Let $ABC$ and $AMN$ be two similar triangles with the same orientation, such that $AB=AC$, $AM=AN$ and having disjoint interiors. Let $O$ be the circumcenter of the triangle $MAB$. Prove that the points $O$, $C$, $N$, $A$ lie on the same circle if and only if the triangle $ABC$ is equilateral. Valentin Vornicu

Click for solution Tell me if anything is wrong with the following: We take a rotation of center $A$ and angle $\angle BAC$. This takes the circumcenter $O$ of $ABM$ into the circumcenter $P$ of $ACN$. Thus, $AO=AP$. The condition that $O,C,N,A$ are cyclic is equivalent to $PA=PO \, \, \Longleftrightarrow \, \, \triangle OAP$ equilateral $\, \, \Longleftrightarrow \, \, \angle BAC = \frac{\pi}{3}$. If the above is correct, then this problem is one of the most trivial problems given at a Romanian TST, in the last three years, at least.

Let $p$ a prime number, $p\geq 5$. Find the number of polynomials of the form \[ x^p + px^k + p x^l + 1, \quad k > l, \quad k, l \in \left\{1,2,\dots,p-1\right\}, \]which are irreducible in $\mathbb{Z}[X]$. Valentin Vornicu

Click for solution Note that, if $k-l$ is odd, then $x+1$ divides $x^p+1$ and also $x^k+x^l$, so $P(x)$ is not irreducible. Now suppose $k=l+2r$, $r$ a positive integer. Then let $Q(x)=P(x-1)$. We'll prove that $Q(x)$ is irreducible, from which it will follow that $P(x)$ is irreducible as well. We have $P(x-1)=(x-1)^p+p(x-1)^k+p(x-1)^l+1$. The binomial expansion of the first term produces multiples of $p$, except for the last term, which is $-1$ and cancels out with the fourth term, $1$. The binomial expansions of the middle two terms produce multiples of $p$ (obviously, since they are multiples of $p$ themselves). The last terms given by each expansion are either both $p$ or both $-p$, since $k, l$ are both even or both odd. Thus $P(x-1)=x^p+pxR(x) \pm 2p$, which is irreducible by Eisenstein's criterion. So we have to count the number of pairs $k, l$ with $k-l$ even, $p>k>l>0$. This gives $2 \times \binom{ \frac{p-1}{2}}{2}$.

Click for solution The main ideas (what exactly do you call a method¿) for mine are: If $p \nmid a,b$ is any prime and we have any $k$ with $p|a^{k}+k$, then $p|a^{k}+k|b^{k}+k \implies p|b^{k}+k$, thus $a^{k}+k \equiv 0 \equiv b^{k}+k \mod p \iff a^{k}\equiv b^{k}\mod p$. If we additionally would have $k \equiv 1 \mod (p-1)$, then Fermats Little Theorem gives $a \equiv a^{k}\equiv b^{k}\equiv b \mod p$. But for $k \equiv 1 \mod (p-1)$ we have also $a^{k}+k \equiv a+k$, so to get $p|a^{k}+k$ we just need $p|a+k$. So lets look for some $k \equiv 1 \mod (p-1)$ and $k \equiv-a \mod p$, which can be found by the Chinese Remainder Theorem or directly as $k=(p-1)(a+1)+1$. Till now we didn't use any special property of $p$ except that it doesn't divide $a,b$. But we know that for any such prime we have $p|(a-b)$ from chossing a suitable $k$ with $p|a^{k}+k$, thus if $a \neq b$ we would get a contradiction by choosing a big prime.

