id stringlengths 12 12 | source stringlengths 76 76 | problem stringlengths 59 2.04k | solutions listlengths 0 11 |
|---|---|---|---|
USAMO-1992-1 | https://artofproblemsolving.com/wiki/index.php/1992_USAMO_Problems/Problem_1 | Find, as a function of \(\, n, \,\) the sum of the digits of
\[
9 \times 99 \times 9999 \times \cdots \times \left( 10^{2^n} - 1 \right),
\]
where each factor has twice as many digits as the previous one. | [
"The answer is \\(9 \\cdot 2^n\\).\n\nLet us denote the quantity \\(\\prod_{k=0}^n \\bigl( 10^{2^k}-1 \\bigr)\\) as \\(P_n\\). We wish to find the sum of the digits of \\(P_n\\).\n\nWe first note that\n\n\\[\nP_{n-1} < \\prod_{k=0}^{n-1} 10^{2^k} = 10^{2^n-1},\n\\]\n\nso \\(P_{n-1}\\) is a number of at most \\(2^n\... |
USAMO-1992-2 | https://artofproblemsolving.com/wiki/index.php/1992_USAMO_Problems/Problem_2 | Prove
\[
\frac{1}{\cos 0^\circ \cos 1^\circ} + \frac{1}{\cos 1^\circ \cos 2^\circ} + \cdots + \frac{1}{\cos 88^\circ \cos 89^\circ} = \frac{\cos 1^\circ}{\sin^2 1^\circ}.
\] | [
"Consider the points \\(M_k = (1, \\tan k^\\circ)\\) in the coordinate plane with origin \\(O=(0,0)\\), for integers \\(0 \\le k \\le 89\\).\n\n\\[\n[asy] size(200); defaultpen(1); pair O=(0,0), a=expi(0), b=expi(1/6), c=expi(2/6), d=expi(3/6), y=expi(32/30), z= expi(34/30); pair A=a, B=b/b.x, C= c/c.x, D=d/d.x, Y=... |
USAMO-1992-3 | https://artofproblemsolving.com/wiki/index.php/1992_USAMO_Problems/Problem_3 | For a nonempty set \(S\) of integers, let \(\sigma(S)\) be the sum of the elements of \(S\). Suppose that \(A = \{a_1, a_2, \ldots, a_{11}\}\) is a set of positive integers with \(a_1 < a_2 < \cdots < a_{11}\) and that, for each positive integer \(n \le 1500\), there is a subset \(S\) of \(A\) for which \(\sigma(S) = n... | [
"Let's a \\(n\\)-\\(p\\) set be a set \\(Z\\) such that \\(Z=\\{a_1,a_2,\\cdots,a_n\\}\\), where \\(\\forall i<n\\), \\(i\\in \\mathbb{Z}^+\\), \\(a_i<a_{i+1}\\), and for each \\(x\\le p\\), \\(x\\in \\mathbb{Z}^+\\), \\(\\exists Y\\subseteq Z\\), \\(\\sigma(Y)=x\\), \\(\\nexists \\sigma(Y)=p+1\\).\n\n(For Example ... |
USAMO-1992-4 | https://artofproblemsolving.com/wiki/index.php/1992_USAMO_Problems/Problem_4 | Chords \(AA'\), \(BB'\), and \(CC'\) of a sphere meet at an interior point \(P\) but are not contained in the same plane. The sphere through \(A\), \(B\), \(C\), and \(P\) is tangent to the sphere through \(A'\), \(B'\), \(C'\), and \(P\). Prove that \(AA'=BB'=CC'\). | [
"Consider the plane through \\(A,A',B,B'\\). This plane, of course, also contains \\(P\\). We can easily find the \\(\\triangle APB\\) is isosceles because the base angles are equal. Thus, \\(AP=BP\\). Similarly, \\(A'P=B'P\\). Thus, \\(AA'=BB'\\). By symmetry, \\(BB'=CC'\\) and \\(CC'=AA'\\), and hence \\(AA'=BB'=... |
USAMO-1992-5 | https://artofproblemsolving.com/wiki/index.php/1992_USAMO_Problems/Problem_5 | Let \(\, P(z) \,\) be a polynomial with complex coefficients which is of degree \(\, 1992 \,\) and has distinct zeros. Prove that there exist complex numbers \(\, a_1, a_2, \ldots, a_{1992} \,\) such that \(\, P(z) \,\) divides the polynomial \(\left( \cdots \left( (z-a_1)^2 - a_2 \right)^2 \cdots - a_{1991} \right)^2 ... | [
"Since the zeros \\(z_{1}, z_{2}, \\ldots, z_{1992}\\) of the polynomial \\(P(z)\\) are distinct, the polynomial \\(P(z)\\) divides the polynomial\n\n\\[\nQ\\left(z\\right) = \\left( \\ldots \\left( \\left(z-a_{1}\\right)^{2}-a_{2}\\right)^{2}\\ldots-a_{1991}\\right)^{2}-a_{1992}\n\\]\n\nif and only if \\(Q\\left(z... |
USAMO-1993-1 | https://artofproblemsolving.com/wiki/index.php/1993_USAMO_Problems/Problem_1 | For each integer \(n\ge 2\), determine, with proof, which of the two positive real numbers \(a\) and \(b\) satisfying
\[
a^n=a+1,\qquad b^{2n}=b+3a
\]
is larger. | [
"Square and rearrange the first equation and also rearrange the second.\n\n\\[\n\\begin{align} a^{2n}-a&=a^2+a+1\\\\ b^{2n}-b&=3a \\end{align}\n\\]\n\nIt is trivial that\n\n\\[\n\\begin{align*} (a-1)^2 > 0 \\tag{3} \\end{align*}\n\\]\n\nsince \\(a-1\\) clearly cannot equal \\(0\\) (Otherwise \\(a^n=1\\neq 1+1\\)). ... |
USAMO-1993-2 | https://artofproblemsolving.com/wiki/index.php/1993_USAMO_Problems/Problem_2 | Let \(ABCD\) be a convex quadrilateral such that diagonals \(AC\) and \(BD\) intersect at right angles, and let \(E\) be their intersection. Prove that the reflections of \(E\) across \(AB\), \(BC\), \(CD\), \(DA\) are concyclic. | [
"## Diagram\n\n\\[\n[asy] import olympiad; defaultpen(0.8pt+fontsize(12pt)); pair E; E=(0,0); label('$E$',E,N); pair A,B,C,D; A=(9,0); B=(0,13); C=(-13,0); D=(0,-11); draw(A--B--C--D--cycle,blue); label('$A$',A,E); label('$B$',B,N); label('$C$',C,W); label('$D$',D,S); pair T,R,S,Q; T=reflect(A, B)*E; R=reflect(C, B... |
USAMO-1993-3 | https://artofproblemsolving.com/wiki/index.php/1993_USAMO_Problems/Problem_3 | Consider functions \(f : [0, 1] \rightarrow \mathbb{R}\) which satisfy
Find, with proof, the smallest constant \(c\) such that
\[
f(x) \le cx
\]
for every function \(f\) satisfying (i)-(iii) and every \(x\) in \([0, 1]\). | [
"My claim: \\(c\\ge2\\)\n\nLemma 1) \\(f\\left(\\left(\\frac{1}{2}\\right)^n\\right)\\le\\left(\\frac{1}{2}\\right)^n\\) for \\(n\\in \\mathbb{Z}, n\\ge0\\)\n\nFor \\(n=0\\), \\(f(1)=1\\) (ii)\n\nAssume that it is true for \\(n-1\\), then \\(f\\left(\\left(\\frac{1}{2}\\right)^{n}\\right)+f\\left(\\left(\\frac{1}{2... |
USAMO-1993-4 | https://artofproblemsolving.com/wiki/index.php/1993_USAMO_Problems/Problem_4 | Let \(a\), \(b\) be odd positive integers. Define the sequence \((f_n)\) by putting \(f_1 = a\), \(f_2 = b\), and by letting \(f_n\) for \(n\ge3\) be the greatest odd divisor of \(f_{n-1} + f_{n-2}\). Show that \(f_n\) is constant for \(n\) sufficiently large and determine the eventual value as a function of \(a\) and ... | [
"Part 1) Prove that \\(f_n\\) is constant for sufficiently large \\(n\\).\n\nNote that if there is some \\(f_n=f_{n-1}\\) for any \\(n\\), then \\(\\frac{f_{n}+f_{n-1}}{2}=f_n\\), which is odd. Thus, \\(f_{n+1}=f_n=f_{n-1}\\) and by induction, all \\(f_p\\) is constant for \\(p\\ge n\\).\n\nAlso note that \\(f_n>0\... |
USAMO-1993-5 | https://artofproblemsolving.com/wiki/index.php/1993_USAMO_Problems/Problem_5 | Let \(a_0, a_1, a_2,\cdots\) be a sequence of positive real numbers satisfying \(a_{i-1}a_{i+1}\le a^2_i\) for \(i = 1, 2, 3,\cdots\) . (Such a sequence is said to be log concave.) Show that for each \(n > 1\),
\[
\frac{a_0+\cdots+a_n}{n+1}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\ge\frac{a_0+\cdots+a_{n-1}}{n}\cdot\frac{a_... | [
"Notice that because\n\n\\[\n(a_0 + a_1 + \\dots + a_{n-1})(a_1 + a_2 + \\dots + a_n) = (a_0 + (a_1 + \\dots + a_{n-1})(-a_0 + (a_0 + a_1 + \\dots + a_n))\n\\]\n\n\\[\n= -a_0^2 + a_0 ((a_0 + a_1 + \\dots + a_n) - (a_1 + a_2 + \\dots + a_{n-1})) + (a_1 + a_2 + \\dots + a_{n-1})(a_0 + a_1 + \\dots + a_n)\n\\]\n\n\\[\... |
USAMO-1994-1 | https://artofproblemsolving.com/wiki/index.php/1994_USAMO_Problems/Problem_1 | Let \(\, k_1 < k_2 < k_3 <\cdots\,\), be positive integers, no two consecutive, and let \(\, s_m = k_1+k_2+\cdots+k_m\,\), for \(\, m = 1,2,3,\ldots\;\;\). Prove that, for each positive integer \(n\), the interval \(\, [s_n, s_{n+1})\,\), contains at least one perfect square. | [
"We want to show that the distance between \\(s_n\\) and \\(s_{n+1}\\) is greater than the distance between \\(s_n\\) and the next perfect square following \\(s_n\\).\n\nGiven \\(s_n=\\sum_{i=1}^{n}k_i\\), where no \\(k_i\\) are consecutive, we can put a lower bound on \\(k_n\\). This occurs when all \\(k_{i+1}=k_i... |
USAMO-1994-2 | https://artofproblemsolving.com/wiki/index.php/1994_USAMO_Problems/Problem_2 | The sides of a \(99\)-gon are initially colored so that consecutive sides are red, blue, red, blue,..., red, blue, yellow. We make a sequence of modifications in the coloring, changing the color of one side at a time to one of the three given colors (red, blue, yellow), under the constraint that no two adjacent sides m... | [
"We proceed by representing the colors as numbers, i.e. Red = 0, Blue = 1, Yellow = 2. Thus, we start with some sequence 0101...012 and are trying to end up with the sequence 1010...0102. Generate a second sequence of terms by subtracting each term by the following term and taking it modulus 3, i.e. (1-0, 0-1, 1-0,... |
USAMO-1994-3 | https://artofproblemsolving.com/wiki/index.php/1994_USAMO_Problems/Problem_3 | A convex hexagon \(ABCDEF\) is inscribed in a circle such that \(AB=CD=EF\) and diagonals \(AD,BE\), and \(CF\) are concurrent. Let \(P\) be the intersection of \(AD\) and \(CE\). Prove that \(\frac{CP}{PE}=(\frac{AC}{CE})^2\). | [
"Let the diagonals \\(AD\\), \\(BE\\), \\(CF\\) meet at \\(Q\\).\n\nFirst, let's show that the triangles \\(\\triangle AEC\\) and \\(\\triangle QED\\) are similar.\n\n\\[\n[asy] pair A,B,C,D,E,F,P,Q; A=(-0.96,0.28); B=(-0.352,0.936); C=(0,1); D=(4/5,3/5); E=(4/5,-3/5); F=(0,-1); P=IntersectionPoint(A--D,C--E); Q=I... |
USAMO-1994-4 | https://artofproblemsolving.com/wiki/index.php/1994_USAMO_Problems/Problem_4 | Let \(\, a_1, a_2, a_3, \ldots \,\) be a sequence of positive real numbers satisfying \(\, \sum_{j = 1}^n a_j \geq \sqrt {n} \,\) for all \(\, n \geq 1\). Prove that, for all \(\, n \geq 1, \,\)
\[
\sum_{j = 1}^n a_j^2 > \frac {1}{4} \left( 1 + \frac {1}{2} + \cdots + \frac {1}{n} \right).
