because \$M\$ and \$N\$ are midpoints of \$BC\$ and \$BD\$ respectively), and from the congruence of \$MAB\$ and \$NAB\$ we have \$\\angle{CBA} = \\angle{MBA} = \\angle{NBA} = \\angle{DBA}\$, and again \$\|AB\|\$ is equal to itself. Thus \$ABC\$ and \$ABD\$ are congruent, thus \$\\mathbf{\|AC\| = \|AD\|}\$ and \$\\mathbf{\|BC\| = \|BD\|}\$.\n\nBy applying the same logic to faces \$BCD\$ and \$ACD\$ we get \$\|AC\| = \|BC\|\$ and \$\|AD\| = \|BD\|\$. Finally, applying the same logic to faces \$ACB\$ and \$ACD\$ we get \$\|AB\| = \|AD\|\$ and \$\|CB\| = \|CD\|\$. Putting all these equalities together, we get that all edges of the tetrahedron are equal, and thus the tetrahedron is regular. \$\\blacksquare\$" ]
USAMO-1980-5	https://artofproblemsolving.com/wiki/index.php/1980_USAMO_Problems/Problem_5	If $x, y, z$ are reals such that $0\le x, y, z \le 1$, show that $\frac{x}{y + z + 1} + \frac{y}{z + x + 1} + \frac{z}{x + y + 1} \le 1 - (1 - x)(1 - y)(1 - z)$	[ "Rewrite the given inequality so that \$1\$ is isolated on the right side. Set the left side to be \$f(x, y, z)\$. Now a routine computation shows\n\n\\[\n\\frac{\\partial^2 f}{\\partial x^2} = \\frac{2y}{(x + z + 1)^3} + \\frac{2z}{(x + y + 1)^3}\\geq 0\n\\]\n\nwhich shows that \$f\$ is convex (concave up) in all three variables. Thus the maxima can only occur at the endpoints, i.e. if and only if \$x, y, z \\in \\{0,1\\}\$. Checking all eight cases shows that the value of the expression cannot exceed 1." ]
USAMO-1981-1	https://artofproblemsolving.com/wiki/index.php/1981_USAMO_Problems/Problem_1	Prove that if $n$ is not a multiple of $3$, then the angle $\frac{\pi}{n}$ can be trisected with ruler and compasses.	[ "Let \$n=3k+1\$. Multiply throughout by \$\\pi/3n\$. We get\n\n\\[\n\\frac{\\pi}{3} = \\frac{\\pi \\times k}{n} + \\frac{\\pi}{3n}\n\\]\n\nRe-arranging, we get\n\n\\[\n\\frac{\\pi}{3} - \\frac{\\pi \\times k}{n} = \\frac{\\pi}{3n}\n\\]\n\nA way to interpret it is that if we know the value \$k\$, then the remainder angle of subtracting \$k\$ times the given angle from \$\\frac{\\pi}{3}\$ gives us \$\\frac{\\pi}{3n}\$, the desired trisected angle.\n\nThis can be extended to the case when \$n=3k+2\$ where now, the equation becomes \$\\frac{\\pi}{3} - \\frac{\\pi \\times k}{n} = \\frac{2\\pi}{3n}\$\n\nHence in this case, we will have to subtract \$k\$ times the original angle from \$\\frac{\\pi}{3}\$ to get twice the the trisected angle. We can bisect it after that to get the trisected angle.\n\n## Generalization\n\nIf regular polygons of \$m\$ sides and \$n\$ sides can be constructed, where \$m\$ and \$n\$ are relatively prime integers greater than or equal to three, then regular polygons of \$mn\$ sides can be constructed. Indeed, such a polygon can be constructed by first constructing an \$m\$-gon, and then creating \$m\$ distinct \$n\$-gons with at least one vertex being a vertex of the \$m\$-gon." ]
USAMO-1981-2	https://artofproblemsolving.com/wiki/index.php/1981_USAMO_Problems/Problem_2	What is the largest number of towns that can meet the following criteria. Each pair is directly linked by just one of air, bus or train. At least one pair is linked by air, at least one pair by bus and at least one pair by train. No town has an air link, a bus link and a train link. No three towns, $A, B, C$ are such that the links between $AB, AC$ and $BC$ are all air, all bus or all train.	[ "Assume \$AB\$, \$AC\$, and \$AD\$ are all rail.\n\nNone of \$BC\$, \$CD\$, or \$CD\$ can be rail, as those cities would form a rail triangle with \$A\$.\n\nIf \$BC\$ is bus, then \$BD\$ is bus as well, as otherwise \$B\$ has all three types.\n\nHowever, \$CD\$ cannot be rail (as \$\\triangle ACD\$ would be a rail triangle), bus (as \$BCD\$ would be a bus triangle), or ferry (as \$C\$ and \$D\$ would have all three types).\n\nTherefore, no city can have three connections of the same type.\n\nAssume there are 5 towns - \$A\$, \$B\$, \$C\$, \$D\$, and \$E\$.\n\nTwo connections from \$A\$ must be of one type, and two of another; otherwise there would be at least three connections of the same type from \$A\$, which has been shown to be impossible.\n\nLet \$AB\$ and \$AC\$ be rail connections, and \$AD\$ and \$AE\$ be bus.\n\nAssume \$CD\$ is air.\n\n\$BC\$ cannot be rail (\$\\triangle ABC\$ would be a rail triangle) or bus (\$C\$ would have all three types), so \$BC\$ must be air.\n\n\$DE\$ cannot be bus (\$\\triangle ADE\$ would be a bus triangle) or rail (\$D\$ would have all three types), so \$DE\$ must be air.\n\n\$BE\$ cannot be rail (\$E\$ would have all three types) or bus (\$B\$ would have all three types), so \$BE\$ must be air.\n\nHowever, \$BD\$ cannot be rail (\$D\$ would have all three types), bus (\$B\$ would have all three types), or air (\$D\$ would have three air connections).\n\nTherefore, the assumption that \$CD\$ is air is false.\n\n\$CD\$ can equally be rail or bus; assume it is bus.\n\n\$BC\$ cannot be rail (\$\\triangle ABC\$ would be a rail triangle) or air (\$C\$ would have all three types), so \$BC\$ must be bus.\n\n\$BD\$ cannot be air (\$B\$ would have all three types) or bus (\$D\$ would have three bus connections), so \$BD\$ must be rail.\n\n\$DE\$ cannot be air (\$D\$ would have all three types) or bus (\$D\$ would have three bus connections), so \$DE\$ must be rail.\n\n\$CE\$ cannot be air (\$C\$ would have all three types) or bus (\$E\$ would have three bus connections), so \$CE\$ must be rail.\n\nThe only connection remaining is \$BE\$, which cannot be orange as both \$B\$ and \$D\$ would have all three types, but this means there are no air connections.\n\nTherefore, it is impossible with five (or more) towns.\n\nA four-town mapping is possible:\n\n\$AB\$, \$BC\$, \$CD\$, and \$DA\$ are connected by bus.\n\n\$AC\$ is connected by rail.\n\n\$BD\$ is connected by air.\n\nTherefore, the maximum number of towns is \$4\$." ]
USAMO-1981-3	https://artofproblemsolving.com/wiki/index.php/1981_USAMO_Problems/Problem_3	Show that for any triangle, $\frac{3\sqrt{3}}{2}\ge \sin(3A) + \sin(3B) + \sin (3C) \ge -2$. When does the equality hold?	[ "Given three angles that add to \$180^\\circ\$, one can construct a triangle from them. However, its angles must all be nonnegative; thus the constraints on the angles of a triangle are \$0^\\circ\\le A,B,C\\le180^\\circ\$ and \$A+B+C=180^\\circ\$.\n\nIn fact, at this point, we only care about \$3A, 3B,\$ and \$3C\$. Let us call them \$x, y,\$ and \$z\$. We have:\n\n\\[\n0 \\le x,y,z \\le 540^\\circ, x+y+z = 540^\\circ\n\\]\n\nand we must prove that\n\n\\[\n\\frac{3\\sqrt{3}}{2}\\ge \\sin(x) + \\sin(y) + \\sin (z) \\ge -2\n\\]\n\nWithout loss of generality, assume \$x \\le y \\le z\$. It follows that \$x \\le 180^\\circ\$ and \$z \\ge 180^\\circ\$ (otherwise, \$x+y+z\$ would be strictly greater than \$180^\\circ\$ or strictly less than \$180^\\circ\$, respectively).\n\nSince \$\\sin(u)\$ is nonnegative over the interval \$[0^\\circ, 180^\\circ]\$ and \$-1 \\le \\sin (u) \\le 1\$ for all real \$u\$, we can immediately use the \$x \\le 180^\\circ\$ result to show that\n\n\\[\n\\sin( x) + \\sin (y) + \\sin (z) \\le 0 + -1 + -1 = -2\n\\]\n\nThis proves the lower bound; equality occurs when \$\\sin(x) = 0\$ and \$\\sin (y) = \\sin (z) = -1\$, and this is reachable only when \$y = z = 270^\\circ\$ and \$x = 0^\\circ\$, which translates into \$A = 0^\\circ\$ and \$B = C = 90^\\circ\$.\n\nNow for the upper bound. It is true that \$\\sin(u)\$ is non-positive over the interval \$[180^\\circ, 360^\\circ]\$. Also, \$\\frac{3\\sqrt{3}}{2}>2\$ (by a simple squaring argument). Then \$z \\ge 180^\\circ\$. If \$z \\le 360^\\circ\$, then \$\\sin (z) \\le 0\$, and then\n\n\\[\n\\sin (x) + \\sin (y) + \\sin (z) \\le 1 + 1 + 0 = 2 < \\frac{3\\sqrt{3}}{2}\n\\]\n\nTherefore, we need to handle the case where \$z \\ge 360^\\circ\$. \$\\sin(u + 360^\\circ) = \\sin(u)\$, so we can subtract \$360^\\circ\$ from \$z\$ due to periodicity. Then the constraints become \$0^\\circ \\le x,y,z \\le 180^\\circ\$ and \$x+y+z = 180^\\circ\$.\n\nLet us suppose that \$(x,y,z)\$ is the selection of \$x,y,z\$ given the above constraints with the largest value of \$\\sin (x) + \\sin (y) + \\sin (z)\$. We shall see that, given these constraints, this value increases when the distance between two of these variables decreases, and it is maximized when \$x=y=z\$.\n\n\$\\textbf{Lemma:}\$ If \$f(x)\$ is well-behaved and \$f'(x)\$ is strictly decreasing over an open interval \$(a, b)\$, then, for any \$x,y\$ selected from that interval with \$x \\ne y\$, then \$2f\\left(\\frac{x + y}{2}\\right) > f(x) + f(y).\$\n\n\$\\textbf{Proof of lemma:}\$ It is true that, for any well-behaved function \$f(x)\$, \$f(x+d) = f(x) + \\int_{x}^{x+d} f'(u)\\,du\$, and likewise, \$f(x-d) = f(x) - \\int_{x-d}^x f'(u)\\,du\$. Without loss of generality, suppose \$x < y\$; let \$d = \\frac{1}{2}(y-x)\$. Since \$f'(u)\$ is given to be strictly decreasing over the relevant interval, we have \$f'(u) > f'(x+d)\$ for all \$u\\in(x, x+d)\$; therefore\n\n\\[\nf(x+d) = f(x) + \\int_x^{x+d} f'(u)\\,du > f(x) + \\int_x^{x+d} f'(x+d)\\,du = f(x) + d\\cdot f'(x+d)\n\\]\n\nLikewise, \$f'(u) < f'(y-d)\$ for all \$u\\in(y-d, y)\$; therefore\n\n\\[\nf(y-d) = f(y) - \\int_{y-d}^y f'(u)\\,du > f(y) - \\int_{y-d}^y f'(y-d)\\,du = f(y) - d\\cdot f'(y-d)\n\\]\n\nTherefore, adding our inequalities together,\n\n\\[\nf(x+d) + f(y-d) > f(x) + d\\cdot f'(x+d) + f(y) - d\\cdot f'(y-d)\n\\]\n\nBut \$x+d = y-d = \\frac{x+y}{2}\$. In that case, the \$d\\cdot f'(x+d)\$ and \$-d\\cdot f'(y-d)\$ terms cancel, and we get our desired result: \$2 f\\left(\\frac{x+y}{2}\\right) > f(x) + f(y)\$.\n\nSince \$\\sin(u)\$ is well-behaved, and \$\\sin'(u) = \\cos(u)\$, which is strictly decreasing over the open interval \$(0^\\circ, 180^\\circ)\$, then we can apply our lemma. Suppose for the sake of contradiction that \$x,y,z\$ are not all the same value, and this produces a maximum. Then, without loss of generality, suppose \$x \\ne y\$. Let \$a = \\frac{x+y}{2}\$. Then the selection \$(a,a,z)\$ also satisfies the constraints \$0^\\circ \\le a,a,z \\le 180^\\circ\$ and \$a+a+z \\le 180^\\circ\$. Furthermore, by the above lemma, \$2\\cdot\\sin (a) > \\sin (x) + \\sin (y)\$, so this new selection \$(a,a,z)\$ has a larger corresponding value \$\\sin (a) + \\sin (a) + \\sin (z)\$ than the selection \$(x,y,z)\$. This contradicts our original assumption that \$(x,y,z)\$ was chosen to have the maximum. Therefore, \$x=y=z\$.\n\nThen, since \$x+y+z = 180^\\circ\$, we conclude \$x=y=z=60^\\circ\$. Then\n\n\\[\n\\sin (x) + \\sin (y) + \\sin (z) = 3 \\cdot \\sin (60^\\circ) = 3 \\cdot \\frac{\\sqrt{3}}{2}\n\\]\n\nSo \$\\frac{\\sqrt{3}}{2}\$ is indeed the maximum value of \$\\sin(x) + \\sin (y) + \\sin (z)\$, proving the bound. Translating back into the previous problem, the maximum occurs when \$x=y=60^\\circ\$ and \$z=420^\\circ\$; translating into the original problem, we have proven the upper bound, and equality with the upper bound occurs when \$A=B=20^\\circ\$ and \$C=140^\\circ\$.\n\n~ \$\\LaTeX\$ by eevee9046" ]
USAMO-1981-4	https://artofproblemsolving.com/wiki/index.php/1981_USAMO_Problems/Problem_4	A convex polygon has $n$ sides. Each vertex is joined to a point $P$ not in the same plane. If $A, B, C$ are adjacent vertices of the polygon take the angle between the planes $PAB$ and $PBC$. The sum of the $n$ such angles equals the sum of the $n$ angles in the polygon. Show that $n=3$.	[]
USAMO-1981-5	https://artofproblemsolving.com/wiki/index.php/1981_USAMO_Problems/Problem_5	Show that for any positive real $x$, $[nx]\ge \sum_{1}^{n}\left(\frac{[kx]}{k}\right)$	[ "First of all we write \$[kx]=kx-\\{kx\\}\$. So, we need to prove that \$\\sum_{1}^{n}\\left(\\frac{\\{kx\\}}{k}\\right)\\geq \\{nx\\}.\$ Let's denote \$a_k=\\{kx\\}\$. It is easy to see that \$a_k+a_m \\geq a_{k+m}\$. We need to prove \$\\sum_{1}^{n}\\left(\\frac{a_k}{k}\\right)\\geq a_n.\$\n\nWe will prove it by induction by \$n\$. The base is obvious, so we need to make a step.\n\nLet's take \$m\$ such that \$\\frac{a_m}{m}\$ is minimal. If \$m=n\$ then our inequality is obvious. So, \$m<n\$. Then, by induction, \$\\sum_{1}^{n-m}\\left(\\frac{a_k}{k}\\right)\\geq a_{n-m}\$ and \$\\sum_{n-m+1}^{n}\\left(\\frac{a_k}{k}\\right)\\geq m\\cdot \\frac{a_m}{m}=a_m\$. Now we can add these two inequalities and get\n\n\\[\n\\sum_{1}^{n}\\left(\\frac{a_k}{k}\\right) \\geq a_{n-m}+a_m \\geq a_n\n\\]" ]
USAMO-1982-1	https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_1	In a party with $1982$ people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else?	[ "We induct on \$n\$ to prove that in a party with \$n\$ people, there must be at least \$(n-3)\$ people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people \$A, B, C\$, who do not know each other.)\n\nBase case: \$n = 4\$ is obvious.\n\nInductive step: Suppose in a party with \$k\$ people (with \$k \\ge 4\$), at least \$(k-3)\$ people know everyone else. Consider a party with \$(k+1)\$ people. Take \$k\$ of the people (leaving another person, \$A\$, out) and apply the inductive step to conclude that at least \$(k-3)\$ people know everyone else in the \$k\$-person group, \$G\$.\n\nNow suppose that everyone in the group \$G\$ knows each other. Then take \$3\$ of these people and \$A\$ to deduce that \$A\$ knows a person \$B \\in G\$, which means \$B\$ knows everyone else. Then apply the inductive step on the remaining \$k\$ people (excluding \$B\$) to find \$(k-3)\$ people out of them that know everyone else (including \$B\$, of course). Then these \$(k-3)\$ people and \$B\$, which enumerate \$(k-2)\$ people, know everyone else.\n\nSuppose that there exist two people \$B, C \\in G\$ who do not know each other. Because \$k-3 \\ge 1\$, there exist at least one person in \$G\$, person \$D\$, who knows everyone else in \$G\$. Now, take \$A, B, C, D\$ and observe that because \$B, C\$ do not know each other, either \$A\$ or \$D\$ knows everyone else of \$A, B, C, D\$ (by the problem condition), so in particular \$A\$ and \$D\$ know each other. Then apply the inductive step on the remaining \$k\$ people (excluding \$D\$) to find \$(k-3)\$ people out of them that know everyone else (including \$D\$, of course). Then these \$(k-3)\$ people and \$D\$, which enumerate \$(k-2)\$ people, know everyone else.\n\nThis completes the inductive step and thus the proof of this stronger result, which easily implies that at least \$1982 - 3 = \\boxed{1979}\$ people know everyone else." ]
USAMO-1982-2	https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_2	Let $S_r=x^r+y^r+z^r$ with $x,y,z$ real. It is known that if $S_1=0$, $()$ $\frac{S_{m+n}}{m+n}=\frac{S_m}{m}\frac{S_n}{n}$ for $(m,n)=(2,3),(3,2),(2,5)$, or $(5,2)$. Determine all other pairs of integers $(m,n)$ if any, so that $()$ holds for all real numbers $x,y,z$ such that $x+y+z=0$.	[ "Claim Both \$m,n\$ can not be even.\n\nProof \$x+y+z=0\$ ,\$\\implies x=-(y+z)\$.\n\nSince \$\\frac{S_{m+n}}{m+n} = \\frac{S_m S_n}{mn}\$,\n\nby equating cofficient of \$y^{m+n}\$ on LHS and RHS ,get\n\n\$\\frac{2}{m+n}=\\frac{4}{mn}\$.\n\n\$\\implies \\frac{m}{2} + \\frac {n}{2} = \\frac{m\\cdot n}{2\\cdot2}\$.\n\nSo we have, \$\\frac{m}{2} \\biggm{\|} \\frac{n}{2}\$ and \$\\frac{n}{2} \\biggm{\|} \\frac{m}{2}\$.\n\n\$\\implies m=n=4\$.\n\nSo we have \$S_8=2(S_4)^2\$.\n\nNow since it will true for all real \$x,y,z,x+y+z=0\$. So choose \$x=1,y=-1,z=0\$.\n\n\$S_8=2\$ and \$S_4=2\$ so \$S_8 \\neq 2 S_4^2\$.\n\nThis is contradiction. So, at least one of \$m,n\$ must be odd. WLOG assume \$n\$ is odd and m is even. The coefficient of \$y^{m+n-1}\$ in \$\\frac{S_{m+n}}{m+n}\$ is \$\\frac{\\binom{m+n}{1} }{m+n} =1\$\n\nThe coefficient of \$y^{m+n-1}\$ in \$\\frac{S_m\\cdot S_n}{m\\cdot n}\$ is \$\\frac{2}{m}\$.\n\nTherefore, \$\\boxed{m=2}\$.\n\nNow choose \$x=y=\\frac1,z=(-2)\$. (sic)\n\nSince \$\\frac{S_{n+2}}{2+n}=\\frac{S_2}{2}\\frac{S_n}{n}\$ holds for all real \$x,y,z\$ such that \$x+y+z=0\$.\n\nWe have \$\\frac{2^{n+2}-2}{n+2} = 3\\cdot\\frac{2^n-2}{n}\$. Therefore,\n\n\\begin{equation} \\label{eq:l2} \\frac{2^{n+1}-1}{n+2} =3\\cdot\\frac{2^{n-1}-1}{n}\\ldots \\tag{} \\end{equation}\n\nClearly \$(\\ref{eq:l2})\$ holds for \$n\\in\\{5,3\\}\$. And one can say that for \$n\\ge 6\$, \$\\text{RHS of (\\ref{eq:l2})}<\\text{LHS of (\\ref{eq:l2})}\$.\n\nSo our answer is \$(m,n)=(5,2),(2,5),(3,2),(2,3)\$.\n\n-ftheftics (edited by integralarefun)" ]
