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eebb93b | 1 2 3 4 5 6 7 | {"year":2014,"label":"1","problem":"Let $A B C$ be an acute triangle, and let $X$ be a variable interior point on the minor $\\operatorname{arc} B C$ of its circumcircle. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $C A$ and $C B$, respectively. Let $R$ be the intersection of line $P Q$ and the perpendicular from $B$ to $A C$. Let $\\ell$ be the line through $P$ parallel to $X R$. Prove that as $X$ varies along minor arc $B C$, the line $\\ell$ always passes through a fixed point.","solution":" The fixed point is the orthocenter, since $\\ell$ is a Simson line. See Lemma 4.4 of Euclidean Geometry in Math Olympiads."}
{"year":2014,"label":"2","problem":"Let $a_{1}, a_{2}, a_{3}, \\ldots$ be a sequence of integers, with the property that every consecutive group of $a_{i}$ 's averages to a perfect square. More precisely, for all positive integers $n$ and $k$, the quantity $$ \\frac{a_{n}+a_{n+1}+\\cdots+a_{n+k-1}}{k} $$ is always the square of an integer. Prove that the sequence must be constant (all $a_{i}$ are equal to the same perfect square).","solution":" Let $\\nu_{p}(n)$ denote the largest exponent of $p$ dividing $n$. The problem follows from the following proposition. ## Proposition Let $\\left(a_{n}\\right)$ be a sequence of integers and let $p$ be a prime. Suppose that every consecutive group of $a_{i}$ 's with length at most $p$ averages to a perfect square. Then $\\nu_{p}\\left(a_{i}\\right)$ is independent of $i$. We proceed by induction on the smallest value of $\\nu_{p}\\left(a_{i}\\right)$ as $i$ ranges (which must be even, as each of the $a_{i}$ are themselves a square). First we prove two claims. Claim - If $j \\equiv k(\\bmod p)$ then $a_{j} \\equiv a_{k}(\\bmod p)$. Claim - If some $a_{i}$ is divisible by $p$ then all of them are. $$ S_{n}=a_{1}+a_{2}+\\cdots+a_{n} \\equiv a_{2}+\\cdots+a_{n} \\quad(\\bmod p) $$ Call an integer $k$ with $2 \\leq k<p$ a pivot if $1-k^{-1}$ is a quadratic nonresidue modulo $p$. We claim that for any pivot $k, S_{k} \\equiv 0(\\bmod p)$. If not, then $$ \\frac{a_{1}+a_{2}+\\cdots+a_{k}}{k} \\text { and } \\frac{a_{2}+\\cdots+a_{k}}{k-1} $$ are both qudaratic residues. Division implies that $\\frac{k-1}{k}=1-k^{-1}$ is a quadratic residue, contradiction. Next we claim that there is an integer $m$ with $S_{m} \\equiv S_{m+1} \\equiv 0(\\bmod p)$, which implies $p \\mid a_{m+1}$. If 2 is a pivot, then we simply take $m=1$. Otherwise, there are $\\frac{1}{2}(p-1)$ pivots, one for each nonresidue (which includes neither 0 nor 1 ), and all pivots lie in [3, $p-1]$, so we can find an $m$ such that $m$ and $m+1$ are both pivots. Repeating this procedure starting with $a_{m+1}$ shows that $a_{2 m+1}, a_{3 m+1}, \\ldots$ must all be divisible by $p$. Combined with the first claim and the fact that $m<p$, we find that all the $a_{i}$ are divisible by $p$. The second claim establishes the base case of our induction. Now assume all $a_{i}$ are divisible by $p$ and hence $p^{2}$. Then all the averages in our proposition (with length at $\\operatorname{most} p$ ) are divisible by $p$ and hence $p^{2}$. Thus the map $a_{i} \\mapsto \\frac{1}{p^{2}} a_{i}$ gives a new sequence satisfying the proposition, and our inductive hypothesis completes the proof. Remark. There is a subtle bug that arises if one omits the condition that $k \\leq p$ in the proposition. When $k=p^{2}$ the average $\\frac{a_{1}+\\cdots+a_{p^{2}}}{p^{2}}$ is not necessarily divisible by $p$ even if all the $a_{i}$ are. Hence it is not valid to divide through by $p$. This is why the condition $k \\leq p$ was added."}
{"year":2014,"label":"3","problem":"Let $n$ be an even positive integer, and let $G$ be an $n$-vertex (simple) graph with exactly $\\frac{n^{2}}{4}$ edges. An unordered pair of distinct vertices $\\{x, y\\}$ is said to be amicable if they have a common neighbor (there is a vertex $z$ such that $x z$ and $y z$ are both edges). Prove that $G$ has at least $2\\binom{n \/ 2}{2}$ pairs of vertices which are amicable.","solution":" First, we prove the following lemma. (https:\/\/en.wikipedia.org\/wiki\/Friendship_ paradox). Lemma (On average, your friends are more popular than you) For a vertex $v$, let $a(v)$ denote the average degree of the neighbors of $v$ (setting $a(v)=0$ if $\\operatorname{deg} v=0)$. Then $$ \\sum_{v} a(v) \\geq \\sum_{v} \\operatorname{deg} v=2 \\# E $$ $$ \\begin{aligned} \\sum_{v} a(v) & =\\sum_{v} \\frac{\\sum_{w \\sim v} \\operatorname{deg} w}{\\operatorname{deg} v} \\\\ & =\\sum_{v} \\sum_{w \\sim v} \\frac{\\operatorname{deg} w}{\\operatorname{deg} v} \\\\ & =\\sum_{\\text {edges } v w}\\left(\\frac{\\operatorname{deg} w}{\\operatorname{deg} v}+\\frac{\\operatorname{deg} v}{\\operatorname{deg} w}\\right) \\\\ & \\stackrel{\\text { AM-GM }}{\\geq} \\sum_{\\text {edges } v w} 2=2 \\# E=\\sum_{v} \\operatorname{deg} v \\end{aligned} $$ as desired. Corollary (On average, your most popular friend is more popular than you) For a vertex $v$, let $m(v)$ denote the maximum degree of the neighbors of $v$ (setting $m(v)=0$ if $\\operatorname{deg} v=0)$. Then $$ \\sum_{v} m(v) \\geq \\sum_{v} \\operatorname{deg} v=2 \\# E \\text {. } $$ We can use this to count amicable pairs by noting that any particular vertex $v$ is in at least $m(v)-1$ amicable pairs. So, the number of amicable pairs is at least $$ \\frac{1}{2} \\sum_{v}(m(v)-1) \\geq \\# E-\\frac{1}{2} \\# V $$ Note that up until now we haven't used any information about $G$. But now if we plug in $\\# E=n^{2} \/ 4, \\# V=n$, then we get exactly the desired answer. (Equality holds for $G=K_{n \/ 2, n \/ 2}$.)"}
{"year":2014,"label":"4","problem":"Let $n$ be a positive even integer, and let $c_{1}, c_{2}, \\ldots, c_{n-1}$ be real numbers satisfying $$ \\sum_{i=1}^{n-1}\\left|c_{i}-1\\right|<1 $$ Prove that $$ 2 x^{n}-c_{n-1} x^{n-1}+c_{n-2} x^{n-2}-\\cdots-c_{1} x^{1}+2 $$ has no real roots.","solution":" We will prove the polynomial is positive for all $x \\in \\mathbb{R}$. As $c_{i}>0$, the result is vacuous for $x \\leq 0$, so we restrict attention to $x>0$. Then letting $c_{i}=1-d_{i}$ for each $i$, the inequality we want to prove becomes $$ x^{n}+1+\\frac{x^{n+1}+1}{x+1}>\\sum_{1}^{n-1} d_{i} x^{i} \\quad \\text { given } \\sum\\left|d_{i}\\right|<1 $$ But obviously $x^{n}+1>x^{i}$ for any $1 \\leq i \\leq n-1$ and $x>0$. So in fact $x^{n}+1>\\sum_{1}^{n-1}\\left|d_{i}\\right| x^{i}$ holds for $x>0$, as needed."}
{"year":2014,"label":"5","problem":"Let $A B C D$ be a cyclic quadrilateral, and let $E, F, G$, and $H$ be the midpoints of $A B, B C, C D$, and $D A$ respectively. Let $W, X, Y$ and $Z$ be the orthocenters of triangles $A H E, B E F, C F G$ and $D G H$, respectively. Prove that the quadrilaterals $A B C D$ and $W X Y Z$ have the same area.","solution":" We begin with: Claim - Point $W$ has coordinates $\\frac{1}{2}(2 a+b+d)$. By symmetry, we have $$ \\begin{aligned} w & =\\frac{1}{2}(2 a+b+d) \\\\ x & =\\frac{1}{2}(2 b+c+a) \\\\ y & =\\frac{1}{2}(2 c+d+b) \\\\ z & =\\frac{1}{2}(2 d+a+c) . \\end{aligned} $$ We see that $w-y=a-c, x-z=b-d$. So the diagonals of $W X Y Z$ have the same length as those of $A B C D$ as well as the same directed angle between them. This implies the areas are equal, too."}
{"year":2014,"label":"6","problem":"For a prime $p$, a subset $S$ of residues modulo $p$ is called a sum-free multiplicative subgroup of $\\mathbb{F}_{p}$ if - there is a nonzero residue $\\alpha$ modulo $p$ such that $S=\\left\\{1, \\alpha^{1}, \\alpha^{2}, \\ldots\\right\\}$ (all considered $\\bmod p$ ), and - there are no $a, b, c \\in S$ (not necessarily distinct) such that $a+b \\equiv c(\\bmod p)$. Prove that for every integer $N$, there is a prime $p$ and a sum-free multiplicative subgroup $S$ of $\\mathbb{F}_{p}$ such that $|S| \\geq N$.","solution":" We first prove the following general lemma. ## Lemma If $f, g \\in \\mathbb{Z}[X]$ are relatively prime nonconstant polynomials, then for sufficiently large primes $p$, they have no common root modulo $p$. $$ a(X) f(X)+b(X) g(X) \\equiv c $$ So, plugging in $X=r$ we get $p \\mid c$, so the set of permissible primes $p$ is finite. With this we can give the construction. ## Claim - Suppose that - $n$ is a positive integer with $n \\not \\equiv 0(\\bmod 3)$; - $p$ is a prime which is $1 \\bmod n$; and - $\\alpha$ is a primitive $n^{\\prime}$ th root of unity modulo $p$. Then $|S|=n$ and, if $p$ is sufficiently large in $n$, is also sum-free. $$ 1+\\alpha^{k} \\equiv \\alpha^{m} \\quad(\\bmod p) $$ for some integers $k, m \\in \\mathbb{Z}$. This means $(X+1)^{n}-1$ and $X^{n}-1$ have common root $X=\\alpha^{k}$. But $$ \\underset{\\mathbb{Z}[x]}{\\operatorname{gcd}}\\left((X+1)^{n}-1, X^{n}-1\\right)=1 \\quad \\forall n \\not \\equiv 0 \\quad(\\bmod 3) $$ because when $3 \\nmid n$ the two polynomials have no common complex roots. (Indeed, if $|\\omega|=|1+\\omega|=1$ then $\\omega=-\\frac{1}{2} \\pm \\frac{\\sqrt{3}}{2} i$.) Thus $p$ is bounded by the lemma, as desired."}
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