\] | [
"Note that if we try to minimize \\((a_j)^2\\), we would try to make the \\(a_j\\) as equal as possible. However, by the condition given in the problem, this isn't possible, the \\(a_j\\)'s have to be an increasing sequence. Thinking of minimizing sequences, we realize that the optimal equation is \\(a_n = \\sqrt{n... |
USAMO-1994-5 | https://artofproblemsolving.com/wiki/index.php/1994_USAMO_Problems/Problem_5 | Let \(\, |U|, \, \sigma(U) \,\) and \(\, \pi(U) \,\) denote the number of elements, the sum, and the product, respectively, of a finite set \(\, U \,\) of positive integers. (If \(\, U \,\) is the empty set, \(\, |U| = 0, \, \sigma(U) = 0, \, \pi(U) = 1\).) Let \(\, S \,\) be a finite set of positive integers. As usual... | [
"Let, for \\(|S|\\geq k \\geq 1\\), \\(A_k\\) be a collection of sets such that \\(\\left|\\bigcap_{k \\in U}A_k\\right| = \\binom{m- \\sigma(U)}{|S|}\\). We'll try to find such a collection to prove \\(\\left|\\bigcup_{k \\in U}A_k\\right| = \\binom{m}{|S|}-\\pi(S)\\), since we can use PIE to show that \\(\\left|\... |
USAMO-1995-1 | https://artofproblemsolving.com/wiki/index.php/1995_USAMO_Problems/Problem_1 | Let \(\, p \,\) be an odd prime. The sequence \((a_n)_{n \geq 0}\) is defined as follows: \(\, a_0 = 0,\) \(a_1 = 1, \, \ldots, \, a_{p-2} = p-2 \,\) and, for all \(\, n \geq p-1, \,\) \(\, a_n \,\) is the least positive integer that does not form an arithmetic sequence of length \(\, p \,\) with any of the preceding t... | [
"I have to make the assumption that \\(a_{n+1}>a_{n}\\) (without this assumption, I can have the sequence\n\n\\({1,\\cdots,p-2,1,1,\\cdots,1,2,1,\\cdots}\\))\n\nAll of the following work are in base \\(p\\) otherwise stated\n\nLemma 1: for an arithmetic sequence of length \\(\\, p \\,\\) to exist, there must be a n... |
USAMO-1995-2 | https://artofproblemsolving.com/wiki/index.php/1995_USAMO_Problems/Problem_2 | A calculator is broken so that the only keys that still work are the \(\, \sin, \; \cos,\) \(\tan, \; \sin^{-1}, \; \cos^{-1}, \,\) and \(\, \tan^{-1} \,\) buttons. The display initially shows 0. Given any positive rational number \(\, q, \,\) show that pressing some finite sequence of buttons will yield \(\, q\). Assu... | [
"We will prove the following, stronger statement : If \\(m\\) and \\(n\\) are relatively prime nonnegative integers such that \\(n>0\\), then the some finite sequence of buttons will yield \\(\\sqrt{m/n}\\).\n\nTo prove this statement, we induct strongly on \\(m+n\\). For our base case, \\(m+n=1\\), we have \\(n=1\... |
USAMO-1995-3 | https://artofproblemsolving.com/wiki/index.php/1995_USAMO_Problems/Problem_3 | Given a nonisosceles, nonright triangle \(ABC,\) let \(O\) denote its circumcenter, and let \(A_1, \, B_1,\) and \(C_1\) be the midpoints of sides \(\overline{BC}\), \(\overline{CA}\)and \(\overline{AB}\) respectively. Point \(A_2\) is located on the ray \(OA_1\) so that \(\triangle OAA_1\) is similar to \(\triangle OA... | [
"LEMMA 1: In \\(\\triangle ABC\\) with circumcenter \\(O\\), \\(\\angle OAC = 90 - \\angle B\\).\n\nPROOF of Lemma 1: The arc \\(AC\\) equals \\(2\\angle B\\) which equals \\(\\angle AOC\\). Since \\(\\triangle AOC\\) is isosceles we have that \\(\\angle OAC = \\angle OCA = 90 - \\angle B\\). QED\n\nDefine \\(H \\i... |
USAMO-1995-4 | https://artofproblemsolving.com/wiki/index.php/1995_USAMO_Problems/Problem_4 | Suppose \(q_1,q_2,...\) is an infinite sequence of integers satisfying the following two conditions:
(a) \(m - n\) divides \(q_m - q_n\) for \(m>n \geq 0\)
(b) There is a polynomial \(P\) such that \(|q_n|<P(n)\) for all \(n\).
Prove that there is a polynomial \(Q\) such that \(q_n = Q(n)\) for each \(n\). | [
"Step 1: Suppose \\(P\\) has degree \\(d\\). Let \\(Q\\) be the polynomial of degree at most \\(d\\) with \\(Q(x)=q_x\\) for \\(0\\leq x\\leq d\\). Since the \\(q_x\\) are all integers, \\(Q\\) has rational coefficients, and there exists \\(k\\) so that \\(kQ\\) has integer coefficients. Then \\(m-n|kQ(m)-kQ(n)\\) ... |
USAMO-1995-5 | https://artofproblemsolving.com/wiki/index.php/1995_USAMO_Problems/Problem_5 | Suppose that in a certain society, each pair of persons can be classified as either amicable or hostile. We shall say that each member of an amicable pair is a friend of the other, and each member of a hostile pair is a foe of the other. Suppose that the society has \(\, n \,\) persons and \(\, q \,\) amicable pairs, a... | [
"Consider the graph with two people joined if they are friends.\n\nFor each person \\(X\\), let \\(A(X)\\) be the set of its friends and \\(B(X)\\) the set of its foes. Note that any edge goes either: from \\(X\\) to \\(A(X)\\) (type 1), from \\(A(X)\\) to \\(B(X)\\) (type 2) or from a point of \\(B(X)\\) to anothe... |
USAMO-1996-1 | https://artofproblemsolving.com/wiki/index.php/1996_USAMO_Problems/Problem_1 | Prove that the average of the numbers \(n\sin n^{\circ}\; (n = 2,4,6,\ldots,180)\) is \(\cot 1^\circ\). | [
"First, as \\(180\\sin{180^\\circ}=0,\\) we omit that term. Now, we multiply by \\(\\sin 1^\\circ\\) to get, after using product to sum, \\((\\cos 1^\\circ-\\cos 3^\\circ)+2(\\cos 3^\\circ-\\cos 5^\\circ)+\\cdots +89(\\cos 177^\\circ-\\cos 179^\\circ)\\). This simplifies to \\(\\cos 1^\\circ+\\cos 3^\\circ +\\cos 5... |
USAMO-1996-2 | https://artofproblemsolving.com/wiki/index.php/1996_USAMO_Problems/Problem_2 | For any nonempty set \(S\) of real numbers, let \(\sigma(S)\) denote the sum of the elements of \(S\). Given a set \(A\) of \(n\) positive integers, consider the collection of all distinct sums \(\sigma(S)\) as \(S\) ranges over the nonempty subsets of \(A\). Prove that this collection of sums can be partitioned into \... | [
"Let set \\(A\\) consist of the integers \\(a_1\\le a_2\\le a_3\\le\\dots\\le a_n\\). For \\(k\\ge 2\\), call \\(a_k\\) greedy if \\(\\sum_{i=1}^{k-1}a_i < a_k\\). Also call \\(a_1\\) greedy. Now put all elements of \\(A\\) into groups of consecutive terms in such a way that each group \\(G\\) begins with a greedy ... |
USAMO-1996-3 | https://artofproblemsolving.com/wiki/index.php/1996_USAMO_Problems/Problem_3 | Let \(ABC\) be a triangle. Prove that there is a line \(l\) (in the plane of triangle \(ABC\)) such that the intersection of the interior of triangle \(ABC\) and the interior of its reflection \(A'B'C'\) in \(l\) has area more than \(\frac{2}{3}\) the area of triangle \(ABC\). | [
"Let the triangle be \\(ABC\\). Assume \\(A\\) is the largest angle. Let \\(AD\\) be the altitude. Assume \\(AB \\le AC\\), so that \\(BD \\le BC/2\\). If \\(BD > \\frac{BC}{3}\\), then reflect in \\(AD\\). If \\(B'\\) is the reflection of \\(B\\), then \\(B'D = BD\\) and the intersection of the two triangles is ju... |
USAMO-1996-4 | https://artofproblemsolving.com/wiki/index.php/1996_USAMO_Problems/Problem_4 | An \(n\)-term sequence \((x_1, x_2, \ldots, x_n)\) in which each term is either 0 or 1 is called a binary sequence of length \(n\). Let \(a_n\) be the number of binary sequences of length n containing no three consecutive terms equal to 0, 1, 0 in that order. Let \(b_n\) be the number of binary sequences of length \(n\... | [
"Given any binary sequence \\(B=(b_1,b_2,b_3,\\dots,b_k)\\), define \\(f(B)=(|b_2-b_1|,|b_3-b_2|,\\dots,|b_k-b_{k-1}|)\\). The operator \\(f\\) basically takes pairs of consecutive terms and returns 0 if the terms are the same and 1 otherwise. Note that for every sequence \\(S\\) of length \\(n\\) there exist exact... |
USAMO-1996-5 | https://artofproblemsolving.com/wiki/index.php/1996_USAMO_Problems/Problem_5 | Let \(ABC\) be a triangle, and \(M\) an interior point such that \(\angle MAB=10^\circ\), \(\angle MBA=20^\circ\) , \(\angle MAC= 40^\circ\) and \(\angle MCA=30^\circ\). Prove that the triangle is isosceles. | [
"Clearly, \\(\\angle AMB = 150^\\circ\\) and \\(\\angle AMC = 110^\\circ\\). Now by the Law of Sines on triangles \\(ABM\\) and \\(ACM\\), we have\n\n\\[\n\\frac{AB}{\\sin 150^\\circ} = \\frac{AM}{\\sin 20^\\circ}\n\\]\n\nand\n\n\\[\n\\frac{AC}{\\sin 110^\\circ} = \\frac{AM}{\\sin 30^\\circ}.\n\\]\n\nCombining thes... |