The real numbers $a_1,a_2,\dots,a_n$ are given such that $|a_i|\leq 1$ for all $i=1,2,\dots,n$ and $a_1+a_2+\cdots+a_n=0$. a) Prove that there exists $k\in\{1,2,\dots,n\}$ such that \[ |a_1+2a_2+\cdots+ka_k|\leq\frac{2k+1}{4}. \] b) Prove that for $n > 2$ the bound above is the best possible. Radu Gologan, Dan Schwarz

Click for solution a) Suppose the contrary. Let $A_k: =a_1+2a_2+\ldots+ka_k$. \[ A_k> \frac{2k+1}4 \] Suppose that $a_1\geq 0$ or else negate all $a_i$s. We say that all $A_k>0$. If it's not true consider the first $A_k$ that is negative. Then $A_{k-1}>0$. So we have $A_k<-\frac{2k+1}4$ and $A_{k-1}>\frac{2k-1}4$. So we have $-A_{k-1}<-\frac{2k-1}4$ and by adding this to $A_k<-\frac{2k+11}4$ we get \[ ka_k<-k \] which is not possible because $a_k\geq -1$. So all $A_k>0$. Now consider $c_i$s defined below: \[ c_n=\frac 1n=\frac 1n-0 \] \[ c_{n-k}=\frac 1{k(k+1)}=\frac 1k-\frac 1{k+1} (k\geq 1) \] Then by the right hand side of these equalities we can see that $c_n+c_{n-1}+\ldots+c_k=\frac 1k$ for $0\leq k\leq n$. Now consider the sum $\sum_{i=1}^n c_iA_i$ the coefficent of $a_{n-k}$ in this sum is \[ \sum_{i=n-k}^n (n-k)c_i=(n-k)\sum_{i=n-k}^n c_i=(n-k)\frac 1{n-k}=1 \] So the sum is equal to $0$ but every term of the sum is positive and that cannot happen! b) Just take $a_1=\frac 34$ and $a_2=\frac 14$ and $a_i=(-1)^i$ for every $i\geq 3$. Then we can check that $A_1=\frac 34$ and $A_2=\frac 54$ and for $i\geq 3$ using induction we can see that \[ A_i=(-1)^i\frac{2k+1}4 \]

Let $\{a_n\}_{n\geq 1}$ be a sequence with $a_1=1$, $a_2=4$ and for all $n>1$, \[ a_{n} = \sqrt{ a_{n-1}a_{n+1} + 1 } . \] a) Prove that all the terms of the sequence are positive integers. b) Prove that $2a_na_{n+1}+1$ is a perfect square for all positive integers $n$. Valentin Vornicu

Let $ABC$ be a triangle with $\angle B = 30^{\circ }$. We consider the closed disks of radius $\frac{AC}3$, centered in $A$, $B$, $C$. Does there exist an equilateral triangle with one vertex in each of the 3 disks? Radu Gologan, Dan Schwarz

Click for solution Denote by $O$ the centre and by $R$ the radius of the circle circumscribed to $\triangle{ABC}$. Since $B=30^\cdot$ we get that $AC=R$, hence $\triangle{OAC}$ is equilateral. Now, it is enough to use the following Lemma (from a Romanian TST 2002 ): Lemma. If $X$ is a variable point on the disk centred at $A$ (with radius $R_1$) and $Y$ is a variable point on the disk centered at $C$ (with radius $R_2$), then the locus of points $Z$ such that $\triangle{XYZ}$ is equilateral, is the disk centred at $O$ and with radius $R_1+R_2$ (and, of course, its reflection in $AC$, but that's redundant here). and the conclusion follows.

For which pairs of positive integers $(m,n)$ there exists a set $A$ such that for all positive integers $x,y$, if $|x-y|=m$, then at least one of the numbers $x,y$ belongs to the set $A$, and if $|x-y|=n$, then at least one of the numbers $x,y$ does not belong to the set $A$? Adapted by Dan Schwarz from A.M.M.

Click for solution IMHO, this is a very nice problem I found something similar, although I'm pretty late: if $k \in A$, then: $k+an+bm \in A$ for $a+b \equiv 0 \pmod 2$; else $k+an+bm \not\in A$. (to get a certain pair $(a,b)$, we first let $b$ oscillate between $0$ and $1$ until $a$ gets the right value, then we do the same thing with $b$) Thus, the only way to get a contradiction is to have $pn=qm$, where $p \not\equiv q \pmod 2$, which leads to the same answer as the one in the previous. Now, if $2^t \| m,n$, then we put $1,2,\ldots,2^t$ in $A$, the others will follow by the above observation applied to each of $1,2,\ldots,2^t$. (when applying it to $1$, we will decide whether some number $\equiv 1 \pmod{2^t}$ gets chosen or not). Note that they can't get into any conflict, because $|x-y|=n$, for example, implies $x \equiv y \pmod{2^t}$. And please tell me if there's something with the previous paragraph.