USAMO-1982-3	https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_3	If a point $A_1$ is in the interior of an equilateral triangle $ABC$ and point $A_2$ is in the interior of $\triangle{A_1BC}$, prove that $I.Q. (A_1BC) > I.Q.(A_2BC)$, where the isoperimetric quotient of a figure $F$ is defined by \[ I.Q.(F) = \frac{\text{Area (F)}}{\text{[Perimeter (F)]}^2} \]	[ "First, an arbitrary triangle \$ABC\$ has isoperimetric quotient (using the notation \$[ABC]\$ for area and \$s = \\frac{a + b + c}{2}\$):\n\n\\[\n\\frac{[ABC]}{4s^2} = \\frac{[ABC]^3}{4s^2 [ABC]^2} = \\frac{r^3 s^3}{4s^2 \\cdot s(s-a)(s-b)(s-c)} = \\frac{r^3}{4(s-a)(s-b)(s-c)}\n\\]\n\n\\[\n= \\frac{1}{4} \\cdot \\frac{r}{s-a} \\cdot \\frac{r}{s-b} \\cdot \\frac{r}{s-c} = \\frac{1}{4} \\tan A/2 \\tan B/2 \\tan C/2.\n\\]\n\nLemma. \$\\tan x \\tan (A - x)\$ is increasing on \$0 < x < \\frac{A}{2}\$, where \$0 < A < 90^\\circ\$.\n\nProof.\n\n\\[\n\\tan x \\tan (A - x) = \\tan x \\cdot \\frac{\\tan A - \\tan x}{1 + \\tan A \\tan x} = 1 - \\frac{1}{\\cos^2 x (1 + \\tan A \\tan x)}\n\\]\n\n\\[\n= 1 - \\frac{2}{1 + \\cos 2x + \\tan A \\sin 2x} = 1 - \\frac{2}{1 + \\sec A \\cos (A - 2x)}\n\\]\n\nis increasing on the desired interval, because \$\\cos (A - 2x)\$ is increasing on \$0 < x < \\frac{A}{2}.\$\n\nLet \$x_1, y_1, z_1\$ and \$x_2, y_2, z_2\$ be half of the angles of triangles \$A_1 BC\$ and \$A_2 BC\$ in that order, respectively. Then it is immediate that \$30^\\circ > y_1 > y_2\$, \$30^\\circ > z_1 > z_2\$, and \$x_1 + y_1 + z_1 = x_2 + y_2 + z_2 = 90^\\circ\$. Hence, by Lemma it follows that\n\n\\[\n\\tan x_1 \\tan y_1 \\tan z_1 = \\tan (90^\\circ - y_1 - z_1) \\tan y_1 \\tan z_1 > \\tan (90^\\circ - y_1 - z_2) \\tan y_1 \\tan z_2\n\\]\n\n\\[\n> \\tan (90^\\circ - y_2 - z_2) \\tan y_2 \\tan z_2 = \\tan x_2 \\tan y_2 \\tan z_2.\n\\]\n\nMultiplying this inequality by \$\\frac{1}{4}\$ gives that \$I.Q[A_1 BC] > I.Q[A_2 BC]\$, as desired." ]
USAMO-1982-4	https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_4	Prove that there exists a positive integer $k$ such that $k\cdot2^n+1$ is composite for every integer $n$.	[ "Indeed, \$\\boxed{k=2935363331541925531}\$ has the requisite property.\n\nTo see why, consider the primes \$3,\\ 5,\\ 17,\\ 257,\\ 65537,\\ 6700417,\\ 641\$, and observe that\n\n\\[\n\\begin{align}k&\\equiv 1\\pmod{3}\\\\ k&\\equiv 1\\pmod{5}\\\\ k&\\equiv 1\\pmod{17}\\\\ k&\\equiv 1\\pmod{257}\\\\ k&\\equiv 1\\pmod{65537}\\\\ k&\\equiv 1\\pmod{6700417}\\\\ k&\\equiv -1\\pmod{641}\\end{align}\n\\]\n\nMoreover,\n\n\\[\n\\begin{align}\\text{ord}_{3}\\left(2\\right)&=2\\\\ \\text{ord}_{5}\\left(2\\right)&=4\\\\ \\text{ord}_{17}\\left(2\\right)&=8\\\\ \\text{ord}_{257}\\left(2\\right)&=16\\\\ \\text{ord}_{65537}\\left(2\\right)&=32\\\\ \\text{ord}_{6700417}\\left(2\\right)&=64\\\\ \\text{ord}_{641}\\left(2\\right)&=64\\end{align}\n\\]\n\nWe conclude that\n\n\\[\n\\begin{align}n\\equiv 1\\pmod{2}&\\implies k\\cdot 2^{n}-1\\equiv k\\cdot\\left(-1\\right)+1\\equiv 0\\pmod{3}\\\\ n\\equiv 2\\pmod{4}&\\implies k\\cdot 2^{n}-1\\equiv k\\cdot\\left(-1\\right)+1\\equiv 0\\pmod{5}\\\\ n\\equiv 4\\pmod{8}&\\implies k\\cdot 2^{n}-1\\equiv k\\cdot\\left(-1\\right)+1\\equiv 0\\pmod{17}\\\\ n\\equiv 8\\pmod{16}&\\implies k\\cdot 2^{n}-1\\equiv k\\cdot\\left(-1\\right)+1\\equiv 0\\pmod{257}\\\\ n\\equiv 16\\pmod{32}&\\implies k\\cdot 2^{n}-1\\equiv k\\cdot\\left(-1\\right)+1\\equiv 0\\pmod{65537}\\\\ n\\equiv 32\\pmod{64}&\\implies k\\cdot 2^{n}-1\\equiv k\\cdot\\left(-1\\right)+1\\equiv 0\\pmod{6700417}\\\\ n\\equiv 0\\pmod{64}&\\implies k\\cdot 2^{n}-1\\equiv k\\cdot1+1\\equiv 0\\pmod{641}\\end{align}\n\\]\n\nAnd \$k>>3,5,17,257,65537,6700417,641\$ so the relevant values will, in fact, always be composite. \$\\blacksquare\$" ]
USAMO-1982-5	https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_5	$A,B$, and $C$ are three interior points of a sphere $S$ such that $AB$ and $AC$ are perpendicular to the diameter of $S$ through $A$, and so that two spheres can be constructed through $A$, $B$, and $C$ which are both tangent to $S$. Prove that the sum of their radii is equal to the radius of $S$.	[ "Let the two tangent spheres be \$S_1\$ and \$S_2\$, and let \$O, O_1, O_2\$ and \$R, R_1, R_2\$ be the origins and radii of \$S, S_1, S_2\$ respectively. Then \$AO\$ stands normal to the plane \$P\$ through \$\\Delta ABC\$. Because both spheres go through \$A\$, \$B\$, and \$C\$, the line \$O_1 O_2\$ also stands normal to \$P\$, meaning \$AO\$ and \$O_1 O_2\$ are both coplanar and parallel. Therefore the problem can be flattened to the plane \$P'\$ through \$A\$, \$O\$, \$O_1\$ and \$O_2\$.\n\nLet \$X, Y, Z, M, N\$ be points on \$P'\$ such that \$X = S \\cap O O_1, \\quad Y = S_1 \\cap S_2 \\neq A, \\quad Z = S \\cap O O_2, \\quad M = O_1 Y \\cap AO, \\quad N = O_2 Y \\cap AO\$\n\nLet \$J\$ be the three circles radical center, meaning \$JX\$ and \$JZ\$ are tangent segments to \$S\$ and \$J \\in AY\$.\n\nBecause \$Y \\in P \\iff \\angle NAY = \\angle YAM = 90 ^{\\circ},\$ we have that \$\\overline{YN}\$ and \$\\overline{YM}\$ are diameters.\n\nThis means that \$\\angle OZN = \\angle ZNY = \\angle JZY\$ and \$\\angle MXO = \\angle YMX= \\angle YXJ\$.\n\nAnd because \$\\angle ZNO = 180^{\\circ} - \\angle AYZ = \\angle ZYJ\$ and \$\\angle OMX = 180^{\\circ} -\\angle XYA = \\angle JYX,\$ we have that \$\\Delta JZY \\sim \\Delta OZN\$ and \$\\Delta JYX \\sim \\Delta OMX\$.\n\nWe then conclude that \$\\overline{OM} = \\overline{OX} \\enspace \\frac {\\overline{JY} }{ \\overline{JX}} = \\overline{OZ} \\enspace \\frac {\\overline{JY} }{ \\overline{JZ}} = \\overline{ON}\$\n\nLet \$a \\top b\$ denote line \$a\$ bisecting line segment \$b\$.\n\nSince \$O_1 O_2 \|\| MN\$ it follows that \$OY \\, \\top \\, \\overline{MN} \\Rightarrow OY \\, \\top \\, \\overline{O_1 O_2}\$. Similarly we have that \$O_1 O_2 \\, \\top \\, \\overline{YM} \\Rightarrow O_1 O_2 \\, \\top \\, \\overline{OY}\$.\n\nAnd so \$O_1 O O_2 Y\$ is a parallelogram because \$\\overline{OY}\$ and \$\\overline{O_1 O_2}\$ bisect each other, meaning \$R = \\overline{OZ} = \\overline{O O_2} + \\overline{O_2 Z} = \\overline{O_1 Y}+ R_2 = R_1 + R_2\$\n\n\\[\nQuod \\enspace Erat \\enspace Demonstrandum\n\\]" ]
USAMO-1983-1	https://artofproblemsolving.com/wiki/index.php/1983_USAMO_Problems/Problem_1	On a given circle, six points $A$, $B$, $C$, $D$, $E$, and $F$ are chosen at random, independently and uniformly with respect to arc length. Determine the probability that the two triangles $ABC$ and $DEF$ are disjoint, i.e., have no common points.	[ "First we give the circle an orientation (e.g., letting the circle be the unit circle in polar coordinates). Then, for any set of six points chosen on the circle, there are exactly \$6!\$ ways to label them one through six. Also, this does not affect the probability we wish to calculate. This will, however, make calculations easier.\n\nNote that, for any unordered set of six points chosen from the circle boundary, the number of ways to number them such that they satisfy this disjoint-triangle property is constant: there are six ways to choose which triangle will be numbered with the numbers one through three, and there are \$(3!)^2\$ ways to arrange the numbers one through three and four through six on these two triangles. Therefore, for any given configuration of points, there are \$6^3=216\$ ways to label them to have this disjoint-triangle property. There are, however, \$6!=720\$ ways to label the points in all, so given any six unordered points, the probability that when we inflict an ordering we produce the disjoint-triangle property is \$216/720=3/10\$.\n\nSince this probability is constant for any configuration of six unordered points we choose, we must have that \$3/10\$ is the probability that we produce the disjoint-triangle property if we choose the points as detailed in the problem statement." ]
USAMO-1983-2	https://artofproblemsolving.com/wiki/index.php/1983_USAMO_Problems/Problem_2	Prove that the zeros of \[ x^5+ax^4+bx^3+cx^2+dx+e=0 \] cannot all be real if $2a^2<5b$.	[ "We prove the contrapositive: if the polynomial in question has the five real roots \$x_1, x_2, x_3, x_4, x_5\$, then \$5b \\le 2a^2\$.\n\nBecause \$a = -(x_1 + x_2 + x_3 + x_4 + x_5)\$ and \$b = x_1x_2 + x_1x_3 + ... + x_4x_5\$ by Vieta's Formulae, we have\n\n\\[\n2b = 2x_1x_2 + 2x_1x_3 + ... + 2x_4x_5 = (x_1 + x_2 + x_3 + x_4 + x_5)^2 - (x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2)\n\\]\n\n\\[\n=a^2 - \\frac{(1+1+1+1+1)(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2)}{5}\n\\]\n\n\\[\n\\le a^2 - \\frac{(x_1 + x_2 + x_3 + x_4 + x_5)^2}{5}\n\\]\n\n(by Cauchy-Schwarz)\n\n\\[\n=\\frac{4a^2}{5},\n\\]\n\nso \$5b \\le 2a^2\$, as desired.", "Lemma:\n\nFor all real numbers \$x_1,x_2,\\cdots x_5\$,\n\n\\[\n2(x_1^2+x_2^2+\\cdots+x_5^2)\\ge\n\\]\n\n\\[\nx_1x_2+x_1x_3+\\cdots+x_4x_5\n\\]\n\nBy the trivial inequality,\n\n\\[\nx^2+y^2\\ge 2xy \\Rightarrow \\frac{x^2}{2} + \\frac{y^2}{2} \\ge xy\n\\]\n\nMaking such an inequality for all the variable pairs and summing them, we find the lemma is true.\n\nNow, let our roots be \$x_1,x_2,\\cdots,x_5\$. By Vieta's, \$a=x_1+x_2+\\cdots+x_5\$ and \$b=x_1x_2+x_1x_3+\\cdots+x_4x_5\$\n\nIf we show that for all real \$x_1,x_2,\\cdots, x_5\$ that \$2a^2\\ge 5b\$, then we have a contradiction and all of \$x_1,x_2,\\cdots, x_5\$ cannot be real. We start by rewriting \$2a^2\\ge 5b\$ as\n\n\\[\n2(x_1+x_2+\\cdots+x_5)^2\\ge 5(x_1x_2+x_1x_3+\\cdots+x_4x_5)\n\\]\n\nWe divide by \$2\$ and find\n\n\\[\n(x_1+x_2+\\cdots+x_5)^2\\ge \\frac{5}{2}(x_1x_2+x_1x_3+\\cdots+x_4x_5)\n\\]\n\nExpanding the LHS, we have\n\n\\[\nx_1^2+x_2^2+\\cdots+x_5^2+2(x_1x_2+x_1x_3+\\cdots+x_4x_5)\\ge\\frac{5}{2}(x_1x_2+x_1x_3+\\cdots+x_4x_5)\n\\]\n\nWe subtract the sum in brackets, and then multiply by \$2\$ to find\n\n\\[\n2x_1^2+2x_2^2+\\cdots+2x_5^2\\ge x_1x_2+x_1x_3+\\cdots+x_4x_5\n\\]\n\nwhich is true by our lemma." ]
USAMO-1983-3	https://artofproblemsolving.com/wiki/index.php/1983_USAMO_Problems/Problem_3	Each set of a finite family of subsets of a line is a union of two closed intervals. Moreover, any three of the sets of the family have a point in common. Prove that there is a point which is common to at least half the sets of the family.	[ "Let us first see that this works for anything less than three sets.\n\nObviously, due to the second condition given, they all share at least one point in common.\n\nNow, let us see that it works for three or more sets.\n\nFirst, take any three of the subsets in the family of sets.\n\nLet them be A, B, and C. We know that they share at least one point in common. Let us take the pair of segments that have the shortest length when you take the union of the pair. Let us denote this segment D. Evidently, if you take any other set of points, it must intersect D at least at one point. In order to minimize the number of intersections between the other varying segments, we must have at close to 1/2 of the remaining segments as possible intersecting each other. However, if we note, putting them that way forces at least 1/2 of the segments in each section." ]
USAMO-1983-4	https://artofproblemsolving.com/wiki/index.php/1983_USAMO_Problems/Problem_4	Six segments $S_1, S_2, S_3, S_4, S_5,$ and $S_6$ are given in a plane. These are congruent to the edges $AB, AC, AD, BC, BD,$ and $CD$, respectively, of a tetrahedron $ABCD$. Show how to construct a segment congruent to the altitude of the tetrahedron from vertex $A$ with straight-edge and compasses.	[ "Throughout this solution, we denote the length of a segment \$S\$ by \$\|S\|\$.\n\nIn this solution, we employ several lemmas. Two we shall take for granted: given any point \$A\$ and a line \$\\ell\$ not passing through \$A\$, we can construct a line \$\\ell'\$ through \$A\$ parallel to \$\\ell\$; and given any point \$A\$ on a line \$\\ell\$, we can construct a line \$\\ell'\$ through \$A\$ perpendicular to \$\\ell\$.\n\nLemma 1: If we have two segments \$S\$ and \$T\$ on the plane with non-zero length, we may construct a circle at either endpoint of \$S\$ whose radius is \$\|T\|\$.\n\nProof: We can construct arbitrarily many copies of \$T\$ by drawing a circle about one of its endpoints through its other endpoint, and then connecting the center of this circle with any other point on the circle. We can construct copies of \$T\$ like this until we create a circle of radius \$\|T\|\$ and center \$P_1\$ that intersects segment \$S\$. We can then take this intersection point \$P_2\$ and draw a line \$\\ell\$ through it perpendicular to \$S\$, and draw a circle with center \$P_2\$ passing through \$P_1\$, and consider its intersection \$P_3\$ with \$\\ell\$. Note that \$P_2P_3\\perp S\$ and \$\|P_2P_3\|=\|T\|\$. Take an endpoint \$P_4\$ of \$S\$: then draw a line through \$P_3\$ parallel to \$S\$, and a line through \$P_4\$ parallel to \$P_2P_3\$. Let these two lines intersect at \$P_5\$. Then \$P_2P_3P_5P_4\$ is a rectangle, so \$\|P_4P_5\|=\|T\|\$. Our desired circle is then a circle centered at \$P_4\$ through \$P_5\$.\n\nLemma 2: Given three collinear points \$A\$, \$B\$, \$C\$ in this order, if \$\|AB\|=a\$ and \$\|BC\|=b\$ with \$a>b\$, then we can construct a segment of length \$\\sqrt{a^2-b^2}\$.\n\nProof: From Lemma 1, we can construct a circle through \$C\$ with radius \$a\$, and then construct a perpendicular through \$B\$ to \$AC\$: these two objects intersect at \$D\$ and \$E\$. Both \$BD\$ and \$DE\$ have length \$\\sqrt{a^2-b^2}\$, from the Pythagorean Theorem.\n\nProof of the original statement: Note that we can construct a triangle \$B'C'D'\$ congruent to triangle \$BCD\$ by applying Lemma 1 to segments \$S_4\$, \$S_5\$, and \$S_6\$. Similarly, we can construct \$A_B\$ and \$A_C\$ outside triangle \$B'C'D'\$ such that \$A_BC'D'\\cong ACD\$ and \$A_CB'D'\\cong ABD\$.\n\nLet \$A'\$ be a point outside of the plane containing \$S_1\$ through \$S_6\$ such that \$A'B'C'D'\\cong ABCD\$. Then the altitudes of triangles \$A_BC'D'\$ and \$A'C'D'\$ to segment \$C'D'\$ are congruent, as are the altitudes of triangles \$A_CB'D'\$ and \$A'B'D'\$ to segment \$B'D'\$. However, if we project the altitudes of \$A'C'D'\$ and \$A'B'D'\$ from \$A'\$ onto the plane, their intersection is the base of the altitude of tetrahedron \$A'B'C'D'\$ from \$A'\$. In addition, these altitude projections are collinear with the altitudes of triangles \$A_BC'D'\$ and \$A_CB'D'\$. Therefore, the altitudes of \$A_BC'D'\$ and \$A_CB'D'\$ from \$A_B\$ and \$A_C\$ intersect at the base \$X'\$ of the altitude of \$A'B'C'D'\$ from \$A'\$. In summary, we can construct \$X'\$ by constructing the perpendiculars from \$A_B\$ and \$A_C\$ to \$C'D'\$ and \$B'D'\$ respectively, and taking their intersection.\n\nLet \$Y'\$ be the intersection of \$A_BX'\$ with \$C'D'\$. Then the altitude length we seek to construct is, from the Pythagorean Theorem, \$\\sqrt{\|A_BY'\|^2-\|X'Y'\|^2}\$. We can directly apply Lemma 2 to segment \$A_BY'X'\$ to obtain this segment. This shows how to construct a segment of length \$A'X'\$." ]