USAMO-1996-6 | https://artofproblemsolving.com/wiki/index.php/1996_USAMO_Problems/Problem_6 | Determine (with proof) whether there is a subset \(X\) of the integers with the following property: for any integer \(n\) there is exactly one solution of \(a + 2b = n\) with \(a,b \in X\). | [
"Start with an incomplete subset \\(S = (S_1, S_2, S_3, ... S_m)\\), such that for any integer n, there is exactly zero or one solutions to \\(a + 2b = n\\) with \\(a,b \\in S\\). Let \\(N\\) be the smallest integer such that for any \\(S_i\\), \\(|S_i| < N\\). Note that \\(|S_i+2S_j| < 3N\\) for any \\(S_i\\) and ... |
USAMO-1997-1 | https://artofproblemsolving.com/wiki/index.php/1997_USAMO_Problems/Problem_1 | Let \(p_1,p_2,p_3,...\) be the prime numbers listed in increasing order, and let \(x_0\) be a real number between \(0\) and \(1\). For positive integer \(k\), define
\[
x_{k}=\begin{cases}0&\text{ if }x_{k-1}=0\\ \left\{\frac{p_{k}}{x_{k-1}}\right\}&\text{ if }x_{k-1}\ne0\end{cases}
\]
where \(\{x\}\) denotes the fra... | [
"All rational numbers between 0 and 1 inclusive will eventually yield some \\(x_k = 0\\). To begin, note that by definition, all rational numbers can be written as a quotient of coprime integers. Let \\(x_0 = \\frac{m}{n}\\), where \\(m,n\\) are coprime positive integers. Since \\(0<x_0<1\\), \\(0<m<n\\). Now\n\n\\... |
USAMO-1997-2 | https://artofproblemsolving.com/wiki/index.php/1997_USAMO_Problems/Problem_2 | \(\triangle ABC\) is a triangle. Take points \(D, E, F\) on the perpendicular bisectors of \(BC, CA, AB\) respectively. Show that the lines through \(A, B, C\) perpendicular to \(EF, FD, DE\) respectively are concurrent. | [
"Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By Carnot's Theorem, The three lines are concurrent if\n\n\\[\nFA'^2-EA'^2+EC'^2-DC'^2+DB'^2-FB'^2 = AF^2-AE^2+CE^2-CD^2+BD^2-BF^2 = 0\n\\]\n\nBut this is clearly true, since D lies on the perpendicular bisector of BC, BD = DC.\n\nQED",
"We... |
USAMO-1997-3 | https://artofproblemsolving.com/wiki/index.php/1997_USAMO_Problems/Problem_3 | Prove that for any integer \(n\), there exists a unique polynomial \(Q(x)\) with coefficients in \(\{0,1,...,9\}\) such that \(Q(-2)=Q(-5)=n\). | [
"The problem implies that \\(Q(x) = (x+2)(x+5) \\cdot H_y(x) + n\\). We define \\(H_y(x) = a_0 + a_1x + ... + a_nx^n\\). Then, after expanding, the coefficient of \\(x^k\\) is \\(10a_k + 7a_{k-1} + a_{k-2}\\) where \\(2 \\le k \\le n\\). The constant term is \\(10a_0 + n\\). Since this has to be within the set \\({... |
USAMO-1997-4 | https://artofproblemsolving.com/wiki/index.php/1997_USAMO_Problems/Problem_4 | To clip a convex \(n\)-gon means to choose a pair of consecutive sides \(AB, BC\) and to replace them by three segments \(AM, MN,\) and \(NC,\) where \(M\) is the midpoint of \(AB\) and \(N\) is the midpoint of \(BC\). In other words, one cuts off the triangle \(MBN\) to obtain a convex \((n+1)\)-gon. A regular hexagon... | [
"\\[\n[asy] size(200); draw((1, 0)--(0.5, 0.866)--(-0.5, 0.866)--(-1, 0)--(-0.5, -0.866)--(0.5, -0.866)--(1, 0)); draw((1, 0)--(-0.5, 0.866)--(-0.5, -0.866)--(1, 0), blue); draw((-1, 0)--(0.5, -0.866)--(0.5, 0.866)--(-1, 0), blue); [/asy]\n\\]\n\n\\(\\textbf{Claim:}\\) It is impossible to choose two non-adjacent s... |
USAMO-1997-5 | https://artofproblemsolving.com/wiki/index.php/1997_USAMO_Problems/Problem_5 | Prove that, for all positive real numbers \(a, b, c,\)
\[
\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}
\]
. | [
"Because the inequality is homogenous (i.e. \\((a, b, c)\\) can be replaced with \\((ka, kb, kc)\\) without changing the inequality other than by a factor of \\(k^n\\) for some \\(n\\)), without loss of generality, let \\(abc = 1\\).\n\nLemma:\n\n\\[\n\\frac{1}{a^3 + b^3 + 1} \\le \\frac{c}{a + b + c}.\n\\]\n\nProo... |
USAMO-1997-6 | https://artofproblemsolving.com/wiki/index.php/1997_USAMO_Problems/Problem_6 | Suppose the sequence of nonnegative integers \(a_1,a_2,...,a_{1997}\) satisfies
\[
a_i+a_j \le a_{i+j} \le a_i+a_j+1
\]
for all \(i, j \ge 1\) with \(i+j \le 1997\). Show that there exists a real number \(x\) such that \(a_n=\lfloor{nx}\rfloor\) (the greatest integer \(\le nx\)) for all \(1 \le n \le 1997\). | [
"Given nonnegative integers \\(a_1,a_2,\\dots, a_{1997}\\) satisfying the given inequalities, let \\(I_n\\) be the set of all \\(x\\) such that \\(a_n=\\lfloor nx\\rfloor\\). Therefore,\n\n\\[\nI_n=\\{x\\,:\\,a_n\\le nx<a_n+1\\}.\n\\]\n\nWe can rewrite this as\n\n\\[\nI_n=\\left[\\frac{a_n}{n},\\frac{a_n+1}{n}\\rig... |
USAMO-1998-1 | https://artofproblemsolving.com/wiki/index.php/1998_USAMO_Problems/Problem_1 | Suppose that the set \(\{1,2,\cdots, 1998\}\) has been partitioned into disjoint pairs \(\{a_i,b_i\}\) (\(1\leq i\leq 999\)) so that for all \(i\), \(|a_i-b_i|\) equals \(1\) or \(6\). Prove that the sum
\[
|a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}|
\]
ends in the digit \(9\). | [
"Notice that \\(|a_i - b_i| \\equiv 1 \\pmod 5\\), so \\(S=|a_1-b_1|+|a_2-b_2|+\\cdots +|a_{999}-b_{999}| \\equiv 1+1+\\cdots + 1 \\equiv 999 \\equiv 4 \\bmod{5}\\).\n\nAlso, for integers \\(M, N\\) we have \\(|M-N| \\equiv M-N \\equiv M+N \\bmod{2}\\).\n\nThus, we also have \\(S \\equiv a_1+b_1+a_2+b_2+\\cdots +a_... |
USAMO-1998-2 | https://artofproblemsolving.com/wiki/index.php/1998_USAMO_Problems/Problem_2 | Let \({\cal C}_1\) and \({\cal C}_2\) be concentric circles, with \({\cal C}_2\) in the interior of \({\cal C}_1\). From a point \(A\) on \({\cal C}_1\) one draws the tangent \(AB\) to \({\cal C}_2\) (\(B\in {\cal C}_2\)). Let \(C\) be the second point of intersection of \(AB\) and \({\cal C}_1\), and let \(D\) be the ... | [
"\\[\n[asy] pair O,A,B,C,D,E,F,DEb,CFb,Fo,M; O=(0,0); A=(1.732,1); B=(0,1); C=(-1.732,1); D=(0.866,1); Fo=(-1,-0.5); path AC,AF,DE,CF,DEbM,CFbM,C1,C2; C1=circle(O,2); C2=circle(O,1); E=intersectionpoints(A--Fo,C2)[0]; F=intersectionpoints(A--Fo,C2)[1]; DEb=((D.x+E.x)/2.0,(D.y+E.y)/2.0); CFb=((C.x+F.x)/2.0,(C.y+F... |
USAMO-1998-3 | https://artofproblemsolving.com/wiki/index.php/1998_USAMO_Problems/Problem_3 | Let \(a_0,\cdots a_n\) be real numbers in the interval \(\left(0,\frac {\pi}{2}\right)\) such that
\[
\tan{\left(a_0 - \frac {\pi}{4}\right)} + \tan{\left(a_1 - \frac {\pi}{4}\right)} + \cdots + \tan{\left(a_n - \frac {\pi}{4}\right)}\ge n - 1
\]
Prove that \(\tan{\left(a_0\right)}\tan{\left(a_1\right)}\cdots \tan{\l... | [
"Let \\(y_i = \\tan{(a_i - \\frac {\\pi}{4})}\\), where \\(0\\le i\\le n\\). Then we have\n\n- \\(y_0 + y_1 + \\cdots + y_n\\ge n - 1\\)\n- \\(1 + y_i\\ge \\sum_{j\\neq i}{(1 - y_j)}\\)\n- \\(\\frac {1 + y_i}{n}\\ge \\frac {1}{n}\\sum_{j\\neq i}{(1 - y_j)}\\)\n\nBy AM-GM,\n\n\\[\n\\begin{align*} \\frac {1}{n}\\sum_... |
USAMO-1998-4 | https://artofproblemsolving.com/wiki/index.php/1998_USAMO_Problems/Problem_4 | A computer screen shows a \(98 \times 98\) chessboard, colored in the usual way. One can select with a mouse any rectangle with sides on the lines of the chessboard and click the mouse button: as a result, the colors in the selected rectangle switch (black becomes white, white becomes black). Find, with proof, the mini... | [
"Answer: \\(98\\).\n\nThere are \\(4\\cdot97\\) adjacent pairs of squares in the border and each pair has one black and one white square. Each move can fix at most \\(4\\) pairs, so we need at least \\(97\\) moves. However, we start with two corners one color and two another, so at least one rectangle must include ... |
USAMO-1998-5 | https://artofproblemsolving.com/wiki/index.php/1998_USAMO_Problems/Problem_5 | Prove that for each \(n\geq 2\), there is a set \(S\) of \(n\) integers such that \((a-b)^2\) divides \(ab\) for every distinct \(a,b\in S\). | [
"Proof by induction. For n=2, the proof is trivial, since \\(S = (1,2)\\) satisfies the condition. Assume now that there is such a set S of n elements, \\(a_1, a_2,...a_n\\) which satisfy the condition. The key is to note that if \\(m=a_1a_2...a_n\\), then if we define \\(b_i=a_i + km\\) for all \\(i\\le n\\), wher... |
USAMO-1998-6 | https://artofproblemsolving.com/wiki/index.php/1998_USAMO_Problems/Problem_6 | Let \(n \geq 5\) be an integer. Find the largest integer \(k\) (as a function of \(n\)) such that there exists a convex \(n\)-gon \(A_{1}A_{2}\dots A_{n}\) for which exactly \(k\) of the quadrilaterals \(A_{i}A_{i+1}A_{i+2}A_{i+3}\) have an inscribed circle. (Here \(A_{n+j} = A_{j}\).) | [