Let $x_i$, $1\leq i\leq n$ be real numbers. Prove that \[ \sum_{1\leq i<j\leq n}|x_i+x_j|\geq\frac{n-2}{2}\sum_{i=1}^n|x_i|. \] Discrete version by Dan Schwarz of a Putnam problem

Click for solution Here's mine. I think it's too short, maybe I'm wrong, but... maybe not \[ \{a_1,a_2,...,a_n\}=\{b_1,b_2,...,b_r\}\cup \{-c_1,-c_2,...,-c_s\} \] (devide set $\{a_1,a_2,...,a_n\}$ to negative and non-negative groups). With $r+s=n,b_i \ge 0,c_j \ge 0$. Denote \[ R=\sum_{i=1}^r b_i,S=\sum_{j=1}^s c_j \] So the inequality can be rewrite as follow \[ 2\sum_{i=1}^r\sum_{j=1}^s|a_i-b_j| \ge (s-r)(R-S) \] Assume that $s \ge r$, so we can suppose that $R \ge S$. In this case, notice that \[ \sum_{i=1}^r\sum_{j=1}^s|a_i-b_j| \ge\sum_{i=1}^r(sa_i-S)=sR-rS \] So we need to prove \[ 2(sR-rS ) \ge (s-r)(R-S) \] But it's immediately true because $s \ge r$ and $R \ge S$.

The circle of center $I$ is inscribed in the convex quadrilateral $ABCD$. Let $M$ and $N$ be points on the segments $AI$ and $CI$, respectively, such that $\angle MBN = \frac 12 \angle ABC$. Prove that $\angle MDN = \frac 12 \angle ADC$.

Click for solution This one is easy. Move $M$ on $AI$. Then map $BM \mapsto BN$ is projective from pencil of lines through $B$ onto it self, since it is rotation for angle ${1\over 2}\angle ABC$ around $B$. There for map $M \mapsto N$ is projective form line $AI$ to line $CI$. Now this map induces projective map $\rho: DM \mapsto DN$ from pencil of lines through $D$ onto itself. Now, when $M= A$ then $N= I$, so $\angle MDN = {1\over 2} \angle ADC$ and when $M= I$ then $N= C$ so again $\angle MDN = {1\over 2} \angle ADC$, and when $M$ is incenter of triangle $ABD$, $N$ is incenter of triangle $BCD$, from where we get again $\angle MDN = {1\over 2} \angle ADC$. This implys that map $\rho$ is actualy rotation around $D$ for fixed angle ${1\over 2}\angle ADC$ since every projective map is exactly determined with 3 points.

Let $A$ be point in the exterior of the circle $\mathcal C$. Two lines passing through $A$ intersect the circle $\mathcal C$ in points $B$ and $C$ (with $B$ between $A$ and $C$) respectively in $D$ and $E$ (with $D$ between $A$ and $E$). The parallel from $D$ to $BC$ intersects the second time the circle $\mathcal C$ in $F$. Let $G$ be the second point of intersection between the circle $\mathcal C$ and the line $AF$ and $M$ the point in which the lines $AB$ and $EG$ intersect. Prove that \[ \frac 1{AM} = \frac 1{AB} + \frac 1{AC}. \]

Click for solution $MG \times ME = MB \times MC = (AB-AM) \times (AC-AM)$ $MG \times ME = AB \times AC - AB \times AM - AC \times AM + AM^2$ Now if we show that $MG \times MF = AM^2$ then the problem is finished, and it is true since the triangles $AMG$ and $AME$ are similar, thus $\frac{MG}{AM} = \frac{AM}{ME}$ equivalent to $AM^2 = MG \times MF$. So we have desired $AB \times AC= AB \times AM + AC \times AM$ which is $\frac{1}{AM} = \frac{1}{AB} + \frac{1}{AC}$. the problem is finished.