USAMO-1983-5	https://artofproblemsolving.com/wiki/index.php/1983_USAMO_Problems/Problem_5	Consider an open interval of length $1/n$ on the real number line, where $n$ is a positive integer. Prove that the number of irreducible fractions $p/q$, with $1\le q\le n$, contained in the given interval is at most $(n+1)/2$.	[ "Let \$I\$ be an open interval of length \$1/n\$ and \$F_n\$ the set of fractions \$p/q\\in I\$ with \$p,q\\in\\mathbb{Z}\$, \$\\gcd(p,q)=1\$ and \$1\\leq q\\leq n\$.\n\nAssume that \$\\frac{p}{q}\\in F_n\$. If \$k\\in\\mathbb{Z}\$ is such that \$1\\leq kq\\leq n\$, and \$p'\\in\\mathbb{Z}\$ is such that \$\\gcd(p',kq)=1\$, then\n\n\\[\n\\left\|\\frac{p}{q}-\\frac{p'}{kq}\\right\|\\geq\\frac{1}{kq}\\geq \\frac{1}{n}\n\\]\n\nTherefore \$\\frac{p'}{kq}\\notin I\\supset F_n\$. This means that \$\\frac{p}{q}\$ is the only fraction in \$F_n\$ with denominator \$q\$ or multiple of \$q\$.\n\nTherefore, from each of the pairs in \$P=\\left\\{(k,2k):\\ 1\\leq k\\leq \\left\\lfloor\\frac{n+1}{2}\\right\\rfloor\\right\\}\$ at most one element from each can be a denominator of a fraction in \$F_n\$.\n\nHence \$\|F_n\|\\leq \|P\|\\leq\\frac{n+1}{2}\$" ]
USAMO-1984-1	https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_1	In the polynomial $x^4 - 18x^3 + kx^2 + 200x - 1984 = 0$, the product of $2$ of its roots is $- 32$. Find $k$.	[ "Using Vieta's formulas, we have:\n\n\\[\n\\begin{align}a+b+c+d &= 18,\\\\ ab+ac+ad+bc+bd+cd &= k,\\\\ abc+abd+acd+bcd &=-200,\\\\ abcd &=-1984.\\\\ \\end{align}\n\\]\n\nFrom the last of these equations, we see that \$cd = \\frac{abcd}{ab} = \\frac{-1984}{-32} = 62\$. Thus, the second equation becomes \$-32+ac+ad+bc+bd+62=k\$, and so \$ac+ad+bc+bd=k-30\$. The key insight is now to factor the left-hand side as a product of two binomials: \$(a+b)(c+d)=k-30\$, so that we now only need to determine \$a+b\$ and \$c+d\$ rather than all four of \$a,b,c,d\$.\n\nLet \$p=a+b\$ and \$q=c+d\$. Plugging our known values for \$ab\$ and \$cd\$ into the third Vieta equation, \$-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)\$, we have \$-200 = -32(c+d) + 62(a+b) = 62p-32q\$. Moreover, the first Vieta equation, \$a+b+c+d=18\$, gives \$p+q=18\$. Thus we have two linear equations in \$p\$ and \$q\$, which we solve to obtain \$p=4\$ and \$q=14\$.\n\nTherefore, we have \$(\\underbrace{a+b}_4)(\\underbrace{c+d}_{14}) = k-30\$, yielding \$k=4\\cdot 14+30 = \\boxed{86}\$.", "We start as before: \$ab=-32\$ and \$cd=62\$. We now observe that a and b must be the roots of a quadratic, \$x^2+rx-32\$, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic \$x^2+sx+62\$.\n\nNow\n\n\\[\n\\begin{align}x^4-18x^3+kx^2+200x-1984 =& (x^2+rx-32)(x^2+sx+62)\\\\ =& x^4+(r+s)x^3+(62-32+rs)x^2\\\\ &+(62s-32r)x-1984.\\end{align}\n\\]\n\nEquating the coefficients of \$x^3\$ and \$x\$ with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of \$x^2\$ and get \$k=\\boxed{86}.\$", "Let \$r_1, r_2, r_3, r_4\$ be the roots of the polynomial \$x^4 - 18x^3 + kx^2 + 200x - 1984 = 0\$. We are given that \$r_1 r_2 = -32\$.\n\nBy Vieta's formulas, we have:\n\n\\[\nr_1+r_2+r_3+r_4 = 18\n\\]\n\n\\[\nr_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4 = k\n\\]\n\n\\[\nr_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4 = -200\n\\]\n\n\\[\nr_1r_2r_3r_4 = -1984\n\\]\n\nSince \$r_1r_2 = -32\$, we have \$(-32)r_3r_4 = -1984\$, so \$r_3r_4 = \\frac{-1984}{-32} = 62\$.\n\nAlso, \$r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4 = -200\$, so \$r_1r_2(r_3+r_4)+r_3r_4(r_1+r_2) = -200\$. Substituting \$r_1r_2 = -32\$ and \$r_3r_4 = 62\$, we have \$-32(r_3+r_4)+62(r_1+r_2) = -200\$.\n\nLet \$A = r_1+r_2\$ and \$B = r_3+r_4\$. Then \$A+B = 18\$, so \$B = 18-A\$. Substituting this into the equation, we have \$-32B+62A = -200\$, so \$-32(18-A)+62A = -200\$. \$-576+32A+62A = -200\$, so \$94A = 376\$, which means \$A = \\frac{376}{94} = 4\$. Then \$B = 18-A = 18-4 = 14\$.\n\nSo we have \$r_1+r_2 = 4\$ and \$r_1r_2 = -32\$. Then \$r_1\$ and \$r_2\$ are roots of \$x^2-4x-32 = 0\$, so \$(x-8)(x+4) = 0\$, which means \$r_1 = 8\$ and \$r_2 = -4\$ (or vice versa).\n\nAlso we have \$r_3+r_4 = 14\$ and \$r_3r_4 = 62\$. Then \$r_3\$ and \$r_4\$ are roots of \$x^2-14x+62 = 0\$. Using the quadratic formula, \$x = \\frac{14 \\pm \\sqrt{14^2-4(62)}}{2} = \\frac{14 \\pm \\sqrt{196-248}}{2} = \\frac{14 \\pm \\sqrt{-52}}{2} = 7 \\pm i\\sqrt{13}\$. Then \$r_3 = 7+i\\sqrt{13}\$ and \$r_4 = 7-i\\sqrt{13}\$.\n\nWe want to find \$k = r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4\$. \$k = r_1r_2+r_3r_4 + (r_1+r_2)(r_3+r_4) = -32+62+(4)(14) = 30+56 = 86\$.\n\nFinal Answer: The final answer is \$\\boxed{86}\$\n\n~avm2023", "Since two of the roots have product \$-32,\$ the equation can be factored in the form\n\n\\[\nx^4 - 18x^3 + kx^2 + 200x - 1984 = (x^2 + ax - 32)(x^2 + bx + c).\n\\]\n\nExpanding, we get\n\n\\[\nx^4 - 18x^3 + kx^2 + 200x - 1984 = x^4 + (a + b) x^3 + (ab + c - 32) x^2 + (ac - 32b) x - 32c = 0.\n\\]\n\nMatching coefficients, we get \\begin{align} a + b &= -18, \\\\ ab + c - 32 &= k, \\\\ ac - 32b &= 200, \\\\ -32c &= -1984. \\end{align}Then \$c = \\frac{-1984}{-32} = 62,\$ so \$62a - 32b = 200.\$ With \$a + b = -18,\$ we can solve to find \$a = -4\$ and \$b = -14.\$ Then\n\n\\[\nk = ab + c - 32 = \\boxed{86}.\n\\]" ]
USAMO-1984-2	https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_2	The geometric mean of any set of $m$ non-negative numbers is the $m$-th root of their product. $\quad (\text{i})\quad$ For which positive integers $n$ is there a finite set $S_n$ of $n$ distinct positive integers such that the geometric mean of any subset of $S_n$ is an integer? $\quad (\text{ii})\quad$ Is there an infinite set $S$ of distinct positive integers such that the geometric mean of any finite subset of $S$ is an integer?	[ "a) We claim that for any numbers \$p_1\$, \$p_2\$, ... \$p_n\$, \$p_1^{n!}, p_2^{n!}, ... p_n^{n!}\$ will satisfy the condition, which holds for any number \$n\$.\n\nSince \$\\sqrt[n] ab = \\sqrt[n] a * \\sqrt[n] b\$, we can separate each geometric mean into the product of parts, where each part is the \$k\$th root of each member of the subset and the subset has \$k\$ members.\n\nAssume our subset has \$k\$ members. Then, we know that the \$k\$th root of each of these members is an integer (namely \$p^{n!/k}\$), because \$k \\leq n\$ and thus \$k \| n!\$. Since each root is an integer, the geometric mean will also be an integer.\n\nb) If we define \$q\$ as an arbitrarily large number, and \$x\$ and \$y\$ as numbers in set \$S\$, we know that \${\\sqrt[q]{\\frac{x}{y}}}\$ is irrational for large enough \$q\$, meaning that it cannot be expressed as the fraction of two integers. However, both the geometric mean of the set of \$x\$ and \$q-1\$ other arbitrary numbers in \$S\$ and the set of \$y\$ and the same other \$q-1\$ numbers are integers, so since the other numbers cancel out, the geometric means divided, or \${\\sqrt[q]{\\frac{x}{y}}}\$, must be rational. This is a contradiction, so no such infinite \$S\$ is possible.\n\n-aops111 (first solution dont bully me)", "(i) The solution is the same as in the first solution\n\n(ii) Let \$p\$ and \$q\$ be some prime numbers, and let \$v_p(n)\$ be the largest number such that \$p^{v_p(n)}\$ divides \$n\$. \$v_p(n)\$ is also the exponent of \$p\$ in the prime factorization of \$n\$ (or \$0\$, if \$p\$ isn't a divisor of \$n\$). Then, because \$S\$ is infinite, we can choose \$q-1\$ numbers \$a_1, a_2, ... a_{q-1}\$ in \$S\$ such that \$v_p(a_i)\$ all give the same remainder when divided by \$q\$. (Otherwise, because \$v_p(n) \\mod q\$ has \$q\$ distinct values, \$S\$ could contain at most \$q(p-2)\$ numbers, which is finite.) Call that remainder \$r\$.\n\nNow take some integer \$b\$ in \$S\$, not part of the other \$q-1\$ values. Together, their product must be a \$q\$th power, meaning \$q\|v_p(a_1a_2...a_{q-1}b)\$ Thus, \$v_p(a_1a_2...a_{q-1}b) \\equiv 0 \\pmod q\$. It can be seen easily that for two positive integers \$x\$ and \$y\$ we have \$v_p(ab) = v_p(a) + v_p(b)\$, and by definition \$v_p(a_i) \\equiv r \\pmod q\$, so we can write \$v_p(b) + (q-1)r \\equiv 0 \\pmod q\$. Adding \$r\$ to both sides we have \$v_p(b) \\equiv v_p(b) + qr \\equiv r \\pmod q\$.\n\nThe choice of \$b\$ was arbitrary, meaning for all numbers \$n\$ in \$S\$, \$v_p(n)\$ modulo \$q\$ is the constant \$r\$, whether part of \$a_1, a_2... a_{q-1}\$ or not. The choice of \$q\$ was also arbitrary, thus for two integers \$m\$ and \$n\$ in \$S\$, \$v_p(m) = v_p(n)\$. Otherwise, we could pick \$q\$ larger than both values, and get a contradiction. Thus for all numbers \$n\$ in \$S\$, \$v_p(n)\$ is constant. The choice of \$p\$ was also arbitrary, meaning any two distinct numbers in \$S\$ have the same prime factorization. By the fundamental theorem of arithmetic, this means the two numbers are the same, contradicting with them being distinct. Thus, no such \$S\$ exists.\n\n-Circling" ]
USAMO-1984-3	https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_3	$P$, $A$, $B$, $C$, and $D$ are five distinct points in space such that $\angle APB = \angle BPC = \angle CPD = \angle DPA = \theta$, where $\theta$ is a given acute angle. Determine the greatest and least values of $\angle APC + \angle BPD$.	[ "Greatest value is achieved when all the points are as close as possible to all being on a plane.\n\nSince \$\\theta < \\frac{\\pi}{2}\$, then \$\\angle APC + \\angle BPD < \\pi\$\n\nSmallest value is achieved when point P is above and the remaining points are as close as possible to colinear when \$\\theta > 0\$, then \$\\angle APC + \\angle BPD > 0\$\n\nand the inequality for this problem is:\n\n\\[\n0 < \\angle APC + \\angle BPD < \\pi\n\\]\n\n~Tomas Diaz. orders@tomasdiaz.com" ]
USAMO-1984-4	https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_4	A difficult mathematical competition consisted of a Part I and a Part II with a combined total of $28$ problems. Each contestant solved $7$ problems altogether. For each pair of problems, there were exactly two contestants who solved both of them. Prove that there was a contestant who, in Part I, solved either no problems or at least four problems.	[]
USAMO-1984-5	https://artofproblemsolving.com/wiki/index.php/1984_USAMO_Problems/Problem_5	$P(x)$ is a polynomial of degree $3n$ such that \[ \begin{eqnarray} P(0) = P(3) = \cdots &=& P(3n) = 2, \\ P(1) = P(4) = \cdots &=& P(3n-2) = 1, \\ P(2) = P(5) = \cdots &=& P(3n-1) = 0, \quad\text{ and }\\ && P(3n+1) = 730.\end{eqnarray} \] Determine $n$.	[ "By Lagrange Interpolation Formula \$f(x) = 2\\sum_{p=0}^{n}\\left ( \\prod_{0\\leq r\\neq3p\\leq 3n}^{{}}\\frac{x-r}{3p-r} \\right )+ \\sum_{p=1}^{n}\\left ( \\prod_{0\\leq r\\neq3p-2\\leq 3n}^{{}} \\frac{x-r}{3p-2-r}\\right )\$\n\nand hence \$f(3n+1) = 2\\sum_{p=0}^{n}\\left ( \\prod_{0\\leq r\\neq3p\\leq 3n}^{{}}\\frac{3n+1-r}{3p-r} \\right )+ \\sum_{p=1}^{n}\\left ( \\prod_{0\\leq r\\neq3p-2\\leq 3n}^{{}} \\frac{3n+1-r}{3p-2-r}\\right )\$\n\nafter some calculations we get \$f(3n+1) =\\left ( \\binom{3n+1}{0}- \\binom{3n+1}{3}+\\binom{3n+1}{6}- ... \\right )\\left ( 2.(-1)^{3n}-1 \\right )+1\$\n\nGiven \$f(3n+1)= 730\$ so we have to find \$n\$ such that \$\\left ( \\binom{3n+1}{0}- \\binom{3n+1}{3}+\\binom{3n+1}{6}- ... \\right )\\left ( 2.(-1)^{3n}-1 \\right )= 729\$\n\nLemma: If \$p\$ is even \$\\binom{p}{0}- \\binom{p}{3}+ \\binom{p}{6}- \\cdots = \\frac{2^{p+1}sin^{p}\\left ( \\frac{\\pi}{3} \\right )(i)^{p}\\left ( cos\\left ( \\frac{p\\pi}{3} \\right ) \\right )}{3}\$\n\nand if \$p\$ is odd \$\\binom{p}{0}- \\binom{p}{3}+ \\binom{p}{6}- \\cdots = \\frac{-2^{p+1}sin^{p}\\left ( \\frac{\\pi}{3} \\right )(i)^{p+1}\\left ( sin\\left ( \\frac{p\\pi}{3} \\right ) \\right )}{3}\$\n\n\$i\$ is \$\\sqrt{-1}\$ Using above lemmas we do not get any solution when \$n\$ is odd, but when \$n\$ is even \$3n+1=13\$ satisfies the required condition, hence \$n=4\$", "The (3n+1)th differences of the polynomial are zero. Call it p(x), so we have p(3n+1) - (3n+1)C1 p(3n) + (3n+1)C2 p(3n-1) - ... + (-1)3n+1 p(0) = 0, where rCs is the binomial coefficient. Hence p(3n+1) = 2( (3n+1)C1 - (3n+1)C4 + ... ) + ( (3n+1)C3 - (3n+1)C6 + ... ). Putting n = 1, we get: p(4) = 2( 4C1 - 4C4) + 4C3 = 6 + 4 = 10. So n is not 1. Putting n = 2, we get: p(7) = 2( 7C1 - 7C4 + 7C7) + ( 7C3 - 7C6) = 2( 7 - 35 + 1) + (35 - 7) = -26. So n is not 2. Putting n = 3, we get: p(10) = 2( 10C1 - 10C4 + 10C7 - 10C10) + ( 10C3 - 10C6 + 10C9) = 2(10 - 210 + 120 - 1) + (120 - 210 + 10) = -162 -100 = -262. So n is not 3. Putting n = 4, we get: p(13) = 2( 13C1 - 13C4 + 13C7 - 13C10 + 13C13) + ( 13C3 - 13C6 + 13C9 - 13C12) = 2(13 - 715 + 1716 - 286 + 1) + (286 - 1716 + 715 - 13) = 1458 - 728 = 730. So n = 3 works." ]
USAMO-1985-1	https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_1	Determine whether or not there are any positive integral solutions of the simultaneous equations \[ \begin{align} x_1^2 +x_2^2 +\cdots +x_{1985}^2 & = y^3,\\ x_1^3 +x_2^3 +\cdots +x_{1985}^3 & = z^2 \end{align} \] with distinct integers $x_1,x_2,\cdots,x_{1985}$.	[ "Lemma: For a positive integer \$n\$, \$1^3+2^3+\\cdots +n^3 = (1+2+\\cdots +n)^2\$ (Also known as Nicomachus's theorem)\n\nProof by induction: The identity holds for \$1\$. Suppose the identity holds for a number \$n\$. It is well known that the sum of first \$n\$ positive integers is \$\\frac{n(n+1)}{2} = \\frac{n^2+n}{2}\$. Thus its square is \$\\frac{n^4+2n^3+n^2}{4}\$. Adding \$(n+1)^3=n^3+3n^2+3n+1\$ to this we get \$\\frac{n^4+6n^3+13n^2+12n+4}{4}\$, which can be rewritten as \$\\frac{(n^4+4n^3+6n^2+4n+1)+2(n^3+3n^2+3n+1)+(n^2+2n+1)}{4}\$ This simplifies to \$\\frac{(n+1)^4+2(n+1)^3+(n+1)^2}{4} = ({\\frac{(n+1)^2+(n+1)}{2}})^2 = (1+2+\\cdots +n+(n+1))^2\$. The induction is complete.\n\nLet \$j\$ be the sum \$1+2+\\cdots 1985\$, and let \$k\$ be the sum \$1^2 + 2^2 + \\cdots + 1985^2\$. Then assign \$x_i\$ the value \$ik^4\$ for each \$i = 1, 2,\\cdots 1985\$. Then:\n\n\\[\n\\begin{align} x_1^2 +x_2^2 +\\cdots +x_{1985}^2 & = 1^2k^8 +2^2k^8+\\cdots +1985^2k^8 = k^8(1^2+2^2+\\cdots +1985^2) = k^9 = {(k^3)}^3\\\\ x_1^3 +x_2^3 +\\cdots +x_{1985}^3 & = 1^3k^{12}+2^3k^{12}+\\cdots 1985^3k^{12}=k^{12}(1^3+2^3+\\cdots 1985^3) = k^{12}j^2 = ({k^6j})^2 \\end{align}\n\\]\n\nThus, a positive integral solution exists.\n\n-Circling" ]