"Lemma: If quadrilaterals \\(A_iA_{i+1}A_{i+2}A_{i+3}\\) and \\(A_{i+2}A_{i+3}A_{i+4}A_{i+5}\\) in an equiangular \\(n\\)-gon are tangential, and \\(A_iA_{i+3}\\) is the longest side quadrilateral \\(A_iA_{i+1}A_{i+2}A_{i+3}\\) for all \\(i\\), then quadrilateral \\(A_{i+1}A_{i+2}A_{i+3}A_{i+4}\\) is not tangential... |
USAMO-1999-1 | https://artofproblemsolving.com/wiki/index.php/1999_USAMO_Problems/Problem_1 | Some checkers placed on an \(n\times n\) checkerboard satisfy the following conditions:
(a) every square that does not contain a checker shares a side with one that does;
(b) given any pair of squares that contain checkers, there is a sequence of squares containing checkers, starting and ending with the given squares... | [
"For the proof let's look at the checkers board as a graph \\(G\\), where the checkers are the vertices and the edges are every pair of checkers sharing a side.\n\nDefine \\(R\\) as the circuit rank of \\(G\\)(the minimum number of edges to remove from a graph to remove all its cycles).\n\nDefine \\(x\\) as the num... |
USAMO-1999-2 | https://artofproblemsolving.com/wiki/index.php/1999_USAMO_Problems/Problem_2 | Let \(ABCD\) be a cyclic quadrilateral. Prove that
\[
|AB - CD| + |AD - BC| \geq 2|AC - BD|.
\] | [
"Let arc \\(AB\\) of the circumscribed circle (which we assume WLOG has radius 0.5) have value \\(2x\\), \\(BC\\) have \\(2y\\), \\(CD\\) have \\(2z\\), and \\(DA\\) have \\(2w\\). Then our inequality reduces to, for \\(x+y+z+w = 180^\\circ\\):\n\n\\[\n|\\sin x - \\sin z| + |\\sin y - \\sin w| \\ge 2|\\sin (x+y) - ... |
USAMO-1999-3 | https://artofproblemsolving.com/wiki/index.php/1999_USAMO_Problems/Problem_3 | Let \(p > 2\) be a prime and let \(a,b,c,d\) be integers not divisible by \(p\), such that
\[
\left\{ \dfrac{ra}{p} \right\} + \left\{ \dfrac{rb}{p} \right\} + \left\{ \dfrac{rc}{p} \right\} + \left\{ \dfrac{rd}{p} \right\} = 2
\]
for any integer \(r\) not divisible by \(p\). Prove that at least two of the numbers \(... | [
"We see that \\(\\biggl\\{\\frac{ra+rb+rc+rd}{p}\\biggr\\}=0\\) means that \\(p|r(a+b+c+d)\\). Now, since \\(p\\) does not divide \\(r\\) and \\(p\\) is prime, their GCD is 1 so \\(p\\mathrel{|}a+b+c+d\\).\n\nSince \\(\\biggl\\{\\frac{ra}{p}\\biggr\\}+\\biggl\\{\\frac{rb}{p}\\biggr\\}+\\biggl\\{\\frac{rc}{p}\\biggr... |
USAMO-1999-4 | https://artofproblemsolving.com/wiki/index.php/1999_USAMO_Problems/Problem_4 | Let \(a_{1}, a_{2}, \dots, a_{n}\) (\(n > 3\)) be real numbers such that
\[
a_{1} + a_{2} + \cdots + a_{n} \geq n \qquad \mbox{and} \qquad a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2} \geq n^{2}.
\]
Prove that \(\max(a_{1}, a_{2}, \dots, a_{n}) \geq 2\). | [
"First, suppose all the \\(a_i\\) are positive. Then\n\n\\[\n\\max(a_1, \\dotsc, a_n) \\ge \\sqrt{\\frac{a_1^2 + \\dotsb + a_n^2}{n}} \\ge \\sqrt{n} \\ge 2 .\n\\]\n\nSuppose, on the other hand, that without loss of generality,\n\n\\[\na_1 \\ge a_2 \\ge \\dotsb \\ge a_k \\ge 0 > a_{k+1} \\ge \\dotsb \\ge a_n,\n\\]\n... |
USAMO-1999-5 | https://artofproblemsolving.com/wiki/index.php/1999_USAMO_Problems/Problem_5 | The Y2K Game is played on a \(1 \times 2000\) grid as follows. Two players in turn write either an S or an O in an empty square. The first player who produces three consecutive boxes that spell SOS wins. If all boxes are filled without producing SOS then the game is a draw. Prove that the second player has a winning st... | [
"Label the squares \\(1\\) through \\(2000\\). The first player writes a letter in any square \\(k\\). The second player now selects any empty square \\(l\\) such that \\(\\text{min}\\,(|l-1|,|l-2000|,|l-k|) > 3\\), and fills in a S. The first player may again play anywhere, say \\(m\\). If the second player can no... |
USAMO-1999-6 | https://artofproblemsolving.com/wiki/index.php/1999_USAMO_Problems/Problem_6 | Let \(ABCD\) be an isosceles trapezoid with \(AB \parallel CD\). The inscribed circle \(\omega\) of triangle \(BCD\) meets \(CD\) at \(E\). Let \(F\) be a point on the (internal) angle bisector of \(\angle DAC\) such that \(EF \perp CD\). Let the circumscribed circle of triangle \(ACF\) meet line \(CD\) at \(C\) and \(... | [
"Quadrilateral \\(ABCD\\) is cyclic since it is an isosceles trapezoid. \\(AD=BC\\). Triangle \\(ADC\\) and triangle \\(BCD\\) are reflections of each other with respect to diameter which is perpendicular to \\(AB\\). Let the incircle of triangle \\(ADC\\) touch \\(DC\\) at \\(K\\). The reflection implies that \\(D... |
USAMO-2000-1 | https://artofproblemsolving.com/wiki/index.php/2000_USAMO_Problems/Problem_1 | Call a real-valued function \(f\) very convex if
\[
\frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|
\]
holds for all real numbers \(x\) and \(y\). Prove that no very convex function exists. | [
"Let \\(y \\ge x\\), and substitute \\(a = x, 2b = y-x\\). Then a function is very convex if \\(\\frac{f(a) + f(a+2b)}{2} \\ge f(a + b) + 2b\\), or rearranging,\n\n\\[\n\\left[\\frac{f(a+2b)-f(a+b)}{b}\\right]-\\left[\\frac{f(a+b)-f(a)}{b}\\right] \\ge 4\n\\]\n\nLet \\(g(a) = \\frac{f(a+b) - f(a)}{b}\\), which is t... |
USAMO-2000-2 | https://artofproblemsolving.com/wiki/index.php/2000_USAMO_Problems/Problem_2 | Let \(S\) be the set of all triangles \(ABC\) for which
\[
5 \left( \dfrac{1}{AP} + \dfrac{1}{BQ} + \dfrac{1}{CR} \right) - \dfrac{3}{\min\{ AP, BQ, CR \}} = \dfrac{6}{r},
\]
where \(r\) is the inradius and \(P, Q, R\) are the points of tangency of the incircle with sides \(AB, BC, CA,\) respectively. Prove that all ... | [
"We let \\(x = AP = s - a, y = BQ = s-b, z = CR = s-c\\), and without loss of generality let \\(x \\le y \\le z\\). Then \\(x + y + z = 3s - (a+b+c) = s\\), so \\(r = \\frac {A}{s} = \\frac{\\sqrt{(x+y+z)xyz}}{x+y+z} = \\sqrt{\\frac{xyz}{x+y+z}}\\). Thus,\n\n\\[\n6\\sqrt{\\frac{x+y+z}{xyz}} = \\frac{2yz + 5xy + 5xz... |
USAMO-2000-3 | https://artofproblemsolving.com/wiki/index.php/2000_USAMO_Problems/Problem_3 | A game of solitaire is played with \(R\) red cards, \(W\) white cards, and \(B\) blue cards. A player plays all the cards one at a time. With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he ... | [
"We claim (inductively) that the minimum is just going to be \\(\\min(BW,2WR,3RB)\\). We'll start our induction with the case where one of the three quantities is zero, in which case we verify that we can indeed get away without any penalty by, for example, discarding blue if we are out of white.\n\nNow, for the in... |
USAMO-2000-4 | https://artofproblemsolving.com/wiki/index.php/2000_USAMO_Problems/Problem_4 | Find the smallest positive integer \(n\) such that if \(n\) squares of a \(1000 \times 1000\) chessboard are colored, then there will exist three colored squares whose centers form a right triangle with sides parallel to the edges of the board. | [
"We claim that \\(n = 1999\\) is the smallest such number. For \\(n \\le 1998\\), we can simply color any of the \\(1998\\) squares forming the top row and the left column, but excluding the top left corner square.\n\n\\[\n[asy] for(int i = 0; i < 10; ++i){ for(int j = 0; j < 10; ++j){ if((i == 0 || j == 9) && !... |
USAMO-2000-5 | https://artofproblemsolving.com/wiki/index.php/2000_USAMO_Problems/Problem_5 | Let \(A_1A_2A_3\) be a triangle and let \(\omega_1\) be a circle in its plane passing through \(A_1\) and \(A_2.\) Suppose there exist circles \(\omega_2, \omega_3, \dots, \omega_7\) such that for \(k = 2, 3, \dots, 7,\) \(\omega_k\) is externally tangent to \(\omega_{k - 1}\) and passes through \(A_k\) and \(A_{k + 1}... | [
"Let the circumcenter of \\(\\triangle ABC\\) be \\(O\\), and let the center of \\(\\omega_k\\) be \\(O_k\\). \\(\\omega_k\\) and \\(\\omega_{k-1}\\) are externally tangent at the point \\(A_k\\), so \\(O_k, A_k, O_{k-1}\\) are collinear.\n\n\\(O\\) is the intersection of the perpendicular bisectors of \\(\\overlin... |
USAMO-2000-6 | https://artofproblemsolving.com/wiki/index.php/2000_USAMO_Problems/Problem_6 | Let \(a_1, b_1, a_2, b_2, \dots , a_n, b_n\) be nonnegative real numbers. Prove that
\[
\sum_{i, j = 1}^{n} \min\{a_ia_j, b_ib_j\} \le \sum_{i, j = 1}^{n} \min\{a_ib_j, a_jb_i\}.