Let $\gamma$ be the incircle in the triangle $A_0A_1A_2$. For all $i\in\{0,1,2\}$ we make the following constructions (all indices are considered modulo 3): $\gamma_i$ is the circle tangent to $\gamma$ which passes through the points $A_{i+1}$ and $A_{i+2}$; $T_i$ is the point of tangency between $\gamma_i$ and $\gamma$; finally, the common tangent in $T_i$ of $\gamma_i$ and $\gamma$ intersects the line $A_{i+1}A_{i+2}$ in the point $P_i$. Prove that a) the points $P_0$, $P_1$ and $P_2$ are collinear; b) the lines $A_0T_0$, $A_1T_1$ and $A_2T_2$ are concurrent.

Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2. \]

Let $r$ and $s$ be two rational numbers. Find all functions $f: \mathbb Q \to \mathbb Q$ such that for all $x,y\in\mathbb Q$ we have \[ f(x+f(y)) = f(x+r)+y+s. \]

Find all non-negative integers $m,n,p,q$ such that \[ p^mq^n = (p+q)^2 +1 . \]

Let $n>1$ be an integer. A set $S \subset \{ 0,1,2, \ldots, 4n-1\}$ is called rare if, for any $k\in\{0,1,\ldots,n-1\}$, the following two conditions take place at the same time (1) the set $S\cap \{4k-2,4k-1,4k, 4k+1, 4k+2 \}$ has at most two elements; (2) the set $S\cap \{4k+1,4k+2,4k+3\}$ has at most one element. Prove that the set $\{0,1,2,\ldots,4n-1\}$ has exactly $8 \cdot 7^{n-1}$ rare subsets.

Let $p$, $q$ be two integers, $q\geq p\geq 0$. Let $n \geq 2$ be an integer and $a_0=0, a_1 \geq 0, a_2, \ldots, a_{n-1},a_n = 1$ be real numbers such that \[ a_{k} \leq \frac{ a_{k-1} + a_{k+1} } 2 , \ \forall \ k=1,2,\ldots, n-1 . \] Prove that \[ (p+1) \sum_{k=1}^{n-1} a_k^p \geq (q+1) \sum_{k=1}^{n-1} a_k^q . \]

Let $n$ be a positive integer of the form $4k+1$, $k\in \mathbb N$ and $A = \{ a^2 + nb^2 \mid a,b \in \mathbb Z\}$. Prove that there exist integers $x,y$ such that $x^n+y^n \in A$ and $x+y \notin A$.

Let $m$ and $n$ be positive integers and $S$ be a subset with $(2^m-1)n+1$ elements of the set $\{1,2,3,\ldots, 2^mn\}$. Prove that $S$ contains $m+1$ distinct numbers $a_0,a_1,\ldots, a_m$ such that $a_{k-1} \mid a_{k}$ for all $k=1,2,\ldots, m$.

Let $x_1=1$, $x_2$, $x_3$, $\ldots$ be a sequence of real numbers such that for all $n\geq 1$ we have \[ x_{n+1} = x_n + \frac 1{2x_n} . \] Prove that \[ \lfloor 25 x_{625} \rfloor = 625 . \]

Let $ABC$ be an acute triangle with $AB \neq AC$. Let $D$ be the foot of the altitude from $A$ and $\omega$ the circumcircle of the triangle. Let $\omega_1$ be the circle tangent to $AD$, $BD$ and $\omega$. Let $\omega_2$ be the circle tangent to $AD$, $CD$ and $\omega$. Let $\ell$ be the interior common tangent to both $\omega_1$ and $\omega_2$, different from $AD$. Prove that $\ell$ passes through the midpoint of $BC$ if and only if $2BC = AB + AC$.