USAMO-1985-2	https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_2	Determine each real root of \[ x^4-(2\cdot10^{10}+1)x^2-x+10^{20}+10^{10}-1=0 \] correct to four decimal places.	[ "The equation can be re-written as\n\n\\[\n\\begin{align}\\label{eqn1} (x+10^5)^2(x-10^5)^2 -(x+10^5)(x-10^5) -x-1=0. \\end{align}\n\\]\n\nWe first prove that the equation has no negative roots. Let \$x\\le 0.\$ The equation above can be further re-arranged as\n\n\\[\n\\begin{align}[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]=x-1.\\end{align}\n\\]\n\nThe right hand side of the equation is negative. Therefore\n\n\\[\n[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2)]<0,\n\\]\n\nand we have \$-1<(x+10^5)(x-10^5) <2.\$ Then the left hand side of the equation is bounded by\n\n\\[\n\|[(x+10^5)(x-10^5)+1][(x+10^5)(x-10^5)-2]\|\\le 3\\times 3.\n\\]\n\nHowever, since \$\|(x+10^5)(x-10^5)\|\\le 2\$ and \$x<0,\$ it follows that \$\|x+10^5\| <\\frac{2}{\|x-10^5\|}<2\\times 10^{-5}\$ for negative \$x.\$ Then \$x<2\\times 10^{-5}-10^5.\$ The right hand side of the equation is then a large negative number. It cannot be equal to the left hand side which is bounded by 9.\n\nNow let \$x>0.\$ When \$x=10^5,\$ the left hand side of equation (1) is negative. Therefore the equation has real roots on both side of \$10^5\$, as its leading coefficient is positive. We will prove that \$x=10^5\$ is a good approximation of the roots (within \$10^{-2}\$). In fact, we can solve the \"quadratic\" equation (1) for \$(x+10^5)(x-10^5)\$:\n\n\\[\n(x+10^5)(x-10^5)=\\frac{1\\pm\\sqrt{1+4(x+1)}}{2}.\n\\]\n\nThen\n\n\\[\nx-10^5=\\frac{1\\pm\\sqrt{1+4(x+1)}}{2(x+10^5)}.\n\\]\n\nEasy to see that \$\|x-10^5\| <1\$ for positve \$x.\$ Therefore, \$10^5-1<x<10^5+1.\$ Then\n\n\\[\n\\begin{align} \|x-10^5\|&=\\left\|\\frac{1\\pm\\sqrt{1+4(x+1)}}{2(x+10^5)}\\right \|\\\\ &\\le \\left \|\\frac{1}{2(x+10^5)}\\right \|+\\left \|\\frac{\\sqrt{1+4(x+1)}}{2(x+10^5)}\\right \|\\\\ &\\le \\frac{1}{2(10^5-1+10^5)} +\\frac{\\sqrt{1+4(10^5+1+1)}}{2(10^5-1+10^5)} \\\\ &<10^{-2}. \\end{align}\n\\]\n\nLet \$x_1\$ be a root of the equation with \$x_1<10^5.\$ Then \$0<10^5-x_1<10^{-2}\$ and\n\n\\[\nx_1-10^5=\\frac{1-\\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}.\n\\]\n\nAn aproximation of \$x_1\$ is defined as follows:\n\n\\[\n\\tilde{x}_1=10^5+\\frac{1-\\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.\n\\]\n\nWe check the error of the estimate:\n\n\\[\n\\begin{align} \|\\tilde{x}_1-x_1\|&=\\left \| \\frac{1-\\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \\frac{1-\\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} \\right \| \\\\ &\\le \\left \|\\frac{1}{2(10^5+10^5)}- \\frac{1}{2(x_1+10^5)}\\right \|+\\left \|\\frac{\\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \\frac{\\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\\right \|. \\end{align}\n\\]\n\nThe first absolute value\n\n\\[\n\\left \|\\frac{1}{2(10^5+10^5)}- \\frac{1}{2(x_1+10^5)}\\right \| =\\frac{\|x_1- 10^5\|}{2(10^5+10^5)(x_1+10^5)}<10^{-12}.\n\\]\n\nThe second absolute value\n\n\\[\n\\begin{align} &\\left \|\\frac{\\sqrt{1+4(10^5+1)}}{2(10^5+10^5)} - \\frac{\\sqrt{1+4(x_1+1)}}{2(x_1+10^5)} \\right \|\\\\ &\\le \\left \|\\frac{\\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}- \\frac{\\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}\\right \|+\\left \|\\frac{\\sqrt{1+4(x_1+1)}}{2(10^5+10^5)}- \\frac{\\sqrt{1+4(x_1+1)}}{2(x_1+10^5)}\\right \|\\\\ &\\le 10^{-7}+10^{-9}, \\end{align}\n\\]\n\nthrough a rationalized numerator.Therefore \$\|\\tilde{x}_1-x_1\|\\le 10^{-6}.\$\n\nFor a real root \$x_2\$ with \$x_2>10^5,\$ we choose\n\n\\[\n\\tilde{x}_2=10^5+\\frac{1+\\sqrt{1+4(10^5+1)}}{2(10^5+10^5)}.\n\\]\n\nWe can similarly prove it has the desired approximation.\n\n## Notes\n\nAnother round of iteration can increase the accuracy to more than 10 decimal places:\n\n\\[\n\\begin{align} \\tilde{x}_1^\\prime=10^5+\\frac{1-\\sqrt{1+4(\\tilde{x}_1+1)}}{2(\\tilde{x}_1+10^5)},\\\\ \\tilde{x}_2^\\prime=10^5+\\frac{1+\\sqrt{1+4(\\tilde{x}_2+1)}}{2(\\tilde{x}_2+10^5)}. \\end{align}\n\\]\n\nJ.Z." ]
USAMO-1985-3	https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_3	Let $A,B,C,D$ denote four points in space such that at most one of the distances $AB,AC,AD,BC,BD,CD$ is greater than $1$. Determine the maximum value of the sum of the six distances.	[ "Suppose that \$AB\$ is the length that is more than \$1\$. Let spheres with radius \$1\$ around \$A\$ and \$B\$ be \$S_A\$ and \$S_B\$. \$C\$ and \$D\$ must be in the intersection of these spheres, and they must be on the circle created by the intersection to maximize the distance. We have \$AC + BC + AD + BD = 4\$.\n\nIn fact, \$CD\$ must be a diameter of the circle. This maximizes the five lengths \$AC\$, \$BC\$, \$AD\$, \$BD\$, and \$CD\$. Thus, quadrilateral \$ACBD\$ is a rhombus.\n\nSuppose that \$\\angle CAD = 2\\theta\$. Then, \$AB + CD = 2\\sin{\\theta} + 2\\cos{\\theta}\$. To maximize this, we must maximize \$\\sin{\\theta} + \\cos{\\theta}\$ on the range \$0^{\\circ}\$ to \$90^{\\circ}\$. However, note that we really only have to solve this problem on the range \$0^{\\circ}\$ to \$45^{\\circ}\$, since \$\\theta > 45\$ is just a symmetrical function.\n\nFor \$\\theta < 45\$, \$\\sin{\\theta} \\leq \\cos{\\theta}\$. We know that the derivative of \$\\sin{\\theta}\$ is \$\\cos{\\theta}\$, and the derivative of \$\\cos{\\theta}\$ is \$-\\sin{\\theta}\$. Thus, the derivative of \$\\sin{\\theta} + \\cos{\\theta}\$ is \$\\cos{\\theta} - \\sin{\\theta}\$, which is nonnegative between \$0^{\\circ}\$ and \$45^{\\circ}\$. Thus, we can conclude that this is an increasing function on this range.\n\nIt must be true that \$2\\sin{\\theta} \\leq 1\$, so \$\\theta \\leq 30^{\\circ}\$. But, because \$\\sin{\\theta} + \\cos{\\theta}\$ is increasing, it is maximized at \$\\theta = 30^{\\circ}\$. Thus, \$AB = \\sqrt{3}\$, \$CD = 1\$, and our sum is \$5 + \\sqrt{3}\$.\n\n~mathboy100" ]
USAMO-1985-4	https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_4	There are $n$ people at a party. Prove that there are two people such that, of the remaining $n-2$ people, there are at least $\lfloor n/2\rfloor -1$ of them, each of whom knows both or else knows neither of the two. Assume that "know" is a symmetrical relation; $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.	[ "Consider the number of pairs (X, {Y, Z}), where X, Y, Z are distinct points such that X is joined to just one of Y, Z. If X is joined to just k points, then there are just k(n - 1 - k) ≤ (n - 1)2/4 such pairs (X, {Y, Z}). Hence in total there are at most \$\\frac{n(n - 1)^2}{4}\$ such pairs. But there are \$\\frac{n(n - 1)}{2}\$ possible pairs {Y, Z}. So we must be able to find one of them {A, B} which belongs to at most \$\\lfloor \\frac{n-1}{2} \\rfloor\$ such pairs.\n\nHence there are at least \$n - 2 - \\lfloor \\frac{n - 1}{2} \\rfloor = \\lfloor \\frac{n}{2} \\rfloor - 1\$ points X which are joined to both of A and B or to neither of A and B. (If confused by the floor, consider n = 2m and n = 2m+1 separately!)\n\n~John Scholes" ]
USAMO-1985-5	https://artofproblemsolving.com/wiki/index.php/1985_USAMO_Problems/Problem_5	Let $a_1,a_2,a_3,\cdots$ be a non-decreasing sequence of positive integers. For $m\ge1$, define $b_m=\min\{n: a_n \ge m\}$, that is, $b_m$ is the minimum value of $n$ such that $a_n\ge m$. If $a_{19}=85$, determine the maximum value of $a_1+a_2+\cdots+a_{19}+b_1+b_2+\cdots+b_{85}$.	[ "We create an array of dots like so: the array shall go out infinitely to the right and downwards, and at the top of the \$i\$th column we fill the first \$a_i\$ cells with one dot each. Then the \$19\$th row shall have 85 dots. Now consider the first 19 columns of this array, and consider the first 85 rows. In row \$j\$, we see that the number of blank cells is equal to \$b_j-1\$. Therefore the number of filled cells in the first 19 columns of row \$j\$ is equal to \$20-b_j\$.\n\nWe now count the number of cells in the first 19 columns of our array, but we do it in two different ways. First, we can sum the number of dots in each column: this is simply \$a_1+\\cdots+a_{19}\$. Alternatively, we can sum the number of dots in each row: this is \$(20-b_1)+\\cdots +(20-b_{85})\$. Since we have counted the same number in two different ways, these two sums must be equal. Therefore\n\n\\[\na_1+\\cdots +a_{19}+b_1+\\cdots +b_{85}=20\\cdot 85=\\boxed{1700}.\n\\]\n\nNote that this shows that the value of the desired sum is constant." ]
USAMO-1986-1	https://artofproblemsolving.com/wiki/index.php/1986_USAMO_Problems/Problem_1	$(\text{a})$ Do there exist 14 consecutive positive integers each of which is divisible by one or more primes $p$ from the interval $2\le p \le 11$? $(\text{b})$ Do there exist 21 consecutive positive integers each of which is divisible by one or more primes $p$ from the interval $2\le p \le 13$?	[ "(a) To solve part (a), we first note that for any 14 consecutive positive integers, exactly 7 are even (divisible by 2) and therefore satisfy the criteria. We can remove these from the problem, and simplify it to the following question, which is equivalent to part (a): \"Do there exist 7 consecutive positive odd integers each of which is divisible by one or more primes \$p\$ from the interval \$3\\le p \\le 11\$?\" Among any 7 consecutive positive odd integers, the following holds:\n\n\\[\n\\text{Either 2 or 3 are divisible by 3}\n\\]\n\n\\[\n\\text{Either 1 or 2 are divisible by 5}\n\\]\n\n\\[\n\\text{Exactly 1 is divisible by 7}\n\\]\n\n\\[\n\\text{Either 0 or 1 are divisible by 11}\n\\]\n\nFor every one of these seven integers to be divisible by one of 3, 5, 7, or 11, there must be 3 multiples of 3, 2 multiples of 5, 1 multiple of 7, and 1 multiple of 11. Additionally, none of these integers may be a multiple of any two of the four aforementioned primes. Otherwise, the Pigeonhole Principle dictates that at least one integer is not divisible by any of the four, thus failing to meet the criteria. But this cannot be. Calling the consecutive odd integers \$a_1, a_2, \\dots, a_7\$, we note that \$a_1\$, \$a_4\$, and \$a_7\$ must be multiples of 3, and therefore cannot be multiples of 5, 7, or 11. But for there to be two multiples of 5, they must be one of the two pairs \$(a_1,a_6)\$ and \$(a_2,a_7)\$. But each of these pairs contains a multiple of 3, and so at least one of the 7 odd integers is divisible by none of the primes 3, 5, 7, or 11. Therefore the answer to part (a) is no.\n\n(b) To solve part (b), we use a strategy similar to the one used in part (a), reducing part (b) to this equivalent question: \"Do there exist 10 consecutive positive odd integers each of which is divisible by one or more primes \$p\$ from the interval \$3\\le p \\le 13\$?\" (We will ignore the case where the first of the 21 integers is odd, resulting in 11 odd integers instead of 10, as it is only true if the weaker, 10-integer argument is also true.) Among any 10 consecutive positive odd integers, the following holds:\n\n\\[\n\\text{Either 3 or 4 are divisible by 3}\n\\]\n\n\\[\n\\text{Exactly 2 are divisible by 5}\n\\]\n\n\\[\n\\text{Either 1 or 2 are divisible by 7}\n\\]\n\n\\[\n\\text{Either 0 or 1 are divisible by 11}\n\\]\n\n\\[\n\\text{Either 0 or 1 are divisible by 13}\n\\]\n\nFor every one of these ten integers to be divisible by one of 3, 5, 7, 11, or 13, there must be 4 multiples of 3, 2 multiples of 5, 2 multiples of 7, 1 multiple of 11, and 1 multiple of 13. As before, none of these integers may be a multiple of any two of the five primes. Calling the consecutive odd integers \$a_1, a_2, \\dots, a_{10}\$, we note that \$a_1\$, \$a_4\$, \$a_7\$, and \$a_{10}\$ must be multiples of 3, and therefore cannot be multiples of 5, 7, 11, or 13. Unlike part (a), however, this stipulation is not a dead end. Let the multiples of 5 be \$a_3\$ and \$a_8\$ and let the multiples of 7 be \$a_2\$ and \$a_9\$. The multiples of 11 and 13 are \$a_5\$ and \$a_6\$, in some order (it doesn't really matter which). An example of a sequence of 21 consecutive positive integers satisfying part (b) is the integers from 9440 to 9460 (inclusive), which can be obtained by solving modular equations that result from these statements. So the answer to part (b) is yes." ]
USAMO-1986-2	https://artofproblemsolving.com/wiki/index.php/1986_USAMO_Problems/Problem_2	During a certain lecture, each of five mathematicians fell asleep exactly twice. For each pair of mathematicians, there was some moment when both were asleep simultaneously. Prove that, at some moment, three of them were sleeping simultaneously.	[ "We shall assume to the contrary that there was never a time when three mathematicians were sleeping simultaneously, and derive a contradiction.\n\nAs a subtle but logically necessary note, we will assume without loss of generality that no two events (an event is one mathematician either falling asleep or waking up) happen at the same time.\n\nWe have (5 choose 2) = 10 sleeping-pairs of mathematicians to account for. Also, since 5 mathematicians each fell asleep 2 times, we have a total of 10 occasions on which a mathematician fell asleep. Now, if two mathematicians are ever asleep simultaneously, then one of them fell asleep while the other was already asleep. However, because of the \"no three asleep at once\" rule, if a mathematician falls asleep, then at most one other mathematician could have been asleep already. Therefore, each occasion when a mathematician falls asleep can account for at most one sleeping-pair. It would appear that we have just enough to make it. However, when the first mathematician falls asleep, no other mathematicians are asleep...\n\nHey, wait a minute. Couldn't one mathematician, or even two, have been already asleep when the lecture started? In fact, the wording of the problem does not forbid this, and this permits us to construct an easy counterexample:\n\nCall the mathematicians A,B,C,D,E. A starts out asleep. B sleeps; A wakes; C sleeps; B wakes; D sleeps; C wakes; E sleeps; D wakes; A sleeps; E wakes; then, C sleeps; A wakes; E sleeps; C wakes; B sleeps; E wakes; D sleeps; B wakes; A sleeps; D wakes. This creates, in order, the sleeping-pairs AB, BC, CD, DE, EA, then AC, CE, EB, BD, DA; each mathematician falls asleep and wakes up exactly twice, and at no time are three mathematicians asleep.\n\nWell, this must be considered a hole in the problem as written. However, if we add the (nontrivial) assumption that the mathematicians were all awake when the lecture began, then it follows that the first occasion when a mathematician falls asleep can only account for zero sleeping-pairs, and the remaining 9 occasions can only account for 9 of 10 sleeping-pairs, and so there must be some pair of mathematicians that are not simultaneously asleep during the lecture. So our assumption to the contrary implies an impossible sequence of events. Therefore, that assumption must be wrong, and there must be a moment when three mathematicians are asleep." ]
USAMO-1986-3	https://artofproblemsolving.com/wiki/index.php/1986_USAMO_Problems/Problem_3	What is the smallest integer $n$, greater than one, for which the root-mean-square of the first $n$ positive integers is an integer? $\mathbf{Note.}$ The root-mean-square of $n$ numbers $a_1, a_2, \cdots, a_n$ is defined to be \[ \left[\frac{a_1^2 + a_2^2 + \cdots + a_n^2}n\right]^{1/2} \]	[ "Let's first obtain an algebraic expression for the root mean square of the first \$n\$ integers, which we denote \$I_n\$. By repeatedly using the identity \$(x+1)^3 = x^3 + 3x^2 + 3x + 1\$, we can write\n\n\\[\n1^3 + 3\\cdot 1^2 + 3 \\cdot 1 + 1 = 2^3,\n\\]\n\n\\[\n1^3 + 3 \\cdot(1^2 + 2^2) + 3 \\cdot (1 + 2) + 1 + 1 = 3^3,\n\\]\n\nand\n\n\\[\n1^3 + 3\\cdot(1^2 + 2^2 + 3^2) + 3 \\cdot (1 + 2 + 3) + 1 + 1 + 1 = 4^3.\n\\]\n\nWe can continue this pattern indefinitely, and thus for any positive integer \$n\$,\n\n\\[\n1 + 3\\sum_{j=1}^n j^2 + 3 \\sum_{j=1}^n j^1 + \\sum_{j=1}^n j^0 = (n+1)^3.\n\\]\n\nSince \$\\sum_{j=1}^n j = n(n+1)/2\$, we obtain\n\n\\[\n\\sum_{j=1}^n j^2 = \\frac{2n^3 + 3n^2 + n}{6}.\n\\]\n\nTherefore,\n\n\\[\nI_n = \\left(\\frac{1}{n} \\sum_{j=1}^n j^2\\right)^{1/2} = \\left(\\frac{2n^2 + 3n + 1}{6}\\right)^{1/2}.\n\\]\n\nRequiring that \$I_n\$ be an integer, we find that\n\n\\[\n(2n+1 ) (n+1) = 6k^2,\n\\]\n\nwhere \$k\$ is an integer. Using the Euclidean algorithm, we see that \$\\gcd(2n+1, n+1) = \\gcd(n+1,n) = 1\$, and so \$2n+1\$ and \$n+1\$ share no factors greater than 1. The equation above thus implies that \$2n+1\$ and \$n+1\$ is each proportional to a perfect square. Since \$2n+1\$ is odd, there are only two possible cases:\n\nCase 1: \$2n+1 = 3 a^2\$ and \$n+1 = 2b^2\$, where \$a\$ and \$b\$ are integers.\n\nCase 2: \$2n+1 = a^2\$ and \$n+1 = 6b^2\$.\n\nIn Case 1, \$2n+1 = 4b^2 -1 = 3a^2\$. This means that \$(4b^2 -1)/3 = a^2\$ for some integers \$a\$ and \$b\$. We proceed by checking whether \$(4b^2-1)/3\$ is a perfect square for \$b=2, 3, 4, \\dots\$. (The solution \$b=1\$ leads to \$n=1\$, and we are asked to find a value of \$n\$ greater than 1.) The smallest positive integer \$b\$ greater than 1 for which \$(4b^2-1)/3\$ is a perfect square is \$b=13\$, which results in \$n=337\$.\n\nIn Case 2, \$2n+1 = 12b^2 - 1 = a^2\$. Note that \$a^2 = 2n+1\$ is an odd square, and hence is congruent to \$1 \\pmod 4\$. But \$12b^2 -1 \\equiv 3 \\pmod 4\$ for any \$b\$, so Case 2 has no solutions.\n\nAlternatively, one can proceed by checking whether \$12b^2 -1\$ is a perfect square for \$b=1, 2 ,3 ,\\dots\$. We find that \$12b^2 -1\$ is not a perfect square for \$b = 1,2, 3, ..., 7, 8\$, and \$n= 383\$ when \$b=8\$. Thus the smallest positive integers \$a\$ and \$b\$ for which \$12b^2- 1 = a^2\$ result in a value of \$n\$ exceeding the value found in Case 1, which was 337.\n\nIn summary, the smallest value of \$n\$ greater than 1 for which \$I_n\$ is an integer is \$\\boxed{337}\$." ]