\] | [
"Credit for this solution goes to Ravi Boppana.\n\nLemma 1: If \\(r_1, r_2, \\ldots , r_n\\) are non-negative reals and \\(x_1, x_2, \\ldots x_n\\) are reals, then\n\n\\[\n\\sum_{i, j}\\min(r_{i}, r_{j}) x_{i}x_{j}\\ge 0.\n\\]\n\nProof: Without loss of generality assume that the sequence \\(\\{r_i\\}\\) is increasi... |
USAMO-2001-1 | https://artofproblemsolving.com/wiki/index.php/2001_USAMO_Problems/Problem_1 | Each of eight boxes contains six balls. Each ball has been colored with one of \(n\) colors, such that no two balls in the same box are the same color, and no two colors occur together in more than one box. Determine, with justification, the smallest integer \(n\) for which this is possible. | [
"We claim that \\(n=23\\) is the minimum. Consider the following construction (replacing colors with numbers) which fulfills this:\n\n\\[\n\\left[ \\begin{array}{cccccccc} 1 & 1 & 1 & 2 & 3 & 4 & 5 & 6 \\\\ 2 & 7 & 12 & 7 & 8 & 9 & 10 & 11 \\\\ 3 & 8 & 13 & 12 & 13 & 14 & 15 & 16 \\\\ 4 & 9 & 14 & 17 & 17 & 17 & 18... |
USAMO-2001-2 | https://artofproblemsolving.com/wiki/index.php/2001_USAMO_Problems/Problem_2 | Let \(ABC\) be a triangle and let \(\omega\) be its incircle. Denote by \(D_1\) and \(E_1\) the points where \(\omega\) is tangent to sides \(BC\) and \(AC\), respectively. Denote by \(D_2\) and \(E_2\) the points on sides \(BC\) and \(AC\), respectively, such that \(CD_2 = BD_1\) and \(CE_2 = AE_1\), and denote by \(P... | [
"It is well known that the excircle opposite \\(A\\) is tangent to \\(\\overline{BC}\\) at the point \\(D_2\\). (Proof: let the points of tangency of the excircle with the lines \\(BC, AB, AC\\) be \\(D_3, F,G\\) respectively. Then \\(AB+BD_3=AB + BF=AF = AG = AC + CG=AC + CD_3\\). It follows that \\(2CD_3 = AB + B... |
USAMO-2001-3 | https://artofproblemsolving.com/wiki/index.php/2001_USAMO_Problems/Problem_3 | Let \(a, b, c \geq 0\) and satisfy
\[
a^2 + b^2 + c^2 + abc = 4.
\]
Show that
\[
0 \le ab + bc + ca - abc \leq 2.
\] | [
"First we prove the lower bound.\n\nNote that we cannot have \\(a, b, c\\) all greater than 1. Therefore, suppose \\(a \\le 1\\). Then\n\n\\[\nab + bc + ca - abc = a(b + c) + bc(1-a) \\ge 0.\n\\]\n\nNote that, by the Pigeonhole Principle, at least two of \\(a,b,c\\) are either both greater than or less than \\(1\\)... |
USAMO-2001-4 | https://artofproblemsolving.com/wiki/index.php/2001_USAMO_Problems/Problem_4 | Let \(P\) be a point in the plane of triangle \(ABC\) such that the segments \(PA\), \(PB\), and \(PC\) are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to \(PA\). Prove that \(\angle BAC\) is acute. | [
"We know that \\(PB^2+PC^2 < PA^2\\) and we wish to prove that \\(AB^2 + AC^2 > BC^2\\). It would be sufficient to prove that\n\n\\[\nPB^2+PC^2+AB^2+AC^2 \\geq PA^2 + BC^2.\n\\]\n\nSet \\(A(0,0)\\), \\(B(1,0)\\), \\(C(x,y)\\), \\(P(p,q)\\). Then, we wish to show\n\n\\[\n(p-1)^2 + q^2 + (p-x)^2 + (q-y)^2 + 1 + x^2 +... |
USAMO-2001-5 | https://artofproblemsolving.com/wiki/index.php/2001_USAMO_Problems/Problem_5 | Let \(S\) be a set of integers (not necessarily positive) such that
(a) there exist \(a,b \in S\) with \(\gcd(a,b) = \gcd(a - 2,b - 2) = 1\);
(b) if \(x\) and \(y\) are elements of \(S\) (possibly equal), then \(x^2 - y\) also belongs to \(S\).
Prove that \(S\) is the set of all integers. | [
"In the solution below we use the expression \\(S\\) is stable under \\(x\\mapsto f(x)\\) to mean that if \\(x\\) belongs to \\(S\\) then \\(f(x)\\) also belongs to \\(S\\). If \\(c,d\\in S\\), then by (B), \\(S\\) is stable under \\(x\\mapsto c^2 - x\\) and \\(x\\mapsto d^2 - x\\), hence stable under \\(x\\mapsto ... |
USAMO-2001-6 | https://artofproblemsolving.com/wiki/index.php/2001_USAMO_Problems/Problem_6 | Each point in the plane is assigned a real number such that, for any triangle, the number at the center of its inscribed circle is equal to the arithmetic mean of the three numbers at its vertices. Prove that all points in the plane are assigned the same number. | [
"We label each upper case point with the corresponding lower case letter as its assigned number. The key step is the following lemma.\n\nLemma: If \\(ABCD\\) is an isosceles trapezoid, then \\(a + c = b + d\\).\n\nProof: Assume without loss of generality that \\(BC\\parallel AD\\), and that rays \\(AB\\) and \\(DC\... |
USAMO-2002-1 | https://artofproblemsolving.com/wiki/index.php/2002_USAMO_Problems/Problem_1 | Let \(S\) be a set with 2002 elements, and let \(N\) be an integer with \(0 \le N \le 2^{2002}\). Prove that it is possible to color every subset of \(S\) either blue or red so that the following conditions hold:
(a) the union of any two red subsets is red;
(b) the union of any two blue subsets is blue;
(c) there ar... | [
"Let a set colored in such a manner be properly colored. We prove that any set with \\(n\\) elements can be properly colored for any \\(0 \\le N \\le 2^n\\). We proceed by induction.\n\nThe base case, \\(n = 0\\), is trivial.\n\nSuppose that our claim holds for \\(n = k\\). Let \\(s \\in S\\), \\(|S| = k + 1\\), an... |
USAMO-2002-2 | https://artofproblemsolving.com/wiki/index.php/2002_USAMO_Problems/Problem_2 | Let \(ABC\) be a triangle such that
\[
\left( \cot \frac{A}{2} \right)^2 + \left( 2 \cot \frac{B}{2} \right)^2 + \left( 3 \cot \frac{C}{2} \right)^2 = \left( \frac{6s}{7r} \right)^2,
\]
where \(s\) and \(r\) denote its semiperimeter and inradius, respectively. Prove that triangle \(ABC\) is similar to a triangle \(T\... | [
"Let \\(a,b,c\\) denote \\(BC, CA, AB\\), respectively. Since the line from a triangle's incenter to one of its vertices bisects the angle at the triangle's vertex, the condition of the problem is equivalent to the\n\n\\[\n\\left( \\frac{s-a}{r} \\right)^2 + 4\\left( \\frac{s-b}{r} \\right)^2 + 9\\left( \\frac{s-c}... |
USAMO-2002-3 | https://artofproblemsolving.com/wiki/index.php/2002_USAMO_Problems/Problem_3 | Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree \(n\) with real coefficients is the average of two monic polynomials of degree \(n\) with \(n\) real roots. | [
"Lemma. For any set of ordered pairs of reals \\(\\{ (x_i , y_i ) \\}_{i=1}^{n}\\) with \\(x_i \\neq x_j\\) for all \\(i \\neq j\\), there exists a unique monic real polynomial \\(P(x)\\) of degree \\(n\\) such that \\(P(x_i) = y_i\\) for all integers \\(i \\in [1,n]\\).\n\nProof 1. By the Lagrange Interpolation Fo... |
USAMO-2002-4 | https://artofproblemsolving.com/wiki/index.php/2002_USAMO_Problems/Problem_4 | Let \(\mathbb{R}\) be the set of real numbers. Determine all functions \(f : \mathbb{R} \rightarrow \mathbb{R}\) such that
\[
f(x^2 - y^2) = xf(x) - yf(y)
\]
for all pairs of real numbers \(x\) and \(y\). | [
"We first prove that \\(f\\) is odd.\n\nNote that \\(f(0) = f(x^2 - x^2) = xf(x) - xf(x) = 0\\), and for nonzero \\(y\\), \\(xf(x) + yf(-y) = f(x^2 - y^2) = xf(x) - yf(y)\\), or \\(yf(-y) = -yf(y)\\), which implies \\(f(-y) = -f(y)\\). Therefore \\(f\\) is odd. Henceforth, we shall assume that all variables are non... |
USAMO-2002-5 | https://artofproblemsolving.com/wiki/index.php/2002_USAMO_Problems/Problem_5 | Let \(a, b\) be integers greater than 2. Prove that there exists a positive integer \(k\) and a finite sequence \(n_1, n_2, \ldots, n_k\) of positive integers such that \(n_1 = a\), \(n_k = b\), and \(n_in_{i+1}\) is divisible by \(n_i + n_{i+1}\) for each \(i\) (\(1 \le i \le k\)). | [
"We may say that two integers \\(a\\) and \\(b\\) are connected (and write \\(a \\leftrightarrow b\\)) if there exists such a sequence of integers as described in the problem. For reference, we note that \\(\\leftrightarrow\\) is an equivalence relation: it is reflexive (\\(a \\leftrightarrow a\\)), symmetric (\\(a... |
USAMO-2002-6 | https://artofproblemsolving.com/wiki/index.php/2002_USAMO_Problems/Problem_6 | I have an \(n \times n\) sheet of stamps, from which I've been asked to tear out blocks of three adjacent stamps in a single row or column. (I can only tear along the perforations separating adjacent stamps, and each block must come out of the sheet in one piece.) Let \(b(n)\) be the smallest number of blocks I can tea... | [
"The upper bound requires an example of a set of \\(\\frac{1}{5}n^2 + dn\\) blocks whose removal makes it impossible to remove any further blocks. It suffices to show that we can tile the plane by tiles containing one block for every five stamps such that no more blocks can be chosen. Two such tilings are shown bel... |
USAMO-2003-1 | https://artofproblemsolving.com/wiki/index.php/2003_USAMO_Problems/Problem_1 | Prove that for every positive integer \(n\) there exists an \(n\)-digit number divisible by \(5^n\) all of whose digits are odd. | [
"We proceed by induction. For our base case, \\(n=1\\), we have the number 5. Now, suppose that there exists some number \\(a \\cdot 5^{n-1}\\) with \\(n-1\\) digits, all of which are odd. It is then sufficient to prove that there exists an odd digit \\(k\\) such that \\(k\\cdot 10^{n-1} + a \\cdot 5^{n-1} = 5^{n-1... |
USAMO-2003-2 | https://artofproblemsolving.com/wiki/index.php/2003_USAMO_Problems/Problem_2 | A convex polygon \(\mathcal{P}\) in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon \(\mathcal{P}\) are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers. | [
"When \\(\\mathcal{P}\\) is a triangle, the problem is trivial. Otherwise, it is sufficient to prove that any two diagonals of the polygon cut each other into rational lengths. Let two diagonals which intersect at a point \\(P\\) within the polygon be \\(AC\\) and \\(BD\\). Since \\(ABCD\\) is a convex quadrilatera... |
USAMO-2003-3 | https://artofproblemsolving.com/wiki/index.php/2003_USAMO_Problems/Problem_3 | Let \(n \neq 0\). For every sequence of integers
\[
A = a_0,a_1,a_2,\dots, a_n
\]
satisfying \(0 \le a_i \le i\), for \(i=0,\dots,n\), define another sequence
\[
t(A)= t(a_0), t(a_1), t(a_2), \dots, t(a_n)
\]
by setting \(t(a_i)\) to be the number of terms in the sequence \(A\) that precede the term \(a_i\) and are... | [
"Consider some sequence \\(C = c_0, \\ldots, c_n\\) as the image of \\(A\\) after \\(t\\) has been applied some finite number of times.\n\nLemma 1. If \\(t(c_k) = c_k = j\\), then \\({} c_j = \\cdots = c_{j+i} = \\cdots = c_k = j\\) (\\(0 \\le i \\le k-j\\)).\n\nProof. Since the \\(j\\) terms \\(c_0, \\ldots, c_{j-... |
USAMO-2003-4 | https://artofproblemsolving.com/wiki/index.php/2003_USAMO_Problems/Problem_4 | Let \(ABC\) be a triangle. A circle passing through \(A\) and \(B\) intersects segments \(AC\) and \(BC\) at \(D\) and \(E\), respectively. Lines \(AB\) and \(DE\) intersect at \(F\), while lines \(BD\) and \(CF\) intersect at \(M\). Prove that \(MF = MC\) if and only if \(MB\cdot MD = MC^2\). | [
"Extend segment \\(DM\\) through \\(M\\) to \\(G\\) such that \\(FG\\parallel CD\\).\n\n\\[\n[asy] defaultpen(fontsize(10)+0.6); size(250); var theta=22, r=0.58; pair B=origin, A=dir(theta), C=A+(rotate(78)*0.8*A), O=IP(CR(B,r),CR(A,r)); path c=CR(O,r); pair D=IP(c,A--C), E=IP(c,B--C), F=extension(A,B,D,E), M=exten... |
USAMO-2003-5 | https://artofproblemsolving.com/wiki/index.php/2003_USAMO_Problems/Problem_5 | Let \(a\), \(b\), \(c\) be positive real numbers. Prove that
\[
\dfrac{(2a + b + c)^2}{2a^2 + (b + c)^2} + \dfrac{(2b + c + a)^2}{2b^2 + (c + a)^2} + \dfrac{(2c + a + b)^2}{2c^2 + (a + b)^2} \le 8.
\] | [
"Since all terms are homogeneous, we may assume WLOG that \\(a + b + c = 3\\).\n\nThen the LHS becomes \\(\\sum \\frac {(a + 3)^2}{2a^2 + (3 - a)^2} = \\sum \\frac {a^2 + 6a + 9}{3a^2 - 6a + 9} = \\sum \\left(\\frac {1}{3} + \\frac {8a + 6}{3a^2 - 6a + 9}\\right)\\).\n\nNotice \\(3a^2 - 6a + 9 = 3(a - 1)^2 + 6 \\ge... |
USAMO-2003-6 | https://artofproblemsolving.com/wiki/index.php/2003_USAMO_Problems/Problem_6 | At the vertices of a regular hexagon are written six nonnegative integers whose sum is 2003. Bert is allowed to make moves of the following form: he may pick a vertex and replace the number written there by the absolute value of the difference between the numbers written at the two neighboring vertices. Prove that Bert... | [
"Assume the original numbers are \\(a,b,c,d,e,f\\). Since \\(a+b+c+d+e+f\\) is odd, either \\(a+c+e\\) or \\(b+d+f\\) must be odd. WLOG let \\(a+c+e\\) be odd and \\(a\\ge c\\ge e \\ge 0\\).\n\n## Case 1\n\n\\(a,c,e>0\\). Define Operation A as the sequence of moves from Step 1 to Step 3, shown below:\n\n\\[\n[asy] ... |
USAMO-2004-1 | https://artofproblemsolving.com/wiki/index.php/2004_USAMO_Problems/Problem_1 | Let \(ABCD\) be a quadrilateral circumscribed about a circle, whose interior and exterior angles are at least 60 degrees. Prove that
\[
\frac {1}{3}|AB^3 - AD^3| \le |BC^3 - CD^3| \le 3|AB^3 - AD^3|.
\]
When does equality hold? | [
"It suffices to show that\n\n\\[\n|BC^3-CD^3| \\leq 3|AB^3-AD^3|,\n\\]\n\nbecause the other side of the inequality follows by symmetry. By Pitot's Theorem, we have \\(AB-AD=BC-CD\\). WLOG, let \\(AB \\geq AD\\). Then we have that \\(BC \\geq CD\\), so we must show \\(BC^3-CD^3 \\leq 3(AB^3-AD^3)\\). If \\(AB=AD\\),... |
USAMO-2004-2 | https://artofproblemsolving.com/wiki/index.php/2004_USAMO_Problems/Problem_2 | Suppose \(a_1, \dots, a_n\) are integers whose greatest common divisor is 1. Let \(S\) be a set of integers with the following properties:
(a) For \(i = 1, \dots, n\), \(a_i \in S\).
(b) For \(i,j = 1, \dots, n\) (not necessarily distinct), \(a_i - a_j \in S\).
(c) For any integers \(x,y \in S\), if \(x + y \in S\),... | [
"Suppose \\(a_i\\) has only one element; then for the greatest common divisor to be 1, \\(1\\) has to be the sole element. Then \\(1\\) is in \\(S\\) by (a), \\(0\\) is in \\(S\\) by (b), \\(0 + 1 = 1\\in S\\Rightarrow 0 - 1 = - 1\\in S\\) by (c), and we can apply (c) analogously to get that \\(n\\cdot 1 \\in S\\) ... |
USAMO-2004-3 | https://artofproblemsolving.com/wiki/index.php/2004_USAMO_Problems/Problem_3 | For what values of \(k > 0\) is it possible to dissect a \(1 \times k\) rectangle into two similar, but incongruent, polygons? | [
"We will show that a dissection satisfying the requirements of the problem is possible if and only if \\(k\\neq 1\\).\n\nWe first show by contradiction that such a dissection is not possible when \\(k = 1\\). Assume that we have such a dissection. The common boundary of the two dissecting polygons must be a single ... |
USAMO-2004-4 | https://artofproblemsolving.com/wiki/index.php/2004_USAMO_Problems/Problem_4 | Alice and Bob play a game on a 6 by 6 grid. On his or her turn, a player chooses a rational number not yet appearing on the grid and writes it in an empty square of the grid. Alice goes first and then the players alternate. When all squares have numbers written in them, in each row, the square with the greatest number ... | [
"Before the game, Bob may select three useless squares per row. He may then move according to the following rules:\n\n- If Alice writes a number in a useless square, then Bob writes a higher number in a non-useless square in the same row on his next turn.\n- If Alice writes a number in a non-useless square, then Bo... |
USAMO-2004-5 | https://artofproblemsolving.com/wiki/index.php/2004_USAMO_Problems/Problem_5 | Let \(a\), \(b\), and \(c\) be positive real numbers. Prove that
\[
(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3
\]
. | [] |
USAMO-2004-6 | https://artofproblemsolving.com/wiki/index.php/2004_USAMO_Problems/Problem_6 | A circle \(\omega\) is inscribed in a quadrilateral \(ABCD\). Let \(I\) be the center of \(\omega\). Suppose that
\[
(AI + DI)^2 + (BI + CI)^2 = (AB + CD)^2.