USAMO-1986-4	https://artofproblemsolving.com/wiki/index.php/1986_USAMO_Problems/Problem_4	Two distinct circles $K_1$ and $K_2$ are drawn in the plane. They intersect at points $A$ and $B$, where $AB$ is the diameter of $K_1$. A point $P$ on $K_2$ and inside $K_1$ is also given. Using only a "T-square" (i.e. an instrument which can produce a straight line joining two points and the perpendicular to a line through a point on or off the line), find a construction for two points $C$ and $D$ on $K_1$ such that $CD$ is perpendicular to $AB$ and $\angle CPD$ is a right angle.	[]
USAMO-1986-5	https://artofproblemsolving.com/wiki/index.php/1986_USAMO_Problems/Problem_5	By a partition $\pi$ of an integer $n\ge 1,$ we mean here a representation of $n$ as a sum of one or more positive integers where the summands must be put in nondecreasing order. (E.g., if $n=4,$ then the partitions $\pi$ are $1+1+1+1,$ $1+1+2,$ $1+3, 2+2,$ and $4$). For any partition $\pi,$ define $A(\pi)$ to be the number of $1$'s which appear in $\pi,$ and define $B(\pi)$ to be the number of distinct integers which appear in $\pi$ (E.g., if $n=13$ and $\pi$ is the partition $1+1+2+2+2+5,$ then $A(\pi)=2$ and $B(\pi) = 3$). Prove that, for any fixed $n,$ the sum of $A(\pi)$ over all partitions of $\pi$ of $n$ is equal to the sum of $B(\pi)$ over all partitions of $\pi$ of $n.$	[ "Let \$S(n) = \\sum\\limits_{\\pi} A(\\pi)\$ and let \$T(n) = \\sum\\limits_{\\pi} B(\\pi).\$ We will use generating functions to approach this problem -- specifically, we will show that the generating functions of \$S(n)\$ and \$T(n)\$ are equal.\n\nLet us start by finding the generating function of \$S(n).\$ This function counts the total number of 1's in all the partitions of \$n.\$ Another way to count this is by counting the number of partitions of \$n\$ that contain \$x\$ 1's and multiplying this by \$x,\$ then summing for \$1\\leq x \\leq n.\$ However, the number of partitions of \$n\$ that contain \$x\$ 1's is the same as the number of partitions of \$n-x\$ that contain no 1's, so\n\n\\[\nS(n) = \\sum_{k=1}^n k\\cdot (\\text{\\# of partitions of }n-k\\text{ with no 1's}).\n\\]\n\nThe number of partitions of \$m\$ with no 1's is the coefficient of \$x^m\$ in\n\n\\[\n\\begin{align} F(x) &= (1+x^2+x^4+\\ldots)(1+x^3+x^6+\\ldots)(1+x^4+x^8+\\ldots)\\ldots \\\\ &= \\prod\\limits_{i=2}^{\\infty}\\frac{1}{1-x^i} \\end{align}\n\\]\n\nNote that there is no \$(1+x+x^2+x^3+\\ldots)\$ term in \$F(x)\$ because we cannot have any 1's in the partition.\n\nLet \$c_m\$ be the coefficient of \$x^m\$ in the expansion of \$F(x),\$ so we can rewrite it as \$F(x)=c_0+c_1x+c_2x^2+c_3x^3+\\ldots.\$ We wish to compute \$S(n)=1\\cdot c_{n-1}+2\\cdot c_{n-2}+\\ldots+n\\cdot c_0.\$\n\nConsider the power series \$G(x)=(x+2x^2+3x^3+4x^4+\\ldots)F(x).\$\n\n\\[\n\\begin{align} G(x) &= (x+2x^2+3x^3+4x^4+\\ldots)(c_0+c_1x+c_2x^2+c_3x^3+\\ldots)\\\\ &= x(1+2x+3x^2+4x^3+\\ldots) \\prod\\limits_{i=2}^{\\infty}\\frac{1}{1-x^i}\\\\ &= x\\cdot\\frac{1}{(1-x)^2} \\prod\\limits_{i=2}^{\\infty}\\frac{1}{1-x^i}\\\\ &= \\frac{x}{1-x} \\prod\\limits_{i=1}^{\\infty}\\frac{1}{1-x^i} \\end{align}\n\\]\n\nIf we expand the first line, we see that the coefficient of \$x^n\$ in \$G(x)\$ for any \$n\$ is \$1\\cdot c_{n-1}+2\\cdot c_{n-2}+\\ldots+n\\cdot c_0,\$ which is exactly \$S(n)\$! So by definition, \$G(x)=\\frac{x}{1-x} \\prod\\limits_{i=1}^{\\infty}\\frac{1}{1-x^i}\$ is the generating function of \$S(n).\$\n\nNow let's find the generating function of \$T(n).\$ Notice that counting the number of distinct elements in each partition and summing over all partitions is equivalent to counting how many partitions of \$n\$ contain i and then summing for all \$1\\leq i \\leq n.\$\n\nSo,\n\n\\[\nT(n) = \\sum_{k=1}^n k\\cdot (\\text{\\# of partitions of }n\\text{ that contain a }k).\n\\]\n\nHowever, the number of partitions of n that contain a \$i\$ is the same as the total number of partitions of \$n-i,\$ so\n\n\\[\nT(n) = \\sum_{k=1}^n k\\cdot (\\text{\\# of partitions of }n-k).\n\\]\n\nThe generating function for the number of partitions is\n\n\\[\n\\begin{align} P(x) &= (1+x+x^2+\\ldots)(1+x^2+x^4+\\ldots)(1+x^3+x^6+\\ldots)\\ldots \\\\ &= \\prod\\limits_{i=1}^{\\infty} \\frac{1}{1-x^i}. \\end{align}\n\\]\n\nLet's write the expansion of \$P(x)\$ as \$P(x)=d_0+d_1x+d_2x^2+d_3x^3+\\ldots,\$ so we wish to find \$T(n)=d_0+d_1+d_2+\\ldots+d_{n-1}.\$\n\nConsider the power series \$H(x)=(x+x^2+x^3+x^4+\\ldots)P(x).\$\n\n\\[\n\\begin{align} H(x) &= (x+x^2+x^3+x^4+\\ldots)(d_0+d_1x+d_2x^2+d_3x^3+\\ldots) \\\\ &= x(1+x+x^2+x^3+\\ldots) \\prod\\limits_{i=1}^{\\infty} \\frac{1}{1-x^i}\\\\ &= x\\cdot \\frac{1}{1-x}\\prod\\limits_{i=1}^{\\infty} \\frac{1}{1-x^i}\\\\ &= \\frac{x}{1-x} \\prod\\limits_{i=1}^{\\infty} \\frac{1}{1-x^i} \\end{align}\n\\]\n\nIf we expand the first line, we see that the coefficient of \$x^n\$ in \$H(x)\$ for any \$n\$ is \$d_{n-1}+d_{n-2}+\\ldots+d_0,\$ which is precisely \$T(n).\$ This means \$H(x)\$ is the generating function of \$T(n).\$\n\nThus, the generating functions of \$S(n)\$ and \$T(n)\$ are the same, so \$S(n)=T(n)\$ for all \$n\$ and we are done.\n\n~Peggy" ]
USAMO-1987-1	https://artofproblemsolving.com/wiki/index.php/1987_USAMO_Problems/Problem_1	Find all solutions to $(m^2+n)(m + n^2)= (m - n)^3$, where m and n are non-zero integers.	[ "Expanding both sides,\n\n\\[\nm^3+mn+m^2n^2+n^3=m^3-3m^2n+3mn^2-n^3\n\\]\n\nNote that \$m^3\$ can be canceled and as \$n \\neq 0\$, \$n\$ can be factored out. Writing this as a quadratic equation in \$n\$:\n\n\\[\n2n^2+(m^2-3m)n+(3m^2+m)=0\n\\]\n\n. The discriminant \$b^2-4ac\$ equals\n\n\\[\n(m^2-3m)^2-8(3m^2+m)\n\\]\n\n\\[\n=m^4-6m^3-15m^2-8m\n\\]\n\n, which we want to be a perfect square. Miraculously, this factors as \$m(m-8)(m+1)^2\$. This is square iff \$m^2-8m\$ is square or \$m+1=0\$. It can be checked that the only nonzero \$m\$ that work are \$-1, 8, 9\$. Finally, plugging this in and discarding extraneous roots gives all possible ordered pairs \$(m, n)\$ as\n\n\\[\n\\{(-1,-1),(8,-10),(9,-6),(9,-21)\\}\n\\]\n\n." ]
USAMO-1987-2	https://artofproblemsolving.com/wiki/index.php/1987_USAMO_Problems/Problem_2	The feet of the angle bisectors of $\Delta ABC$ form a right-angled triangle. If the right-angle is at $X$, where $AX$ is the bisector of $\angle A$, find all possible values for $\angle A$.	[]
USAMO-1987-3	https://artofproblemsolving.com/wiki/index.php/1987_USAMO_Problems/Problem_3	$X$ is the smallest set of polynomials $p(x)$ such that: Show that if $r(x)$ and $s(x)$ are distinct elements of $X$, then $r(x) \neq s(x)$ for any $0 < x < 1$.	[ "Let \$s(x)\$ be an arbitrary polynomial in \$X.\$ Then \$0<s(x)<1\$ when \$0<x<1.\$ Define \$X_1=\\{s(x)\\in X:s(x)=x\\cdot s_1(x)\$ for some \$s_1(x)\\in X\\},\$ and \$X_2=\\{t(x)\\in X: t(x)=x+(1-x)t_1(x)\$ for some \$t_1(x)\\in X\\}.\$\n\nIf \$s(x) \\in X_1\$ and \$t(x)\\in X_2,\$ we have \$s(x) <x<t(x)\$ for all \$x\$ with \$0<x<1.\$ Therefore \$s(x)\\ne t(x)\$ for any \$0<x<1.\$\n\nFor any \$s(x), t(x) \\in X_1\$, Let \$s(x)=x\\cdot s_1(x)\$ and \$t(x)=x\\cdot t_1(x)\$ for \$s_1(x), t_1(x) \\in X.\$ If \$s_1(x) \\ne t_1(x)\$ for \$0<x<1,\$ then \$s(x)-t(x)=x(s_1(x)-t_1(x))\\ne 0\$ for \$0<x<1.\$\n\nSimilarly, for any \$s(x), t(x) \\in X_2\$, Let \$s(x)=x+(1-x) s_1(x)\$ and \$t(x)=x+(1-x) t_1(x)\$ for \$s_1(x), t_1(x) \\in X.\$ If \$s_1(x) \\ne t_1(x)\$ for \$0<x<1,\$ then \$s(x)-t(x)=(1-x)(s_1(x)-t_1(x))\\ne 0\$ for \$0<x<1.\$\n\nThe proof is done by an induction.\n\nJ.Z." ]
USAMO-1987-4	https://artofproblemsolving.com/wiki/index.php/1987_USAMO_Problems/Problem_4	M is the midpoint of XY. The points P and Q lie on a line through Y on opposite sides of Y, such that $\|XQ\| = 2\|MP\|$ and $\frac{\|XY\|}2 < \|MP\| < \frac{3\|XY\|}2$. For what value of $\frac{\|PY\|}{\|QY\|}$ is $\|PQ\|$ a minimum?	[]
USAMO-1987-5	https://artofproblemsolving.com/wiki/index.php/1987_USAMO_Problems/Problem_5	$a_1, a_2, \cdots, a_n$ is a sequence of 0's and 1's. T is the number of triples $(a_i, a_j, a_k)$ with $i<j<k$ which are not equal to (0, 1, 0) or (1, 0, 1). For $1\le i\le n$, $f(i)$ is the number of $j<i$ with $a_j = a_i$ plus the number of $j>i$ with $a_j\neq a_i$. Show that $T=\sum_{i=1}^n f(i)\cdot\left(\frac{f(i)-1}2\right)$. If n is odd, what is the smallest value of T?	[]
USAMO-1988-1	https://artofproblemsolving.com/wiki/index.php/1988_USAMO_Problems/Problem_1	The repeating decimal $0.ab\cdots k\overline{pq\cdots u}=\frac mn$, where $m$ and $n$ are relatively prime integers, and there is at least one decimal before the repeating part. Show that $n$ is divisble by 2 or 5 (or both). (For example, $0.011\overline{36}=0.01136363636\cdots=\frac 1{88}$, and 88 is divisible by 2.)	[ "First, split up the nonrepeating parts and the repeating parts of the decimal, so that the nonrepeating parts equal to \$\\frac{a}{b}\$ and the repeating parts of the decimal is equal to \$\\frac{c}{d}\$.\n\nSuppose that the length of \$0.ab\\cdots k\$ is \$p\$ digits. Then \$\\frac{a}{b} = \\frac{0.ab\\cdots k}{10^{p+1}}\$ Since \$0.ab\\cdots k < 10^{p+1}\$, after reducing the fraction, there MUST be either a factor of 2 or 5 remaining in the denominator. After adding the fractions \$\\frac{a}{b}+\\frac{c}{d}\$, the simplified denominator \$n\$ will be \$\\\\lcm(b,d)\$ and since \$b\$ has a factor of \$2\$ or \$5\$, \$n\$ must also have a factor of 2 or 5.\n\n\\[\n\\blacksquare\n\\]", "It is well-known that \$0.ab...k \\overline{pq...u} = \\frac{ab...u - ab...k}{99...900...0}\$, where there are a number of 9s equal to the count of digits in \$pq...u\$, and there are a number of 0s equal to the count of digits \$c\$ in \$ab...k\$. Obviously \$ab...k\$ is different from \$pq...u\$ (which is itself the repeating part), so the numerator cannot have \$c\$ consecutive terminating zeros, and hence the denominator still possesses a factor of 2 or 5 not canceled out. This completes the proof." ]
USAMO-1988-2	https://artofproblemsolving.com/wiki/index.php/1988_USAMO_Problems/Problem_2	The cubic polynomial $x^3+ax^2+bx+c$ has real coefficients and three real roots $r\ge s\ge t$. Show that $k=a^2-3b\ge 0$ and that $\sqrt k\le r-t$.	[ "By Vieta's Formulas, \$a=-r-s-t\$, \$b=rs+st+rt\$, and \$c=-rst\$. Now we know \$k=a^2-3b\$; in terms of r, s, and t, then,\n\n\\[\nk=(-r-s-t)^2-3(rs+st+rt)\n\\]\n\n\\[\nk=r^2+s^2+t^2-rs-st-rt\n\\]\n\nNow notice that we can multiply both sides by 2, and rearrange terms to get \$2k=(r-s)^2+(s-t)^2+(r-t)^2\$. But since \$r, s, t\\in \\mathbb{R}\$, the three terms of the RHS are all non-negative (as the square of a real number is always non-negative), and therefore their sum is also non-negative -- that is, \$2k\\ge 0 \\Rightarrow k\\ge 0\$.\n\nNow, we will show that \$\\sqrt k\\le r-t\$. We can square both sides, and the inequality will hold since they are both non-negative (it is given that \$r\\ge t\$, therefore \$r-t\\ge 0\$). This gives \$k \\le r^2-2rt+t^2\$. Now we already have \$k=r^2+s^2+t^2-rs-st-rt\$, so substituting this for k gives\n\n\\[\nr^2+s^2+t^2-rs-st-rt \\le r^2-2rt+t^2\n\\]\n\n\\[\ns^2-rs-st+rt \\le 0\n\\]\n\n\\[\ns^2-(r+t)s+rt \\le 0\n\\]\n\nNote that this is a quadratic. Since its leading coefficient is positive, its value is less than 0 when s is between the two roots. Using the quadratic formula:\n\n\\[\ns=\\frac {r+t\\pm \\sqrt {(-r-t)^2-4rt} } 2\n\\]\n\n\\[\ns=\\frac {r+t\\pm \\sqrt {r^2-2rt+t^2} } 2\n\\]\n\n\\[\ns=\\frac {r+t\\pm (r-t) } 2\n\\]\n\n\\[\ns \\in \\{r, t\\}\n\\]\n\nThe quadratic is 0 when s is equal to r or t, and the inequality holds when its value is less than or equal to 0 -- that is, \$r\\ge s\\ge t\$. (Its value is less than or equal to 0 when s is between the roots, since the graph of the quadratic opens upward.) In fact, the problem tells us this is true. Q.E.D.", "From Vieta's Formula (which tells us that \$a = -(r+s+t)\$ and \$b = rs + st + rt\$), we have that\n\n\\[\nk = a^2 - 3b = r^2 + s^2 + t^2 - rs - st - rt = \\frac{1}{2} ((r-s)^2 + (s-t)^2 + (r-t)^2),\n\\]\n\nclearly non-negative. To prove \$\\sqrt{k} \\le r - t\$, it suffices to prove the square of this relation, or\n\n\\[\nr^2 + s^2 + t^2 - rs - st - rt \\le r^2 - 2rt + t^2.\n\\]\n\nThis in turn simplifies to\n\n\\[\nrs + st - rt - s^2 \\ge 0,\n\\]\n\nor\n\n\\[\n(r - s)(s - t) \\ge 0,\n\\]\n\nwhich is clearly true as \$r \\ge s \\ge t\$. This completes the proof.", "By Vieta's Formulas, \$a = -(r+s+t)\$ and \$b = rs + st + rt\$. \$k = (r+s+t)^2 - 3(rs + st + rt) = r^2+s^2+t^2-rs-st-rt = \\frac{(r-s)^2 + (s-t)^2 + (r-t)^2}{2}\$.\n\nTo show that \$k \\ge 0\$, simply note that by the trivial inequality, all three squares are greater than \$0\$ as they are the squares of real numbers.\n\nTo show that \$\\sqrt{k} \\le r-t\$, since both are positive, it is sufficient to show that \$k \\le (r-t)^2\$. \$\\frac{(r-s)^2 + (s-t)^2 + (r-t)^2}{2} \\le (r-t)^2\$ implies that \$k \\le (r-t)^2\$. \$\\frac{(r-s)^2 + (s-t)^2 - (r-t)^2}{2} \\le 0\$. Let \$y = r-s\$ and \$z = s-t\$. We then have \$\\frac{y^2 + z^2 - (y+z)^2}{2} \\le 0 \\implies -2yz \\le 0\$, which is clearly true as both \$y\$ and \$z\$ are positive." ]
USAMO-1988-3	https://artofproblemsolving.com/wiki/index.php/1988_USAMO_Problems/Problem_3	Let $X$ be the set $\{ 1, 2, \cdots , 20\}$ and let $P$ be the set of all 9-element subsets of $X$. Show that for any map $f: P\mapsto X$ we can find a 10-element subset $Y$ of $X$, such that $f(Y-\{k\})\neq k$ for any $k$ in $Y$.	[]