\]
Prove that \(ABCD\) is an isosceles trapezoid. | [
"Our proof is based on the following key Lemma.\n\nLemma: If a circle \\(\\omega\\), centered at \\(I\\), is inscribed in a quadrilateral \\(ABCD\\), then\n\n\\[\nBI^2 + \\frac{AI}{DI}\\cdot BI\\cdot CI = AB\\cdot BC.\\qquad\\qquad (1)\n\\]\n\nProof: Since circle \\(\\omega\\) is inscribed in \\(ABCD\\), we get \\(... |
USAMO-2005-1 | https://artofproblemsolving.com/wiki/index.php/2005_USAMO_Problems/Problem_1 | Determine all composite positive integers \(n\) for which it is possible to arrange all divisors of \(n\) that are greater than 1 in a circle so that no two adjacent divisors are relatively prime. | [
"No such circular arrangement exists for \\(n=pq\\), where \\(p\\) and \\(q\\) are distinct primes. In that case, the numbers to be arranged are \\(p\\); \\(q\\) and \\(pq\\), and in any circular arrangement, \\(p\\) and \\(q\\) will be adjacent. We claim that the desired circular arrangement exists in all other ca... |
USAMO-2005-2 | https://artofproblemsolving.com/wiki/index.php/2005_USAMO_Problems/Problem_2 | Prove that the system
\[
\begin{align*}x^6 + x^3 + x^3y + y &= 147^{157} \\ x^3 + x^3y + y^2 + y + z^9 &= 157^{147}\end{align*}
\]
has no solutions in integers \(x\), \(y\), and \(z\). | [
"It suffices to show that there are no solutions to this system in the integers mod 19. We note that \\(152 = 8 \\cdot 19\\), so \\(157 \\equiv -147 \\equiv 5 \\pmod{19}\\). For reference, we construct a table of powers of five:\n\n\\[\n\\begin{array}{c|c||c|c} n& 5^n &n & 5^n \\\\\\hline 1 & 5 & 6 & 7 \\\\ 2 & 6 &... |
USAMO-2005-3 | https://artofproblemsolving.com/wiki/index.php/2005_USAMO_Problems/Problem_3 | Let \(ABC\) be an acute-angled triangle, and let \(P\) and \(Q\) be two points on side \(BC\). Construct point \(C_1\) in such a way that convex quadrilateral \(APBC_1\) is cyclic, \(QC_1 \parallel CA\), and \(C_1\) and \(Q\) lie on opposite sides of line \(AB\). Construct point \(B_1\) in such a way that convex quadri... | [
"Let \\(B_1'\\) be the second intersection of the line \\(C_1A\\) with the circumcircle of \\(APC\\), and let \\(Q'\\) be the second intersection of the circumcircle of \\(B_1' C_1P\\) and line \\(BC\\). It is enough to show that \\(B_1'=B_1\\) and \\(Q' =Q\\). All our angles will be directed, and measured mod \\(\... |
USAMO-2005-4 | https://artofproblemsolving.com/wiki/index.php/2005_USAMO_Problems/Problem_4 | Legs \(L_1, L_2, L_3, L_4\) of a square table each have length \(n\), where \(n\) is a positive integer. For how many ordered 4-tuples \((k_1, k_2, k_3, k_4)\) of nonnegative integers can we cut a piece of length \(k_i\) from the end of leg \(L_i \; (i = 1,2,3,4)\) and still have a stable table?
(The table is stable i... | [
"First flip the table upside down and observe the plane that the ends of the legs form. Let this plane be \\(ax+by+c=z\\). Assume WLOG that the table is the square with endpoints \\((0,0), (0,1), (1,0), (1,1)\\) on the xy plane. Thus the lengths of the legs will be \\(c, c+b, c+a, c+a+b\\) respectively. In other wo... |
USAMO-2005-5 | https://artofproblemsolving.com/wiki/index.php/2005_USAMO_Problems/Problem_5 | Let \(n\) be an integer greater than 1. Suppose \(2n\) points are given in the plane, no three of which are collinear. Suppose \(n\) of the given \(2n\) points are colored blue and the other \(n\) colored red. A line in the plane is called a balancing line if it passes through one blue and one red point and, for each s... | [
"Consider the convex hull of the the \\(2n\\) points, or the points that would form the largest convex polygon. If the points in the convex hull contain both red and blue points, then there must be at least 2 edges of the graph of the convex hull such that the edge connects a blue and a red point. Drawing a line th... |
USAMO-2005-6 | https://artofproblemsolving.com/wiki/index.php/2005_USAMO_Problems/Problem_6 | For \(m\) a positive integer, let \(s(m)\) be the sum of the digits of \(m\). For \(n\ge 2\), let \(f(n)\) be the minimal \(k\) for which there exists a set \(S\) of \(n\) positive integers such that \(s\left(\sum_{x\in X} x\right) = k\) for any nonempty subset \(X\subset S\). Prove that there are constants \(0 < C_1 <... | [
"For the upper bound, let \\(p\\) be the smallest integer such that \\(10^p\\geq n(n+1)/2\\) and let\n\n\\[\nS = \\{10^p - 1, 2(10^p - 1), \\ldots, n(10^p - 1)\\}.\n\\]\n\nThe sum of any nonempty set of elements of \\(S\\) will have the form \\(k(10^p - 1)\\) for some \\(1\\leq k\\leq n(n+1)/2\\). Write \\(k(10^p -... |
USAMO-2006-1 | https://artofproblemsolving.com/wiki/index.php/2006_USAMO_Problems/Problem_1 | Let \(p\) be a prime number and let \(s\) be an integer with \(0 < s < p\). Prove that there exist integers \(m\) and \(n\) with \(0 < m < n < p\) and
\[
\left\{ \frac{sm}{p} \right\} < \left\{ \frac{sn}{p} \right\} < \frac{s}{p}
\]
if and only if \(s\) is not a divisor of \(p-1\).
Note: For \(x\) a real number, let... | [
"We proceed by contradiction. Assume that \\(s|(p-1)\\). Then for some positive integer \\(k\\), \\(sk=p-1\\). The conditions given are equivalent to stating that \\(sm \\bmod p < sn \\bmod p< s\\bmod p\\). Now consider the following array modulo p:\n\n\\[\n\\begin{array}{ccccc} \\mbox{Row 1}& s& 2s&\\hdots& ks\\\\... |
USAMO-2006-2 | https://artofproblemsolving.com/wiki/index.php/2006_USAMO_Problems/Problem_2 | For a given positive integer \(k\) find, in terms of \(k\), the minimum value of \(N\) for which there is a set of \(2k+1\) distinct positive integers that has sum greater than \(N\) but every subset of size \(k\) has sum at most \(N/2\). | [
"Let one optimal set of integers be \\(\\{a_1,\\dots,a_{2k+1}\\}\\) with \\(a_1 > a_2 > \\cdots > a_{2k+1} > 0\\).\n\nThe two conditions can now be rewritten as \\(a_1+\\cdots + a_k \\leq N/2\\) and \\(a_1+\\cdots +a_{2k+1} > N\\). Subtracting, we get that \\(a_{k+1}+\\cdots + a_{2k+1} > N/2\\), and hence \\(a_{k+1... |
USAMO-2006-3 | https://artofproblemsolving.com/wiki/index.php/2006_USAMO_Problems/Problem_3 | For integral \(m\), let \(p(m)\) be the greatest prime divisor of \(m\). By convention, we set \(p(\pm 1)=1\) and \(p(0)=\infty\). Find all polynomials \(f\) with integer coefficients such that the sequence \(\{ p(f(n^2))-2n) \}_{n \in \mathbb{Z} \ge 0}\) is bounded above. (In particular, this requires \(f(n^2)\neq 0\)... | [
"Let \\(f(x)\\) be a non-constant polynomial in \\(x\\) of degree \\(d\\) with integer coefficients, suppose further that no prime divides all the coefficients of \\(f\\) (otherwise consider the polynomial obtained by dividing \\(f\\) by the gcd of its coefficients). We further normalize \\(f\\) by multiplying by \... |
USAMO-2006-4 | https://artofproblemsolving.com/wiki/index.php/2006_USAMO_Problems/Problem_4 | Find all positive integers \(n\) such that there are \(k\ge 2\) positive rational numbers \(a_1, a_2, \ldots, a_k\) satisfying \(a_1 + a_2 + \cdots + a_k = a_1\cdot a_2\cdots a_k = n\). | [
"First, consider composite numbers. We can then factor \\(n\\) into \\(p_1p_2.\\) It is easy to see that \\(p_1+p_2\\le n\\), and thus, we can add \\((n-p_1-p_2)\\) 1s in order to achieve a sum and product of \\(n\\). For \\(p_1+p_2=n\\), which is only possible in one case, \\(n=4\\), we consider \\(p_1=p_2=2\\).\n... |
USAMO-2006-5 | https://artofproblemsolving.com/wiki/index.php/2006_USAMO_Problems/Problem_5 | A mathematical frog jumps along the number line. The frog starts at 1, and jumps according to the following rule: if the frog is at integer \(n\), then it can jump either to \(n+1\) or to \(n+2^{m_n+1}\) where \(2^{m_n}\) is the largest power of 2 that is a factor of \(n\). Show that if \(k\ge 2\) is a positive integer... | [
"For \\(i\\geq 0\\) and \\(k\\geq 1\\), let \\(x_{i,k}\\) denote the minimum number of jumps needed to reach the integer \\(n_{i,k} = 2^i k\\). We must prove that\n\n\\[\nx_{i,k} > x_{i,1}\\qquad\\qquad (1)\n\\]\n\nfor all \\(i\\geq 0\\) and \\(k\\geq 2\\). We prove this using the method of descent.\n\nFirst note t... |
USAMO-2006-6 | https://artofproblemsolving.com/wiki/index.php/2006_USAMO_Problems/Problem_6 | Let \(ABCD\) be a quadrilateral, and let \(E\) and \(F\) be points on sides \(AD\) and \(BC\), respectively, such that \(AE/ED = BF/FC\). Ray \(FE\) meets rays \(BA\) and \(CD\) at \(S\) and \(T\) respectively. Prove that the circumcircles of triangles \(SAE\), \(SBF\), \(TCF\), and \(TDE\) pass through a common point. | [
"Let the intersection of the circumcircles of \\(SAE\\) and \\(SBF\\) be \\(X\\), and let the intersection of the circumcircles of \\(TCF\\) and \\(TDE\\) be \\(Y\\).\n\n\\(BXF=BSF=AXE\\) because \\(BSF\\) tends both arcs \\(AE\\) and \\(BF\\). \\(BFX=XSB=XEA\\) because \\(XSB\\) tends both arcs \\(XA\\) and \\(XB\... |
USAMO-2007-1 | https://artofproblemsolving.com/wiki/index.php/2007_USAMO_Problems/Problem_1 | Let \(n\) be a positive integer. Define a sequence by setting \(a_1 = n\) and, for each \(k>1\), letting \(a_k\) be the unique integer in the range \(0 \le a_k \le k-1\) for which \(a_1 + a_2 + \cdots + a_k\) is divisible by \(k\). For instance, when \(n=9\) the obtained sequence is \(9, 1, 2, 0, 3, 3, 3, \ldots\). Pro... | [
"Let \\(S_k = a_1 + a_2 + \\cdots + a_k\\) and \\(b_k = \\frac{S_k}{k}\\). Thus, because \\(S_{k+1} = S_k + a_{k+1}\\),\n\n\\[\nb_{k+1} = \\frac{b_k \\cdot k + a_{k+1}}{k+1} = \\left(\\frac{k}{k+1}\\right) \\cdot b_k + \\frac{a_{k+1}}{k+1}\n\\]\n\n\\(\\frac{k}{k+1} < 1\\), and by definition, \\(\\frac{a_{k+1}}{k+1}... |
USAMO-2007-2 | https://artofproblemsolving.com/wiki/index.php/2007_USAMO_Problems/Problem_2 | A square grid on the Euclidean plane consists of all points \((m,n)\), where \(m\) and \(n\) are integers. Is it possible to cover all grid points by an infinite family of discs with non-overlapping interiors if each disc in the family has radius at least 5? | [
"Lemma. Among 3 tangent circles with radius greater than or equal to 5, one can always fit a circle with radius greater than \\(\\frac{1}{\\sqrt{2}}\\) between those 3 circles.\n\nProof. Descartes' Circle Theorem states that if \\(a\\) is the curvature of a circle (\\(a=\\frac 1{r}\\), positive for externally tange... |
USAMO-2007-3 | https://artofproblemsolving.com/wiki/index.php/2007_USAMO_Problems/Problem_3 | Let \(S\) be a set containing \(n^2+n-1\) elements, for some positive integer \(n\). Suppose that the \(n\)-element subsets of \(S\) are partitioned into two classes. Prove that there are at least \(n\) pairwise disjoint sets in the same class. | [
"Claim: If we have instead \\(k(n+1)-1\\) elements, then we must have k disjoint subsets in a class. We proceed by Induction\n\nBase Case: \\(k=1\\) is trivial.\n\nCase 1: If there exists an \\(n+1\\) element subset \\(X\\) s.t. \\(X\\) has 2 n-element subsets in different classes. Consider the \\((k-1)(n+1)-1\\) e... |
USAMO-2007-4 | https://artofproblemsolving.com/wiki/index.php/2007_USAMO_Problems/Problem_4 | An animal with \(n\) cells is a connected figure consisting of \(n\) equal-sized square cells.\({}^1\) The figure below shows an 8-cell animal.