USAMO-1988-4	https://artofproblemsolving.com/wiki/index.php/1988_USAMO_Problems/Problem_4	$\Delta ABC$ is a triangle with incenter $I$. Show that the circumcenters of $\Delta IAB$, $\Delta IBC$, and $\Delta ICA$ lie on a circle whose center is the circumcenter of $\Delta ABC$.	[ "Let the circumcenters of \$\\Delta IAB\$, \$\\Delta IBC\$, and \$\\Delta ICA\$ be \$O_c\$, \$O_a\$, and \$O_b\$, respectively. It then suffices to show that \$A\$, \$B\$, \$C\$, \$O_a\$, \$O_b\$, and \$O_c\$ are concyclic.\n\nWe shall prove that quadrilateral \$ABO_aC\$ is cyclic first. Let \$\\angle BAC=\\alpha\$, \$\\angle CBA=\\beta\$, and \$\\angle ACB=\\gamma\$. Then \$\\angle ICB=\\gamma/2\$ and \$\\angle IBC=\\beta/2\$. Therefore minor arc \$\\overarc{BIC}\$ in the circumcircle of \$IBC\$ has a degree measure of \$\\beta+\\gamma\$. This shows that \$\\angle CO_aB=\\beta+\\gamma\$, implying that \$\\angle BAC+\\angle BO_aC=\\alpha+\\beta+\\gamma=180^{\\circ}\$. Therefore quadrilateral \$ABO_aC\$ is cyclic.\n\nThis shows that point \$O_a\$ is on the circumcircle of \$\\Delta ABC\$. Analagous proofs show that \$O_b\$ and \$O_c\$ are also on the circumcircle of \$ABC\$, which completes the proof. \$\\blacksquare\$", "Let \$M\$ denote the midpoint of arc \$AC\$. It is well known that \$M\$ is equidistant from \$A\$, \$C\$, and \$I\$ (to check, prove \$\\angle IAM = \\angle AIM = \\frac{\\angle BAC + \\angle ABC}{2}\$), so that \$M\$ is the circumcenter of \$AIC\$. Similar results hold for \$BIC\$ and \$CIA\$, and hence \$O_c\$, \$O_a\$, and \$O_b\$ all lie on the circumcircle of \$ABC\$.", "Extend \$CI\$ to point \$L\$ on \$(ABC)\$. By The Incenter-Excenter Lemma, B, I, A are all concyclic. Thus, L is the circumcenter of triangle \$IAB\$. In other words, \$L=O_c\$, so \$O_c\$ is on \$(ABC)\$. Similarly, we can show that \$O_a\$ and \$O_b\$ are on \$(ABC)\$, and thus, \$A,B,C,O_a,O_b,O_c\$ are all concyclic. It follows that the circumcenters are equal.", "Let the centers be \$T, R, S\$. We want to show that \$ABC\$ and \$TRS\$ have the same circumcircle. By Fact 5 we know that \$CATB\$ lie on a circle and similarly with the others. Thus the two triangles have the same circumcircle. ~coolmath_2018" ]
USAMO-1988-5	https://artofproblemsolving.com/wiki/index.php/1988_USAMO_Problems/Problem_5	Let $p(x)$ be the polynomial $(1-x)^a(1-x^2)^b(1-x^3)^c\cdots(1-x^{32})^k$, where $a, b, \cdots, k$ are integers. When expanded in powers of $x$, the coefficient of $x^1$ is $-2$ and the coefficients of $x^2$, $x^3$, ..., $x^{32}$ are all zero. Find $k$.	[ "First, note that if we reverse the order of the coefficients of each factor, then we will obtain a polynomial whose coefficients are exactly the coefficients of \$p(x)\$ in reverse order. Therefore, if\n\n\\[\np(x)=(1-x)^{a_1}(1-x^2)^{a_2}(1-x^3)^{a_3}\\cdots(1-x^{32})^{a_{32}},\n\\]\n\nwe define the polynomial \$q(x)\$ to be\n\n\\[\nq(x)=(x-1)^{a_1}(x^2-1)^{a_2}(x^3-1)^{a_3}\\cdots(x^{32}-1)^{a_{32}},\n\\]\n\nnoting that if the polynomial has degree \$n\$, then the coefficient of \$x^{n-1}\$ is \$-2\$, while the coefficients of \$x^{n-k}\$ for \$k=2,3,\\dots, 32\$ are all \$0\$.\n\nLet \$P_n\$ be the sum of the \$n\$th powers of the roots of \$q(x)\$. In particular, by Vieta's formulas, we know that \$P_1=2\$. Also, by Newton's Sums, as the coefficients of \$x^{n-k}\$ for \$k=2,3,\\dots,32\$ are all \$0\$, we find that\n\n\\[\n\\begin{align} P_2-2P_1&=0\\\\ P_3-2P_2&=0\\\\ P_4-2P_3&=0\\\\ &\\vdots\\\\ P_{32}-2P_{31}&=0. \\end{align}\n\\]\n\nThus \$P_n=2^n\$ for \$n=1,2,\\dots, 32\$. Now we compute \$P_{32}\$. Note that the roots of \$(x^n-1)^{a_n}\$ are all \$n\$th roots of unity. If \$\\omega=e^{2\\pi i/n}\$, then the sum of \$32\$nd powers of these roots will be\n\n\\[\na_n(1+\\omega^{32}+\\omega^{32\\cdot 2}+\\cdots+\\omega^{32\\cdot(n-1)}).\n\\]\n\nIf \$\\omega^{32}\\ne 1\$, then we can multiply by \$(\\omega^{32}-1)/(\\omega^{32}-1)\$ to obtain\n\n\\[\n\\frac{a_n(1-\\omega^{32n})}{1-\\omega^{32}}.\n\\]\n\nBut as \$\\omega^n=1\$, this is just \$0\$. Therefore the sum of the \$32\$nd powers of the roots of \$q(x)\$ is the same as the sum of the \$32\$nd powers of the roots of\n\n\\[\n(x-1)^{a_1}(x^2-1)^{a_2}(x^4-1)^{a_4}(x^{8}-1)^{a_4}(x^{16}-1)^{a_{16}}(x^{32}-1)^{a_{32}}.\n\\]\n\nThe \$32\$nd power of each of these roots is just \$1\$, hence the sum of the \$32\$nd powers of the roots is\n\n\\[\nP_{32}=2^{32}=a_1+2a_2+4a_4+8a_8+16a_{16}+32a_{32}.\\tag{1}\n\\]\n\nOn the other hand, we can use the same logic to show that\n\n\\[\nP_{16}=2^{16}=a_1+2a_2+4a_4+8a_8+16a_{16}.\\tag{2}\n\\]\n\nSubtracting (2) from (1) and dividing by 32, we find\n\n\\[\na_{32}=\\frac{2^{32}-2^{16}}{2^5}.\n\\]\n\nTherefore, \$a_{32}=2^{27}-2^{11}\$.", "By a limiting process, we can extend the problem to that of finding a sequence \$b_1, b_2, \\ldots\$ of integers such that\n\n\\[\n(1 - z)^{b_1}(1 - z^2)^{b_2}(1 - z^3)^{b_3}\\cdots = 1 - 2z.\n\\]\n\n(The notation comes from the Alcumus version of this problem.)\n\nIf we take logarithmic derivatives on both sides, we get\n\n\\[\n\\sum_{n = 1}^{\\infty}\\frac{b_n\\cdot (-nz^{n - 1})}{1 - z^n} = \\frac{-2}{1 - 2z},\n\\]\n\nand upon multiplying both sides by \$-z\$, this gives us the somewhat simple form\n\n\\[\n\\sum_{n = 1}^{\\infty} nb_n\\cdot\\frac{z^n}{1 - z^n} = \\frac{2z}{1 - 2z}.\n\\]\n\nExpanding all the fractions as geometric series, we get\n\n\\[\n\\sum_{n = 1}^{\\infty} nb_n\\sum_{k = 1}^{\\infty} z^{nk} = \\sum_{n = 1}^{\\infty} 2^nz^n.\n\\]\n\nComparing coefficients, we get\n\n\\[\n\\sum_{d\\mid n} db_d = 2^n\n\\]\n\nfor all positive integers \$n\$. In particular, as in Solution 1, we get\n\n\\[\n\\begin{array}{ll} b_1 + 2b_2 + 4b_4 + 8b_8 + 16b_{16} + 32b_{32} &= 2^{32}, \\\\ b_1 + 2b_2 + 4b_4 + 8b_8 + 16b_{16}\\phantom{ + 32b_{32}} &= 2^{16}, \\end{array}\n\\]\n\nfrom which the answer \$b_{32} = 2^{27} - 2^{11}\$ follows.\n\nRemark: To avoid the question of what an infinite product means in the context of formal power series, we could instead view the problem statement as saying that\n\n\\[\n(1 - z)^{b_1}(1 - z^2)^{b_2}\\cdots (1 - z^{32})^{b_{32}}\\equiv 1 - 2z\\pmod{z^{33}};\n\\]\n\nmodular arithmetic for polynomials can be defined in exactly the same way as modular arithmetic for integers. Uniqueness of the \$b_n\$'s comes from the fact that we have\n\n\\[\n(1 - z)^{b_1}\\cdots (1 - z^{n - 1})^{b_{n - 1}}\\equiv 1 - 2z\\pmod{z^n}\n\\]\n\nfor all \$n\\leq 33\$ by further reduction modulo \$z^n\$ (as \$z^n\\mid z^{33}\$ for \$n\\leq 33\$), so we could uniquely solve for the \$b_n\$'s one at a time. (This idea can be pushed further to explain why it's fine to pass to the infinite product version of the problem.)\n\nTo convert the above solution to one that works with polynomials modulo \$z^{33}\$, note that the derivative is not well-defined, as for instance, \$1\$ and \$1 + z^{33}\$ are equivalent modulo \$z^{33}\$, but their derivatives, \$0\$ and \$33z^{32}\$, are not. However, the operator \$f(z)\\mapsto zf'(z)\$ is well-defined. The other key idea is that for any \$n\$, modulo \$z^n\$, polynomials of the form \$1 - zf(z)\$ are invertible, with inverse\n\n\\[\n\\frac{1}{1 - zf(z)}\\equiv\\frac{1 - (zf(z))^n}{1 - zf(z)} = 1 + zf(z) + \\cdots + (zf(z))^{n - 1}).\n\\]\n\nTherefore, for the polynomial in the problem, call it \$g(z)\$, we can still form the expression \$zg'(z)/g(z)\$, which is what we originally got by taking the logarithmic derivative and multiplying by \$z\$, and expand it to eventually get\n\n\\[\n\\sum_{n = 1}^{32} nb_n\\sum_{k = 1}^{32} z^{nk}\\equiv\\sum_{n = 1}^{32} 2^nz^n\\pmod{z^{33}},\n\\]\n\nwhich gets us the same relations (for \$n\\leq 32\$).", "From the starting point of Solution 2,\n\n\\[\n(1 - z)^{b_1}(1 - z^2)^{b_2}(1 - z^3)^{b_3}\\cdots = 1 - 2z,\n\\]\n\ntaking reciprocals and expanding with geometric series gives us\n\n\\[\n\\prod_{n = 1}^{\\infty}\\left(\\sum_{k = 0}^{\\infty} z^{kn}\\right)^{b_n} = \\sum_{n = 0}^{\\infty} 2^nz^n.\n\\]\n\nOn the right, we have the generating function for the number of monic polynomials of degree \$n\$ over the field \$\\mathbb{F}_2\$ of two elements, and on the left, we have the factorisation of this generating function that considers the breakdown of any given monic polynomial into monic irreducible factors. As such, we have the interpretation\n\n\\[\nb_n = \\text{number of monic irreducible polynomials of degree }n\\text{ over }\\mathbb{F}_2.\n\\]\n\nFrom here, to determine \$b_n\$, we analyse the elements of \$\\mathbb{F}_{2^n}\$, of which there are \$2^{n}\$ in total. Given \$\\alpha\\in\\mathbb{F}_{2^n}\$, if the minimal polynomial \$f_{\\alpha}\$ of \$\\alpha\$ has degree \$d\$, then \$d\\mid n\$ and all other roots of \$f_{\\alpha}\$ appear in \$\\mathbb{F}_{2^n}\$. Moreover, if \$d\\mid n\$ and \$f\$ is an irreducible polynomial of degree \$d\$, then all roots of \$f\$ appear in \$\\mathbb{F}_{2^n}\$. (These statements are all well-known in the theory of finite fields.) As such, for each \$d\\mid n\$, there are precisely \$db_d\$ elements of \$\\mathbb{F}_{2^n}\$ of degree \$d\$, and we obtain the same equation as in Solution 2,\n\n\\[\n\\sum_{d\\mid n} db_d = 2^n.\n\\]\n\nThe rest is as before." ]
USAMO-1989-1	https://artofproblemsolving.com/wiki/index.php/1989_USAMO_Problems/Problem_1	For each positive integer $n$, let \[ \begin{align} S_n &= 1 + \frac 12 + \frac 13 + \cdots + \frac 1n \\ T_n &= S_1 + S_2 + S_3 + \cdots + S_n \\ U_n &= \frac{T_1}{2} + \frac{T_2}{3} + \frac{T_3}{4} + \cdots + \frac{T_n}{n+1}. \end{align} \] Find, with proof, integers $0 < a,\ b,\ c,\ d < 1000000$ such that $T_{1988} = a S_{1989} - b$ and $U_{1988} = c S_{1989} - d$.	[ "We note that for all integers \$n \\ge 2\$,\n\n\\[\n\\begin{align} T_{n-1} &= 1 + \\left(1 + \\frac 12\\right) + \\left(1 + \\frac 12 + \\frac 13\\right) + \\ldots + \\left(1 + \\frac 12 + \\frac 13 + \\ldots + \\frac 1{n-1}\\right) \\\\ &= \\sum_{i=1}^{n-1} \\left(\\frac {n-i}i\\right) = n\\left(\\sum_{i=1}^{n-1} \\frac{1}{i}\\right) - (n-1) = n\\left(\\sum_{i=1}^{n} \\frac{1}{i}\\right) - n \\\\ &= n \\cdot S_{n} - n . \\end{align}\n\\]\n\nIt then follows that\n\n\\[\n\\begin{align} U_{n-1} &= \\sum_{i=2}^{n} \\frac{T_{i-1}}{i} = \\sum_{i=2}^{n}\\ (S_{i} - 1) = T_{n-1} + S_n - (n-1) - S_1 \\\\ &= \\left(nS_n - n\\right) + S_n - n = (n + 1)S_n - 2n . \\end{align}\n\\]\n\nIf we let \$n=1989\$, we see that \$(a,b,c,d) = (1989,1989,1990, 2\\cdot 1989)\$ is a suitable solution. \$\\blacksquare\$\n\nNotice that it is also possible to use induction to prove the equations relating \$T_n\$ and \$U_n\$ with \$S_n\$." ]
USAMO-1989-2	https://artofproblemsolving.com/wiki/index.php/1989_USAMO_Problems/Problem_2	The 20 members of a local tennis club have scheduled exactly 14 two-person games among themselves, with each member playing in at least one game. Prove that within this schedule there must be a set of 6 games with 12 distinct players.	[ "Consider a graph with \$20\$ vertices and \$14\$ edges. The sum of the degrees of the vertices is \$28\$; by the Pigeonhole Principle at least \$12\$ vertices have degrees of \$1\$ and at most \$8\$ vertices have degrees greater than \$1\$. If we keep deleting edges of vertices with degree greater than \$1\$ (a maximum of \$8\$ such edges), then we are left with at least \$6\$ edges, and all of the vertices have degree either \$0\$ or \$1\$. These \$6\$ edges represent the \$6\$ games with \$12\$ distinct players.", "Let a slot be a place we can put a member in a game, so there are two slots per game, and 28 slots total. We begin by filling exactly 20 slots each with a distinct member since each member must play at least one game. Let there be \$m\$ games with both slots filled and \$n\$ games with only one slot filled, so \$2m+n=20\$. Since there are only 14 games, \$m+n \\leq 14 \\Longrightarrow 2m+n \\leq 14+m \\Longleftrightarrow 20 \\leq 14+m \\Longrightarrow m \\geq 6\$, so there must be at least 6 games with two distinct members each, and we must have our desired set of 6 games.", "Assume the contrary.\n\nConsider the largest set of disjoint edges \$E\$. By assumption it has less than \$6\$ edges, i.e. maximum \$10\$ vertices. Call it a vertex set \$V\$.\n\n\$10\$ vertices remain outside \$V\$ and each has to be attached to at least one edge. Now, if any two vertices outside \$V\$ are connected by, say, edge \$e\$, we could have included \$e\$ in \$E\$ and gotten a larger disjoint set, so - a contradiction. Therefore the only option would be that all vertices outside \$V\$ are connected each by one edge to some vertices inside \$V\$. That would take \$10\$ edges, but \$E\$ already includes \$5\$ - again a contradiction.\n\nAll possibilities yield a contradiction, so our assumption can not be correct.\n\n(Cases when largest set \$E\$ is smaller than \$6\$ are equivalent and weaker)", "Assume, for the sake of contradiction, that there is no set of 6 games with 12 distinct players. Then the largest possible collection of games with all distinct players must consist of at most 5 games, involving at most 10 players. So suppose we have 5 such games with 10 distinct players. This leaves 10 other players who must still appear in the schedule. Since there are a total of 14 games, there are \$14-5=9\$ games remaining to cover these 10 players.\n\nEach game has 2 players, so the 9 games account for 18 ``player slots. If every one of the 10 unused players is to appear at least once, then by the pigeonhole principle, some game among these 9 must contain two of the 10 unused players. But then that game, together with the original 5 disjoint games, would give 6 games with 12 distinct players. This contradicts our assumption that no such collection exists. Therefore, there must in fact exist a set of 6 games with 12 distinct players." ]
USAMO-1989-3	https://artofproblemsolving.com/wiki/index.php/1989_USAMO_Problems/Problem_3	Let $P(z)= z^n + c_1 z^{n-1} + c_2 z^{n-2} + \cdots + c_n$ be a polynomial in the complex variable $z$, with real coefficients $c_k$. Suppose that $\|P(i)\| < 1$. Prove that there exist real numbers $a$ and $b$ such that $P(a + bi) = 0$ and $(a^2 + b^2 + 1)^2 < 4 b^2 + 1$.	[ "Let \$z_1, \\dotsc, z_n\$ be the (not necessarily distinct) roots of \$P\$, so that\n\n\\[\nP(z) = \\prod_{j=1}^n (z- z_j) .\n\\]\n\nSince all the coefficients of \$P\$ are real, it follows that if \$w\$ is a root of \$P\$, then \$P( \\overline{w}) = \\overline{ P(w)} = 0\$, so \$\\overline{w}\$, the complex conjugate of \$w\$, is also a root of \$P\$.\n\nSince\n\n\\[\n\\lvert i- z_1 \\rvert \\cdot \\lvert i - z_2 \\rvert \\dotsm \\lvert i - z_n \\rvert = \\lvert P(i) \\rvert < 1,\n\\]\n\nit follows that for some (not necessarily distinct) conjugates \$z_i\$ and \$z_j\$,\n\n\\[\n\\lvert z_i-i \\rvert \\cdot \\lvert z_j-i \\rvert < 1.\n\\]\n\nLet \$z_i = a+bi\$ and \$z_j = a-bi\$, for real \$a,b\$. We note that\n\n\\[\n(a+b+1)^2 - (a+b-1)^2 = 4a + 4b .\n\\]\n\nThus\n\n\\[\n\\begin{align} (a^2+b^2+1)^2 &= (a^2+b^2-1)^2 + 4a^2 + 4b^2 = \\lvert a^2 + b^2 - 1 - 2ai \\rvert ^2 + 4b^2 \\\\ &= \\lvert (a-i)^2 - (bi)^2 \\rvert^2 + 4b^2 \\\\ &= \\bigl( \\lvert a+bi - i \\rvert \\cdot \\lvert a-bi -i \\rvert \\bigr)^2 + 4b^2 \\\\ &= \\bigl( \\lvert z_i - i \\rvert \\cdot \\lvert z_j - i \\rvert \\bigr)^2 + 4b^2 < 1+4b^2. \\end{align}\n\\]\n\nSince \$P(a+bi) = P(z_i) = 0\$, these real numbers \$a,b\$ satisfy the problem's conditions. \$\\blacksquare\$" ]
USAMO-1989-4	https://artofproblemsolving.com/wiki/index.php/1989_USAMO_Problems/Problem_4	Let $ABC$ be an acute-angled triangle whose side lengths satisfy the inequalities $AB < AC < BC$. If point $I$ is the center of the inscribed circle of triangle $ABC$ and point $O$ is the center of the circumscribed circle, prove that line $IO$ intersects segments $AB$ and $BC$.	[ "Consider the lines that pass through the circumcenter \$O\$. Extend \$AO\$, \$BO\$, \$CO\$ to \$D\$,\$E\$,\$F\$ on \$a\$,\$b\$,\$c\$, respectively.\n\nWe notice that \$IO\$ passes through sides \$a\$ and \$c\$ if and only if \$I\$ belongs to either regions \$AOF\$ or \$COD\$.\n\nSince \$AO = BO = CO = R\$, we let \$\\alpha = \\angle OAC = \\angle OCA\$, \$\\beta = \\angle BAO = \\angle ABO\$, \$\\gamma = \\angle BCO = \\angle CBO\$.\n\nWe have \$c < b < a\\implies C < B < A\\implies\$ \$\\gamma+\\alpha <\\beta+\\gamma <\\alpha+\\beta\\implies\\gamma <\\alpha <\\beta\$\n\nSince \$IA\$ divides angle \$A\$ into two equal parts, it must be in the region marked by the \$\\beta\$ of angle \$A\$, so \$I\$ is in \$ABD\$.\n\nSimilarly, \$I\$ is in \$ACF\$ and \$ABE\$. Thus, \$I\$ is in their intersection, \$AOF\$. From above, we have \$IO\$ passes through \$a\$ and \$c\$. \$\\blacksquare\$" ]
USAMO-1989-5	https://artofproblemsolving.com/wiki/index.php/1989_USAMO_Problems/Problem_5	Let $u$ and $v$ be real numbers such that \[ (u + u^2 + u^3 + \cdots + u^8) + 10u^9 = (v + v^2 + v^3 + \cdots + v^{10}) + 10v^{11} = 8. \] Determine, with proof, which of the two numbers, $u$ or $v$, is larger.	[ "The answer is \$v\$.\n\nWe define real functions \$U\$ and \$V\$ as follows:\n\n\\[\n\\begin{align} U(x) &= (x+x^2 + \\dotsb + x^8) + 10x^9 = \\frac{x^{10}-x}{x-1} + 9x^9 \\\\ V(x) &= (x+x^2 + \\dotsb + x^{10}) + 10x^{11} = \\frac{x^{12}-x}{x-1} + 9x^{11} . \\end{align}\n\\]\n\nWe wish to show that if \$U(u)=V(v)=8\$, then \$u <v\$.\n\nWe first note that when \$x \\le 0\$, \$x^{12}-x \\ge 0\$, \$x-1 < 0\$, and \$9x^9 \\le 0\$, so\n\n\\[\nU(x) = \\frac{x^{10}-x}{x-1} + 9x^9 \\le 0 < 8 .\n\\]\n\nSimilarly, \$V(x) \\le 0 < 8\$.\n\nWe also note that if \$x \\ge 9/10\$, then\n\n\\[\n\\begin{align} U(x) &= \\frac{x-x^{10}}{1-x} + 9x^9 \\ge \\frac{9/10 - 9^9/10^9}{1/10} + 9 \\cdot \\frac{9^{9}}{10^9} \\\\ &= 9 - 10 \\cdot \\frac{9^9}{10^9} + 9 \\cdot \\frac{9^9}{10^9} = 9 - \\frac{9^9}{10^9} > 8. \\end{align}\n\\]\n\nSimilarly \$V(x) > 8\$. It then follows that \$u, v \\in (0,9/10)\$.\n\nNow, for all \$x \\in (0,9/10)\$,\n\n\\[\n\\begin{align} V(x) &= U(x) + V(x)-U(x) = U(x) + 10x^{11}+x^{10} -9x^9 \\\\ &= U(x) + x^9 (10x -9) (x+1) < U(x) . \\end{align}\n\\]\n\nSince \$V\$ and \$U\$ are both strictly increasing functions over the nonnegative reals, it then follows that\n\n\\[\nV(u) < U(u) = 8 = V(v),\n\\]\n\nso \$u<v\$, as desired. \$\\blacksquare\$" ]
USAMO-1990-1	https://artofproblemsolving.com/wiki/index.php/1990_USAMO_Problems/Problem_1	A certain state issues license plates consisting of six digits (from 0 through 9). The state requires that any two plates differ in at least two places. (Thus the plates $\boxed{027592}$ and $\boxed{020592}$ cannot both be used.) Determine, with proof, the maximum number of distinct license plates that the state can use.	[ "Consider license plates of \$n\$ digits, for some fixed \$n\$, issued with the same criteria.\n\nWe first note that by the pigeonhole principle, we may have at most \$10^{n-1}\$ distinct plates. Indeed, if we have more, then there must be two plates which agree on the first \$n-1\$ digits; these plates thus differ only on one digit, the last one.\n\nWe now show that it is possible to issue \$10^{n-1}\$ distinct license plates which satisfy the problem's criteria. Indeed, we issue plates with all \$10^{n-1}\$ possible combinations for the first \$n-1\$ digit, and for each plate, we let the last digit be the sum of the preceding digits taken mod 10. This way, if two plates agree on the first \$n-1\$ digits, they agree on the last digit and are thus the same plate, and if two plates differ in only one of the first \$n-1\$ digits, they must differ as well in the last digit.\n\nIt then follows that \$10^{n-1}\$ is the greatest number of license plates the state can issue. For \$n=6\$, as in the problem, this number is \$10^5\$. \$\\blacksquare\$", "\\[\n\\text{Constructive Induction for } n=6\n\\]" ]
USAMO-1990-2	https://artofproblemsolving.com/wiki/index.php/1990_USAMO_Problems/Problem_2	A sequence of functions $\, \{f_n(x) \} \,$ is defined recursively as follows: \[ \begin{align} f_1(x) &= \sqrt {x^2 + 48}, \quad \text{and} \\ f_{n + 1}(x) &= \sqrt {x^2 + 6f_n(x)} \quad \text{for } n \geq 1. \end{align} \] (Recall that $\sqrt {\makebox[5mm]{}}$ is understood to represent the positive square root.) For each positive integer $n$, find all real solutions of the equation $\, f_n(x) = 2x \,$.	[ "We define \$f_0(x) = 8\$. Then the recursive relation holds for \$n=0\$, as well.\n\nSince \$f_n (x) \\ge 0\$ for all nonnegative integers \$n\$, it suffices to consider nonnegative values of \$x\$.\n\nWe claim that the following set of relations hold true for all natural numbers \$n\$ and nonnegative reals \$x\$:\n\n\\[\n\\begin{align} f_n(x) &< 2x \\text{ if }x>4 ; \\\\ f_n(x) &= 2x \\text{ if }x=4 ; \\\\ f_n(x) &> 2x \\text{ if }x<4 . \\end{align}\n\\]\n\nTo prove this claim, we induct on \$n\$. The statement evidently holds for our base case, \$n=0\$.\n\nNow, suppose the claim holds for \$n\$. Then\n\n\\[\n\\begin{align} f_{n+1}(x) &= \\sqrt{x^2 + 6f_n(x)} < \\sqrt{x^2+12x} < \\sqrt{4x^2} = 2x, \\text{ if } x>4 ; \\\\ f_{n+1}(x) &= \\sqrt{x^2 + 6f_n(x)} = \\sqrt{x^2 + 12x} = \\sqrt{4x^2} = 2x, \\text{ if } x=4 ; \\\\ f_{n+1}(x) &= \\sqrt{x^2 + 6f_n(x)} > \\sqrt{x^2+12x} > \\sqrt{4x^2} = 2x, \\text{ if } x<4 . \\end{align}\n\\]\n\nThe claim therefore holds by induction. It then follows that for all nonnegative integers \$n\$, \$x=4\$ is the unique solution to the equation \$f_n(x) = 2x\$. \$\\blacksquare\$", "We claim that the only solution is \$\\boxed{x=4}\$ for all such \$n\$. To show this, we consider all \$x\$ and find solutions \$n\$.\n\nBefore we consider solutions, we show that for all \$x\$, \$f_n(x)\$ is positive for all positive integers \$n\$ by induction. For our base case:\n\n\\[\nf_1(x)=\\sqrt{x^2+48}\\ge\\sqrt{48}>0\n\\]\n\nso it is positive. Next, for our inductive step, assume for some \$n=k-1\$ that \$f_{k-1}(x)\$ is positive; thus \$f_{k-1}(x)>0\$, and we show that \$f_k(x)\$ is positive:\n\n\\[\nf_k(x)=\\sqrt{x^2+6f_{k-1}(x)}\\ge\\sqrt{0+6f_{k-1}(x)}>\\sqrt{0+0}=0\n\\]\n\nthus \$f_k(x)>0\$, so we have proven the claim by induction.\n\nFirst, we consider negative \$x\$. We know that for negative \$x\$, if the equation were to have a solution, we would have \$f_n(x)=2x<0\$ for some \$n\$. However, \$f_n(x)\$ is always nonnegative since \$f_n(x)\$ is always the square root of some number, which can never be negative; thus there are no solutions in this case.\n\nNext, we consider \$x=0\$. Solutions would have to satisfy \$f_n(0)=0\$; we have previously shown that all \$f_n(x)\$ are positive, so there are no such \$n\$.\n\nFinally, we consider positive \$x\$. We divide this set into three groups: \$x<4\$, \$x=4\$, and \$x>4\$. We claim that for increasing \$n\$, \$f_n(x)\$ is decreasing, constant, and increasing, respectively, and we prove this using induction. First, we analyze \$x<4\$. Then\n\n\\[\nf_1(x)<\\sqrt{4^2+48}=\\sqrt{64}=8\n\\]\n\nFor our base case:\n\n\\[\nf_2(x)=\\sqrt{x^2+6f_1(x)}<\\sqrt{x^2+6\\cdot8}=f_1(x)\n\\]\n\nThus \$f_2(x)<f_1(x)\$. For our inductive step, if for some \$n=k-1\$ we have \$f_k(x)<f_{k-1}(x)\$, we show that \$f_{k+1}(x)<f_k(x)\$:\n\n\\[\nf_{k+1}(x)=\\sqrt{x^2+6f_k(x)}<\\sqrt{x^2+6f_{k-1}(x)}=f_k(x)\n\\]\n\nThus \$f_1(x)>f_2(x)>f_3(x)>\\ldots\$ and the conclusion follows.\n\nNext, for \$x=4\$, \$f_1(4)=8\$ by substituting. Then, if \$f_{n-1}(4)=8\$, we show that \$f_n(4)=8\$ as well: \$f_n(4)=\\sqrt{x^2+6f_{n-1}(x)}=\\sqrt{16+6\\cdot8}=\\sqrt{64}=8\$ so the sequence is constant. Notice that in this case, the equation we must solve becomes \$f_n(4)=8\$, which is true for all positive integers \$n\$.\n\nFinally, we analyze \$x>4\$. Then\n\n\\[\nf_1(x)>\\sqrt{4^2+48}=\\sqrt{64}=8\n\\]\n\nFor our base case:\n\n\\[\nf_2(x)=\\sqrt{x^2+6f_1(x)}>\\sqrt{x^2+6\\cdot8}=f_1(x)\n\\]\n\nThus \$f_2(x)>f_1(x)\$. For our inductive step, if for some \$n=k-1\$ we have \$f_k(x)>f_{k-1}(x)\$, we show that \$f_{k+1}(x)>f_k(x)\$:\n\n\\[\nf_{k+1}(x)=\\sqrt{x^2+6f_k(x)}>\\sqrt{x^2+6f_{k-1}(x)}=f_k(x)\n\\]\n\nThus \$f_1(x)<f_2(x)<f_3(x)<\\ldots\$ and the conclusion follows.\n\nIf for some positive integer \$n\$ we have \$f_n(x)=2x\$, then:\n\n\\begin{align} f_n(x)&=\\sqrt{x^2+6f_{n-1}(x)}\\\\ 2x&=\\sqrt{x^2+6f_{n-1}(x)}\\\\ 4x^2&=x^2+6f_{n-1}(x)\\\\ 3x^2&=6f_{n-1}(x)\\\\ \\frac{1}{2}x^2&=f_{n-1}(x) \\end{align} However, if \$0<x<4\$ is a solution, then we must have \$f_{n-1}(x)>f_n(x)\$; substituting values yields \$\\frac{1}{2}x^2>2x\$, which implies \$x>4\$; this is a contradiction. Similarly, if \$x>4\$ is a solution, then we must have \$f_{n-1}(x)<f_n(x)\$; substituting values yields \$\\frac{1}{2}x^2<2x\$, which implies \$x<4\$; this is also a contradiction.\n\nAs a result, since we have already established that \$x=4\$ is a solution for all \$n\$ and no other \$x\$ can be solutions, we conclude that for each positive integer \$n\$, the only real solution of the equation is \$x=4\$.\n\n~ eevee9406" ]
USAMO-1990-3	https://artofproblemsolving.com/wiki/index.php/1990_USAMO_Problems/Problem_3	Suppose that necklace $\, A \,$ has 14 beads and necklace $\, B \,$ has 19. Prove that for any odd integer $n \geq 1$, there is a way to number each of the 33 beads with an integer from the sequence \[ \{ n, n+1, n+2, \dots, n+32 \} \] so that each integer is used once, and adjacent beads correspond to relatively prime integers. (Here a "necklace" is viewed as a circle in which each bead is adjacent to two other beads.)	[ "Lemma. For every positive odd integer \$n\$, there exists a nonnegative integer \$k \\le 17\$ such that \$n+k\$ is relatively prime to \$n+k+15\$, \$n+k+1\$ is relatively prime to \$n+k+14\$.\n\nProof. Consider the positive integers \$n, n+1, n+2\$. Note that at most one of these is divisible by 3, and at most one is divisible by 5. Therefore one of these, say \$n+a\$, is divisible by neither, and is therefore relatively prime to 15. Furthermore, \$n+a+1\$ and \$n+(a+15)+1\$ have different residues mod 13, so one of them, say \$n+b+1\$, is relatively prime to 13. Since \$n+b \\equiv n+a \\pmod{15}\$, \$n+b\$ is relatively prime to 15. But\n\n\\[\n\\gcd(n+b,n+b+15) = \\gcd\\bigl[ n+b,(n+b+15) - (n+b) \\bigr] = \\gcd(n+b,15) = 1,\n\\]\n\nso \$n+b\$ is relatively prime to \$n+b+15\$. Also,\n\n\\[\n\\gcd(n+b+1, n+b+14) = \\gcd \\bigl[ n+b+1, (n+b+14)-(n+b+1) \\bigr] = \\gcd( n+b+1, 13) ,\n\\]\n\nso \$n+b+1\$ is relatively prime to \$n+b+14\$. Finally, \$b \\le a+15 \\le 17\$. It follows that setting \$k=b\$ satisfies the lemma. \$\\blacksquare\$\n\nLet \$k\$ be an integer as described in the lemma. We place the integers \$n, \\dotsc, n+k, n+k+15, \\dotsc, n+32\$ on the necklace with 19 beads, and the integers \$n+k+1, \\dotsc, n+k+14\$ on the necklace with 14 beads, in those orders. Since \$n\$ is odd, \$n\$ is relatively prime to \$n+32 = n+2^5\$. By definition, \$n+k\$ and \$n+k+15\$ are relatively prime, as are \$n+k+1\$ and \$n+k+14\$; finally, \$a\$ and \$a+1\$ are relatively prime for all integers \$a\$. It follows that each bead in this arrangement is relatively prime to its neighbors. \$\\blacksquare\$" ]
USAMO-1990-4	https://artofproblemsolving.com/wiki/index.php/1990_USAMO_Problems/Problem_4	Find, with proof, the number of positive integers whose base-$n$ representation consists of distinct digits with the property that, except for the leftmost digit, every digit differs by $\pm 1$ from some digit further to the left. (Your answer should be an explicit function of $n$ in simplest form.)	[ "Let a \$k\$-good sequence be a sequence of distinct integers \$\\{ a_i \\}_{i=1}^k\$ such that for all integers \$2\\le i \\le k\$, \$a_i\$ differs from some preceding term by \$\\pm 1\$.\n\nLemma. Let \$a\$ be an integer. Then there are \$2^{k-1}\$ \$k\$-good sequences starting on \$a\$, and furthermore, the terms of each of these sequences constitute a permutation of \$k\$ consecutive integers.\n\nProof. We induct on \$k\$. For \$k=1\$, the lemma is trivially true. Now, suppose the lemma holds for \$k\$. If \$\\{ a_i \\}_{i=1}^{k+1}\$ is a \$(k+1)\$-good sequence, then \$\\{ a_i \\}_{i=1}^k\$ is a \$k\$-good sequence which starts on \$a\$, so it is a permutation of \$k\$ consecutive integers, say \$m, \\dotsc, M\$. Then the only possibilities for \$a_{k+1}\$ are \$m-1\$ and \$M+1\$; either way, \$\\{ a_i \\}_{i=1}^{k+1}\$ constitutes a permutation of \$k+1\$ consecutive integers. Since there are \$2^k\$ possible sequences \$\\{a_i\\}_{i=1}^k\$, and 2 choices of \$a_{k+1}\$ for each of these sequences, it also follows that there are \$2^k \\cdot 2 = 2^{k+1}\$ \$(k+1)\$-good sequences which start on \$a\$. Thus the lemma holds by induction. \$\\blacksquare\$\n\nWe now consider the number of desired positive integers with \$k\$ digits. Evidently, \$k\$ must be less than or equal to \$n\$. We also note that the digits of such an integer must constitute a \$k\$-good sequence. Since the minimum of this sequence can be any of the digits \$0, \\dotsc, n-k\$, unless the minimum is 0 and is the first digit (in which case the only possible sequence is an increasing arithmetic sequence), and there are \$2^{k-1}\$ \$k\$-good sequences up to translation, it follows that there are \$(n-k+1) 2^{k-1}-1\$ desired positive integers with \$k\$ digits. Thus the total number of desired positive integers is\n\n\\[\n\\begin{align} \\sum_{k=1}^n \\bigl[ (n-k+1) 2^{k-1}-1 \\bigr] &= -n + \\sum_{k=1}^n \\sum_{j=k}^n 2^{k-1} = -n + \\sum_{j=1}^n \\sum_{k=1}^j 2^{k-1} \\\\ &= -n + \\sum_{j=1}^n (2^k-1) = - 2n -1 + \\sum_{j=0}^n 2^k, \\end{align}\n\\]\n\nwhich is equal to \$2^{n+1} - 2(n+1)\$, our answer. \$\\blacksquare\$" ]
USAMO-1990-5	https://artofproblemsolving.com/wiki/index.php/1990_USAMO_Problems/Problem_5	An acute-angled triangle $ABC$ is given in the plane. The circle with diameter $\, AB \,$ intersects altitude $\, CC' \,$ and its extension at points $\, M \,$ and $\, N \,$, and the circle with diameter $\, AC \,$ intersects altitude $\, BB' \,$ and its extensions at $\, P \,$ and $\, Q \,$. Prove that the points $\, M, N, P, Q \,$ lie on a common circle.	[ "Let \$A'\$ be the intersection of the two circles (other than \$A\$). \$AA'\$ is perpendicular to both \$BA'\$, \$CA'\$ implying \$B\$, \$C\$, \$A'\$ are collinear. Since \$A'\$ is the foot of the altitude from \$A\$: \$A\$, \$H\$, \$A'\$ are concurrent, where \$H\$ is the orthocentre.\n\nNow, \$H\$ is also the intersection of \$BB'\$, \$CC'\$ which means that \$AA'\$, \$MN\$, \$PQ\$ are concurrent. Since \$A\$, \$M\$, \$N\$, \$A'\$ and \$A\$, \$P\$, \$Q\$, \$A'\$ are cyclic, \$M\$, \$N\$, \$P\$, \$Q\$ are cyclic by the radical axis theorem.", "Define \$A'\$ as the foot of the altitude from \$A\$ to \$BC\$. Then, \$AA' \\cap BB' \\cap CC'\$ is the orthocenter. We will denote this point as \$H\$. Since \$\\angle AA'C\$ and \$\\angle AA'B\$ are both \$90^{\\circ}\$, \$A'\$ lies on the circles with diameters \$AC\$ and \$AB\$.\n\nNow we use the Power of a Point theorem with respect to point \$H\$. From the circle with diameter \$AB\$ we get \$AH \\cdot A'H = MH \\cdot NH\$. From the circle with diameter \$AC\$ we get \$AH \\cdot A'H = PH \\cdot QH\$. Thus, we conclude that \$PH \\cdot QH = MH \\cdot NH\$, which implies that \$P\$, \$Q\$, \$M\$, and \$N\$ all lie on a circle.", "Let \$\\omega_1\$ be the circumcircle with diameter \$AB\$ and \$\\omega_2\$ be the circumcircle with diameter \$AC\$. We claim that the second intersection of \$\\omega_1\$ and \$\\omega_2\$ other than \$A\$ is \$A'\$, where \$A'\$ is the feet of the perpendicular from \$A\$ to segment \$BC\$. Note that\n\n\\[\n\\angle AA'B=90^{\\circ}=\\angle AB'B\n\\]\n\nso \$A'\$ lies on \$\\omega_1.\$ Similarly, \$A'\$ lies on \$\\omega_2\$. Hence, \$AA'\$ is the radical axis of \$\\omega_1\$ and \$\\omega_2\$. By the Radical Lemma, it suffices to prove that the intersection of lines \$MN\$ and \$PQ\$ lie on \$AA'\$. But, \$MN\$ is the same line as \$CC'\$ and \$PQ\$ is the same line as \$BB'\$. Since \$AA', BB'\$, and \$CC'\$ intersect at the orthocenter \$H\$, \$H\$ lies on the radical axis \$AA'\$ and we are done. \$\\blacksquare\$", "We know that \$BCB'C'\$ is a cyclic quadrilateral. Hence,\n\n\\[\nHB \\cdot HB' = HC \\cdot HC'\n\\]\n\n\\[\n\\implies Pow_{\\omega_{1}} = Pow_{\\omega_{2}}\n\\]\n\n\\[\n\\implies HM \\cdot HN = HP \\cdot HQ\n\\]\n\n\$\\implies MPNQ\$ is cyclic \$\\raggedright\\blacksquare\$." ]