A dinosaur is an animal with at least 2007 cells. It is said to be primitive if its cells cannot be partitioned into two or more dinosaurs. Find with proof the maximum number ... | [
"Let a \\(n\\)-dino denote an animal with \\(n\\) or more cells.\n\nWe show by induction that an \\(n\\)-dino with \\(4n-2\\) or more animal cells is not primitive. (Note: if it had more, we could just take off enough until it had \\(4n-2\\), which would have a partition, and then add the cells back on.)\n\nBase Ca... |
USAMO-2007-5 | https://artofproblemsolving.com/wiki/index.php/2007_USAMO_Problems/Problem_5 | Prove that for every nonnegative integer \(n\), the number \(7^{7^n}+1\) is the product of at least \(2n+3\) (not necessarily distinct) primes. | [
"The proof is by induction. The base is provided by the \\(n = 0\\) case, where \\(7^{7^0} + 1 = 7^1 + 1 = 2^3\\). To prove the inductive step, it suffices to show that if \\(x = 7^{2m - 1}\\) for some positive integer \\(m\\) then \\((x^7 + 1)/(x + 1)\\) is composite. As a consequence, \\(x^7 + 1\\) has at least t... |
USAMO-2007-6 | https://artofproblemsolving.com/wiki/index.php/2007_USAMO_Problems/Problem_6 | Let \(ABC\) be an acute triangle with \(\omega\), \(\Omega\), and \(R\) being its incircle, circumcircle, and circumradius, respectively. The circle \(\omega_A\) is tangent internally to \(\Omega\) at \(A\) and externally tangent to \(\omega\). Circle \(\Omega_A\) is internally tangent to \(\Omega\) at \(A\) and tangen... | [
"\\[\n[asy] size(400); defaultpen(fontsize(8)); pair A=(2,8), B=(0,0), C=(13,0), I=incenter(A,B,C), O=circumcenter(A,B,C), p_a, q_a, X, Y, X1, Y1, D, E, F; real r=abs(I-foot(I,A,B)), R=abs(A-O), a=abs(B-C), b=abs(A-C), c=abs(A-B), x=(((b+c-a)/2)^2)/(r^2+4*r*R+((b+c-a)/2)^2), y=((b+c-a)/2)^2/(r^2+((b+c-a)/2)^2); p_a... |
USAMO-2008-1 | https://artofproblemsolving.com/wiki/index.php/2008_USAMO_Problems/Problem_1 | Prove that for each positive integer \(n\), there are pairwise relatively prime integers \(k_0, k_1 \dotsc, k_n\), all strictly greater than 1, such that \(k_0 k_1 \dotsm k_n -1\) is the product of two consecutive integers. | [
"We will prove the problem for each nonnegative integer \\(n\\). We wish to show that\n\n\\[\nk_0 k_1 \\dotsc k_n = a^2+a+1,\n\\]\n\nfor some integer \\(a\\). We induct on \\(n\\). For our base case, \\(n=0\\), we may let \\(a\\) be positive integer.\n\nFor the inductive step, suppose that \\(k_0, \\dotsc, k_{n-1}\... |
USAMO-2008-2 | https://artofproblemsolving.com/wiki/index.php/2008_USAMO_Problems/Problem_2 | Let \(ABC\) be an acute, scalene triangle, and let \(M\), \(N\), and \(P\) be the midpoints of \(\overline{BC}\), \(\overline{CA}\), and \(\overline{AB}\), respectively. Let the perpendicular bisectors of \(\overline{AB}\) and \(\overline{AC}\) intersect ray \(AM\) in points \(D\) and \(E\) respectively, and let lines ... | [
"Without Loss of Generality, assume \\(AB >AC\\). It is sufficient to prove that \\(\\angle OFA = 90^{\\circ}\\), as this would immediately prove that \\(A,P,O,F,N\\) are concyclic. By applying the Menelaus' Theorem in the Triangle \\(\\triangle BFC\\) for the transversal \\(E,M,D\\), we have (in magnitude)\n\n\\[\... |
USAMO-2008-3 | https://artofproblemsolving.com/wiki/index.php/2008_USAMO_Problems/Problem_3 | Let \(n\) be a positive integer. Denote by \(S_n\) the set of points \((x, y)\) with integer coordinates such that
\[
\left|x\right| + \left|y + \frac {1}{2}\right| < n
\]
A path is a sequence of distinct points \((x_1 , y_1 ), (x_2 , y_2 ), \ldots , (x_\ell, y_\ell)\) in \(S_n\) such that, for \(i = 2, \ldots , \ell... | [
"Color all the points in \\(S_{n}\\) red or black such that each row of points starts and ends with a red point, and alternates between red and black for each point in the row. This creates a checkerboard pattern, except that the middle two rows are identical.\n\n\\[\n[asy] /* variable declarations */ int n = 4; ... |
USAMO-2008-4 | https://artofproblemsolving.com/wiki/index.php/2008_USAMO_Problems/Problem_4 | Let \(\mathcal{P}\) be a convex polygon with \(n\) sides, \(n\ge3\). Any set of \(n - 3\) diagonals of \(\mathcal{P}\) that do not intersect in the interior of the polygon determine a triangulation of \(\mathcal{P}\) into \(n - 2\) triangles. If \(\mathcal{P}\) is regular and there is a triangulation of \(\mathcal{P}\)... | [
"We label the vertices of \\(\\mathcal{P}\\) as \\(P_0, P_1, P_2, \\ldots, P_n\\). Consider a diagonal \\(d = \\overline{P_a\\,P_{a+k}},\\,k \\le n/2\\) in the triangulation. We show that \\(k\\) must have the form \\(2^{m}\\) for some nonnegative integer \\(m\\).\n\nThis diagonal partitions \\(\\mathcal{P}\\) into... |
USAMO-2008-5 | https://artofproblemsolving.com/wiki/index.php/2008_USAMO_Problems/Problem_5 | Three nonnegative real numbers \(r_1\), \(r_2\), \(r_3\) are written on a blackboard. These numbers have the property that there exist integers \(a_1\), \(a_2\), \(a_3\), not all zero, satisfying \(a_1r_1 + a_2r_2 + a_3r_3 = 0\). We are permitted to perform the following operation: find two numbers \(x\), \(y\) on the ... | [
"Every time we perform an operation on the numbers on the blackboard \\(R = \\left < r_1, r_2, r_3 \\right >\\), we perform the corresponding operation on the integers \\(A = \\left < a_1, a_2, a_3 \\right >\\) so that \\(R \\cdot A = 0\\) continues to hold. (For example, if we replace \\(r_1\\) with \\(r_1 - r_2\\... |
USAMO-2008-6 | https://artofproblemsolving.com/wiki/index.php/2008_USAMO_Problems/Problem_6 | At a certain mathematical conference, every pair of mathematicians are either friends or strangers. At mealtime, every participant eats in one of two large dining rooms. Each mathematician insists upon eating in a room which contains an even number of his or her friends. Prove that the number of ways that the mathemati... | [
"Make the obvious re-interpretation as a graph. Let \\(f : G \\to \\{0,1\\}\\) be an indicator function with \\(f(v) = 0\\) if a vertex is in the first partition and \\(f(v) = 1\\) otherwise (this corresponds, in the actual problem, to putting a mathematician in the first or second room). Then look at \\(f\\) as a ... |
USAMO-2009-1 | https://artofproblemsolving.com/wiki/index.php/2009_USAMO_Problems/Problem_1 | Given circles \(\omega_1\) and \(\omega_2\) intersecting at points \(X\) and \(Y\), let \(\ell_1\) be a line through the center of \(\omega_1\) intersecting \(\omega_2\) at points \(P\) and \(Q\) and let \(\ell_2\) be a line through the center of \(\omega_2\) intersecting \(\omega_1\) at points \(R\) and \(S\). Prove t... | [
"Let \\(\\omega_3\\) be the circumcircle of \\(PQRS\\), \\(r_i\\) to be the radius of \\(\\omega_i\\), and \\(O_i\\) to be the center of the circle \\(\\omega_i\\), where \\(i \\in \\{1,2,3\\}\\). Note that \\(SR\\) and \\(PQ\\) are the radical axises of \\(O_1\\) , \\(O_3\\) and \\(O_2\\) , \\(O_3\\) respectively.... |
USAMO-2009-2 | https://artofproblemsolving.com/wiki/index.php/2009_USAMO_Problems/Problem_2 | Let \(n\) be a positive integer. Determine the size of the largest subset of \(\{ - n, - n + 1, \ldots , n - 1, n\}\) which does not contain three elements \(a, b, c\) (not necessarily distinct) satisfying \(a + b + c = 0\). | [
"Let \\(S\\) be a subset of \\(\\{-n,-n+1,\\dots,n-1,n\\}\\) of largest size satisfying \\(a+b+c\\neq 0\\) for all \\(a,b,c\\in S\\). First, observe that \\(0\\notin S\\). Next note that \\(|S|\\geq \\lceil n/2\\rceil\\), by observing that the set of all the odd numbers in \\(\\{-n,-n+1,\\dots,n-1,n\\}\\) works. To... |
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