USAMO-1991-1	https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_1	In triangle $ABC$, angle $A$ is twice angle $B$, angle $C$ is obtuse, and the three side lengths $a, b, c$ are integers. Determine, with proof, the minimum possible perimeter.	[ "\\[\n[asy] import olympiad; pair A, B, C, D, extensionAC; real angleABC; path braceBC; A = (0, 0); B = (2, 0); D = (1, .5); angleABC = atan(.5); //y = 4x/3 and x+2y = 2 (sides AC and BC, respectively) intersect here: C = (6/11, 8/11); braceBC = brace(C, B, .1); label(\"$\\mathsf{A}$\", A, W); label(\"$\\mathsf{B}$\", B, E); label(\"$\\mathsf{C}$\", C, N); label(\"$\\mathsf{D}$\", D, S); label(\"$\\mathsf{a}$\", braceBC, NE); label(\"$\\mathsf{b}$\", A--C, NW); label(\"$\\mathsf{c}$\", A--B, S); label(\"$\\mathsf{x}$\", A--D, N); draw(A--B--C--cycle); draw(A--D); draw(anglemark(C, B, A)); draw(anglemark(B, A, D)); draw(anglemark(D, A, C)); draw(braceBC); [/asy]\n\\]\n\n(diagram by integralarefun)\n\nAfter drawing the triangle, also draw the angle bisector of \$\\angle A\$, and let it intersect \$\\overline{BC}\$ at \$D\$. Notice that \$\\triangle ADC\\sim \\triangle BAC\$, and let \$AD=x\$. Now from similarity,\n\n\\[\nx=\\frac{bc}{a}\n\\]\n\nHowever, from the angle bisector theorem, we have\n\n\\[\nBD=\\frac{ac}{b+c}\n\\]\n\nbut \$\\triangle ABD\$ is isosceles, so\n\n\\[\nx=BD\\Longrightarrow \\frac{bc}{a}=\\frac{ac}{b+c}\\Longrightarrow a^2=b(b+c)\n\\]\n\nso all sets of side lengths which satisfy the conditions also meet the boxed condition.\n\nNotice that \$\\text{gcd}(a, b, c)=1\$ or else we can form a triangle by dividing \$a, b, c\$ by their greatest common divisor to get smaller integer side lengths, contradicting the perimeter minimality. Since \$a\$ is squared, \$b\$ must also be a square because if it isn't, then \$b\$ must share a common factor with \$b+c\$, meaning it also shares a common factor with \$c\$, which means \$a, b, c\$ share a common factor—a contradiction. Thus we let \$b = x^2, b+c = y^2\$, so \$a = xy\$, and we want the minimal pair \$(x,y)\$.\n\nBy the Law of Cosines,\n\n\\[\nb^2 = a^2 + c^2 - 2ac\\cos B\n\\]\n\nSubstituting \$a^2 = b^2 + bc\$ yields \$\\cos B = \\frac{b+c}{2a} = \\frac{y}{2x}\$. Since \$\\angle C > 90^{\\circ}\$, \$0^{\\circ} < \\angle B < 30^{\\circ} \\Longrightarrow \\sqrt{3} < \\frac{y}{x} < 2\$. For \$x \\le 3\$ there are no integer solutions. For \$x = 4\$, we have \$y = 7\$ that works, so the side lengths are \$(a, b, c)=(28, 16, 33)\$ and the minimal perimeter is \$\\boxed{77}\$.", "In \$\\triangle ABC\$ let \$\\angle B = \\beta, \\angle A = 2\\beta, \\angle C = 180^{\\circ} - 3\\beta\$. From the law of sines, we have\n\n\\[\n\\frac{a}{\\sin 2\\beta} = \\frac{b}{\\sin \\beta} = \\frac{c} {\\sin (180^{\\circ} - 3\\beta)} = \\frac{c}{\\sin 3\\beta}\n\\]\n\nThus the ratio\n\n\\[\nb : a : c = \\sin\\beta : \\sin 2\\beta : \\sin 3\\beta\n\\]\n\nWe can simplify\n\n\\[\n\\frac{\\sin 2\\beta}{\\sin\\beta} = \\frac{2\\sin\\beta\\cos\\beta}{\\sin\\beta} = 2\\cos\\beta\n\\]\n\nLikewise,\n\n\\[\n\\frac{\\sin 3\\beta}{\\sin\\beta} = \\frac{\\sin 2\\beta\\cos\\beta + \\sin\\beta\\cos 2\\beta}{\\sin\\beta} = \\frac{2\\sin\\beta\\cos^2\\beta + \\sin\\beta(\\cos^2\\beta - \\sin^2\\beta)}{\\sin\\beta}\n\\]\n\n\\[\n= {2 \\cos^2 \\beta + \\cos^2 \\beta - \\sin^2 \\beta} = 4\\cos^2 \\beta - 1\n\\]\n\nLetting \$\\gamma = \\cos\\beta\$, rewrite\n\n\\[\nb : a : c = 1 : 2\\gamma : 4\\gamma^2 - 1\n\\]\n\nWe find that to satisfy the conditions for an obtuse triangle, \$\\beta \\in (0^\\circ, 30^\\circ)\$ and therefore \$\\gamma \\in \\left(\\frac{\\sqrt{3}}{2}, 1\\right)\$.\n\nThe rational number with minimum denominator (in order to minimize scaling to obtain integer solutions) above \$\\frac{\\sqrt{3}}{2}\$ is \$\\frac{7}{8}\$, which also has a denominator divisible by 2 (to take advantage of the coefficients of 2 and 4 in the ratio and further minimize scaling).\n\nInserting \$\\gamma = \\frac{7}{8}\$ into the ratio, we find \$b : a : c = 1 : \\frac{7}{4} : \\frac{33}{16}\$. When scaled minimally to obtain integer side lengths, we find\n\n\\[\nb, a, c = 16, 28, 33\n\\]\n\nand that the perimeter is \$\\boxed{77}\$.\n\n(note by integralarefun: The part of the solution about finding \$\\gamma\$ is not rigorous and would likely require further proof in an actual test.)" ]
USAMO-1991-2	https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_2	For any nonempty set $\,S\,$ of numbers, let $\,\sigma(S)\,$ and $\,\pi(S)\,$ denote the sum and product, respectively, of the elements of $\,S\,$. Prove that \[ \sum \frac{\sigma(S)}{\pi(S)} = (n^2 + 2n) - \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \right) (n+1), \] where "$\Sigma$" denotes a sum involving all nonempty subsets $S$ of $\{1,2,3, \ldots,n\}$.	[ "Let \$N(m)\$ denote the set \$\\{1, \\dotsc, m\\}\$. Since \$\\sigma(\\varnothing)\$, the empty sum, is equal to zero, and \$\\pi(\\varnothing)\$, the empty product, is equal to 1, the equation\n\n\\[\n\\sum_{S \\subseteq N(n)} \\frac{\\sigma(S)}{\\pi(S)} = (n+1)^2 -1 - (n+1)\\sum_{k=1}^n \\frac{1}{k}\n\\]\n\nis equivalent to the desired equation when \$n >0\$. We will prove this equation, but first, we prove a lemma.\n\nLemma. For all nonnegative integers \$n\$, \$\\sum_{S \\subseteq N(n)} \\frac{1}{\\pi(S)} = n+1\$.\n\nProof. Evidently,\n\n\\[\n\\sum_{S \\subseteq N(n)} \\frac{1}{\\pi(S)} = \\sum_{k=0}^n \\sum_{S \\subseteq N(n), \|S\| =k} \\frac{1}{\\pi(S)} .\n\\]\n\nBut the terms of this sum are the coefficients of the polynomial \$P(x) = \\prod_{k=1}^n (x + 1/k)\$, and the sum of the coefficients of this polynomial is\n\n\\[\nP(1) = \\prod_{k=1}^n(1 + 1/k) = \\prod_{k=1}^n \\frac{k+1}{k} = \\frac{n+1}{1} = n+1,\n\\]\n\nas desired. \$\\blacksquare\$\n\nWe now prove our equation by induction on \$n\$. For \$n=0\$, we have the simple equation 0=0.\n\nSuppose the equation holds when \$n=a\$. Then\n\n\\[\n\\begin{align} \\sum_{S \\subseteq N(a+1)} \\frac{\\sigma(S)}{\\pi(S)} &= \\sum_{S \\subseteq N(a)} \\frac{\\sigma(S)}{\\pi(S)} + \\sum_{S \\subseteq N(a)} \\frac{\\sigma(S \\cup \\{a+1\\})}{\\pi(S \\cup \\{a+1\\})} \\\\ &= \\sum_{S \\subseteq N(a)} \\frac{\\sigma(S)}{\\pi(S)} + \\sum_{S \\subseteq N(a)} \\frac{\\sigma(S) + (a+1)}{(a+1) \\cdot \\pi(S)} \\\\ &= \\frac{a+2}{a+1} \\sum_{S\\subseteq N(a)} \\frac{\\sigma(S)}{\\pi(S)} + \\sum_{S \\subseteq N(a)} \\frac{1}{\\pi(S)} . \\end{align}\n\\]\n\nBy inductive hypothesis,\n\n\\[\n\\begin{align} \\frac{a+2}{a+1} \\sum_{S \\subseteq N(a)} \\frac{\\sigma(S)}{\\pi(S)} &= \\frac{a+2}{a+1} \\left[ (a+1)^2 -1 - (a+1) \\sum_{k=1}^a \\frac{1}{k} \\right] \\\\ &= (a+2)(a+1) - \\frac{a+2}{a+1} - (a+2) \\sum_{k=1}^a \\frac{1}{k} \\\\ &= (a+2)(a+1) - (a+2) \\sum_{k=1}^{a+1} \\frac{1}{k}, \\end{align}\n\\]\n\nand by the lemma,\n\n\\[\n\\sum_{S \\subseteq N(a)} \\frac{1}{\\pi(S)} = a+1 .\n\\]\n\nSince \$(a+2)(s+1) = (a+3)(a+1) = (a+2)^2 -1\$, our equation thus holds by induction. Thus the problem statement is proven. \$\\blacksquare\$" ]
USAMO-1991-3	https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_3	Show that, for any fixed integer $\,n \geq 1,\,$ the sequence \[ 2, \; 2^2, \; 2^{2^2}, \; 2^{2^{2^2}}, \ldots \pmod{n} \] is eventually constant. [The tower of exponents is defined by $a_1 = 2, \; a_{i+1} = 2^{a_i}$. Also $a_i \pmod{n}$ means the remainder which results from dividing $\,a_i\,$ by $\,n$.]	[ "Suppose that the problem statement is false for some integer \$n \\ge 1\$. Then there is a least \$n\$, which we call \$b\$, for which the statement is false.\n\nSince all integers are equivalent mod 1, \$b\\neq 1\$.\n\nNote that for all integers \$b\$, the sequence \$2^0, 2^1, 2^2, \\dotsc\$ eventually becomes cyclic mod \$b\$. Let \$k\$ be the period of this cycle. Since there are \$k-1\$ nonzero residues mod \$b\$. \$1 \\le k\\le b-1 < b\$. Since\n\n\\[\n2, 2^2, 2^{2^2}, 2^{2^{2^2}}, \\dotsc\n\\]\n\ndoes not become constant mod \$b\$, it follows the sequence of exponents of these terms, i.e., the sequence\n\n\\[\n1, 2, 2^2, 2^{2^{2}}, \\dotsc\n\\]\n\ndoes not become constant mod \$k\$. Then the problem statement is false for \$n=k\$. Since \$k<b\$, this is a contradiction. Therefore the problem statement is true. \$\\blacksquare\$\n\nNote that we may replace 2 with any other positive integer, and both the problem and this solution remain valid.", "We'll prove by strong induction that for every natural number \$n\$, the sequence \$a_1, a_2, \\ldots\$ is eventually constant. Since every term of the sequence is \$0 \\mathrm{\\ mod\\ } 1\$, the claim is true when \$n = 1\$. Assuming that it's true for \$1, \\ldots, n\$, we'll now show that it's true for \$n + 1\$ as well.\n\nSuppose first that \$n + 1\$ is odd. Since \$\\varphi(n + 1) < n + 1\$, by our inductive hypothesis there exists an \$m\$ such that\n\n\\[\na_m = a_{m + 1} = a_{m + 2} = \\cdots \\pmod{\\varphi(n + 1)}.\n\\]\n\nSince \$n + 1\$ is coprime to powers of \$2\$, it follows by Euler's theorem that\n\n\\[\n2^{a_m} = 2^{a_{m + 1}} = 2^{a_{m + 2}} = \\cdots \\pmod{n + 1},\n\\]\n\nor equivalently\n\n\\[\na_{m + 1} = a_{m + 2} = a_{m + 3} = \\cdots \\pmod{n + 1},\n\\]\n\nwhich is what we wanted to show.\n\nNow suppose that \$n + 1\$ is even. Write \$n + 1 = 2^{k} \\cdot s\$, where \$1 \\leq s < n + 1\$ is odd. The series must eventually be constant \$\\textrm{mod\\ } 2^k\$, since \$a_m = 0 \\textrm{\\ mod\\ } {2^k}\$ for large enough \$m\$. And by our inductive hypothesis, the series must also eventually be constant \$\\textrm{mod\\ } s\$. So for large enough \$m\$,\n\n\\[\na_m = a_{m + 1} = a_{m + 2} = \\cdots \\pmod{2^k},\n\\]\n\n\\[\na_m = a_{m + 1} = a_{m + 2} = \\cdots \\pmod{s}.\n\\]\n\nSince \$2^k\$ and \$s\$ are coprime, these equations are also true modulo \$2^k \\cdot s = n + 1\$. So\n\n\\[\na_m = a_{m + 1} = a_{m + 2} = \\cdots \\pmod{n + 1},\n\\]\n\nwhich completes the proof." ]
USAMO-1991-4	https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_4	Let $\, a =\frac{m^{m+1} + n^{n+1}}{m^m + n^n}, \,$ where $\,m\,$ and $\,n\,$ are positive integers. Prove that $\,a^m + a^n \geq m^m + n^n$. [You may wish to analyze the ratio $\,\frac{a^N - N^N}{a-N},$ for real $\, a \geq 0 \,$ and integer $\, N \ge 1$.]	[ "Let us assume without loss of generality that \$m\\ge n\$. We then note that\n\n\\[\nm-a = \\frac{m^{m+1} + m \\cdot n^n}{m^m+n^n} - \\frac{m^{m+1} - n^{n+1}}{m^m+n^n} = n^n \\frac{m-n}{m^m+n^n} \\qquad () .\n\\]\n\nSimilarly,\n\n\\[\na-n = m^m \\frac{m-n}{m^m+n^n} \\qquad () .\n\\]\n\nWe note that equations \$()\$ and \$(*)\$ imply that \$n \\le a \\le m\$. Then \$a/m \\le 1 \\le a/n\$, so\n\n\\[\n\\frac{1}{m} \\sum_{i=0}^{m-1} (a/m)^i \\le 1 \\le \\frac{1}{n} \\sum_{i=0}^{n-1} (a/n)^i .\n\\]\n\nMultiplying this inequality by \$m^m n^n(m-n)/(m^m+n^n)\$, we have\n\n\\[\nn^n \\frac{(m-n)}{m^m+n^n} \\sum_{i=0}^{m-1} a^i m^{m-1-i} \\le m^m \\frac{(m-n)}{m^m+n^n} \\sum_{i=0}^{n-1} a^i n^{n-1-i} .\n\\]\n\nIt then follows that\n\n\\[\n\\begin{align} m^m - a^m &= (m-a) \\sum_{i=0}^{m-1} a^i m^{m-1-i} = n^n \\frac{m-n}{m^m+n^n} \\sum_{i=0}^{m-1} a^i m^{m-1-i} \\\\ &\\le m^m \\frac{m-n}{m^m+n^n} \\sum_{i=0}^{n-1} a^i n^{n-1-i} = (a-n) \\sum_{i=0}^{n-1} a^i n^{n-1-i} \\\\ &= a^n - n^n . \\end{align*}\n\\]\n\nRearranging this inequality, we find that \$a^m + a^n \\ge m^m + n^n\$, as desired. \$\\blacksquare\$" ]
USAMO-1991-5	https://artofproblemsolving.com/wiki/index.php/1991_USAMO_Problems/Problem_5	Let $\, D \,$ be an arbitrary point on side $\, AB \,$ of a given triangle $\, ABC, \,$ and let $\, E \,$ be the interior point where $\, CD \,$ intersects the external common tangent to the incircles of triangles $\, ACD \,$ and $\, BCD$. As $\, D \,$ assumes all positions between $\, A \,$ and $\, B \,$, prove that the point $\, E \,$ traces the arc of a circle. \[ [asy] size(220); defaultpen(1); pair A=(0,0), B=(220,0), C=(18.7723,118.523); pair D=(72.6,0); pair Ia=incenter(A,D,C), Ib=incenter(B,D,C); pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129); pair E=IntersectionPoint((Ta--Tb),(C--D)); path Oa=circle(Ia,inradius(A,D,C)); path Ob=circle(Ib,inradius(B,D,C)); pair Da=IP(Oa,A--B), Db=IP(Ob,A--B); draw(D--C--A--B--C); draw(Ta--Tb); draw(Oa); draw(Ob); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(E,linewidth(4)); dot(Ta,linewidth(4)); dot(Tb,linewidth(4)); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,W); label("$D$",D,S); label("$E$",E,NNE); [/asy] \]	[ "Let the incircle of \$ACD\$ and the incircle of \$BCD\$ touch line \$AB\$ at points \$D_a,D_b\$, respectively; let these circles touch \$CD\$ at \$C_a\$, \$C_b\$, respectively; and let them touch their common external tangent containing \$E\$ at \$T_a,T_b\$, respectively, as shown in the diagram below.\n\n\\[\n[asy] size(220); defaultpen(1); pair A=(0,0), B=(220,0), C=(18.7723,118.523); pair D=(72.6,0); pair Ia=incenter(A,D,C), Ib=incenter(B,D,C); pair Ta=(24.9758,52.5775),Tb=(86.6196,67.4129); pair E=IntersectionPoint((Ta--Tb),(C--D)); path Oa=circle(Ia,inradius(A,D,C)); path Ob=circle(Ib,inradius(B,D,C)); pair Da=IP(Oa,A--B), Db=IP(Ob,A--B); pair Ca=IP(Oa,C--D), Cb=IP(Ob,C--D); draw(D--C--A--B--C); draw(Ta--Tb); draw(Oa); draw(Ob); dot(A,linewidth(4)); dot(B,linewidth(4)); dot(C,linewidth(4)); dot(D,linewidth(4)); dot(E,linewidth(4)); dot(Ta,linewidth(4)); dot(Tb,linewidth(4)); dot(Ca,linewidth(4)); dot(Cb,linewidth(4)); dot(Da,linewidth(4)); dot(Db,linewidth(4)); label(\"\$A\$\",A,SW); label(\"\$B\$\",B,SE); label(\"\$C\$\",C,W); label(\"\$D\$\",D,S); label(\"\$E\$\",E,NNE); label(\"\$T_a\$\",Ta,N); label(\"\$T_b\$\",Tb,WNW); label(\"\$D_a\$\",Da,S); label(\"\$D_b\$\",Db,S); label(\"\$C_a\$\",Ca,WSW); label(\"\$C_b\$\",Cb,ENE); [/asy]\n\\]\n\nWe note that\n\n\\[\nCE = CC_a - EC_a = CC_b - EC_b = \\frac{CC_a + CC_b - (EC_a + EC_b)}{2} .\n\\]\n\nOn the other hand, since \$EC_a\$ and \$ET_a\$ are tangents from the same point to a common circle, \$EC_a = T_aE\$, and similarly \$EC_b = ET_b\$, so\n\n\\[\nEC_a + EC_b = T_aE + ET_b = T_a T_b .\n\\]\n\nOn the other hand, the segments \$T_a T_b\$ and \$D_a D_b\$ evidently have the same length, and \$D_a D_b = D_aD + DD_b\$, so \$EC_a + EC_b = D_aD + DD_b\$. Thus\n\n\\[\nCE = \\frac{CC_a + CC_b - (EC_a + EC_b)}{2} = \\frac{CC_a + CC_b - D_aD - DD_b}{2} .\n\\]\n\nIf we let \$s_a\$ be the semiperimeter of triangle \$ACD\$, then \$CC_a = s_a - AD\$, and \$D_aD = s_a - AC\$, so\n\n\\[\nCC_a - D_aD = (s_a - AD) - (s_a - AC) = AC - AD .\n\\]\n\nSimilarly,\n\n\\[\nCC_b - DD_b = BC - DB,\n\\]\n\nso that\n\n\\[\n\\begin{align} CE &= \\frac{CC_a + CC_b - D_aD - DD_b}{2} = \\frac{AC + BC - (AD+DB)}{2} \\\\ &= \\frac{AC + BC - AB}{2} . \\end{align}\n\\]\n\nThus \$E\$ lies on the arc of the circle with center \$C\$ and radius \$(AB+BC-AB)/2\$ intercepted by segments \$CA\$ and \$CB\$. If we choose an arbitrary point \$X\$ on this arc and let \$D\$ be the intersection of lines \$CX\$ and \$AB\$, then \$X\$ becomes point \$E\$ in the diagram, so every point on this arc is in the locus of \$E\$. \$\\blacksquare\$", "Define the same points as in the first solution. First extend \$T_aT_b\$ to intersect \$AB\$ at a point \$P\$; without loss of generality let \$A\$ lie in between \$B\$ and \$P\$. Then the incircle of \$\\triangle ACD\$ is also the incircle of \$\\triangle PED\$, while the incircle of \$\\triangle BCD\$ is the \$P\$-excircle of \$\\triangle PED\$. It follows that \$EC_a = C_bD\$; denote this equality by \$()\$.\n\nNow remark that\n\n\\[\nCC_a + CC_b = \\frac{AC+AD-CD}2 + \\frac{BC+DC-BD}2 = \\frac{AC+BC-AB}2 + CD.\n\\]\n\nHence\n\n\\[\nCC_a + CC_b - CD = CC_a - C_bD \\stackrel{()}= CC_a - C_aE = CE\n\\]\n\nis a constant equal to \$r := \\tfrac{AC+BC-AB}2\$, and so \$C\$ lies on the circle with center \$C\$ and radius \$r\$." ]

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