Let $ABC$ be an isosceles triangle with $AB=AC$. Let $D$ be a point on the segment $BC$ such that $BD=2DC$. Let $P$ be a point on the segment $AD$ such that $\angle BAC=\angle BPD$. Prove that $\angle BAC=2\angle DPC$.

Let $a\ge b$ and $c\ge d$ be real numbers. Prove that the equation \[(x+a)(x+d)+(x+b)(x+c)=0\] has real roots.

How many $6$-tuples $(a, b, c, d, e, f)$ of natural numbers are there for which $a>b>c>d>e>f$ and $a+f=b+e=c+d=30$ are simultaneously true?

Let $ABCD$ be a trapezium with $AB\parallel CD$. Let $P$ be a point on $AC$ such that $C$ is between $A$ and $P$; and let $X, Y$ be the midpoints of $AB, CD$ respectively. Let $PX$ intersect $BC$ in $N$ and $PY$ intersect $AD$ in $M$. Prove that $MN\parallel AB$.

Let $0<x<y<z<p$ be integers where $p$ is a prime. Prove that the following statements are equivalent: $(a) x^3\equiv y^3\pmod p\text{ and }x^3\equiv z^3\pmod p$ $(b) y^2\equiv zx\pmod p\text{ and }z^2\equiv xy\pmod p$

Let $f:\mathbb{R}\longrightarrow \mathbb{R}$ be a function such that $f(x+y+xy)=f(x)+f(y)+f(xy)$ for all $x, y\in\mathbb{R}$. Prove that $f$ satisfies $f(x+y)=f(x)+f(y)$ for all $x, y\in\mathbb{R}$.

The cirumcentre of the cyclic quadrilateral $ABCD$ is $O$. The second intersection point of the circles $ABO$ and $CDO$, other than $O$, is $P$, which lies in the interior of the triangle $DAO$. Choose a point $Q$ on the extension of $OP$ beyond $P$, and a point $R$ on the extension of $OP$ beyond $O$. Prove that $\angle QAP=\angle OBR$ if and only if $\angle PDQ=\angle RCO$.

Let $P(z)=a_nz^n+a_{n-1}z^{n-1}+\ldots+a_mz^m$ be a polynomial with complex coefficients such that $a_m\neq 0, a_n\neq 0$ and $n>m$. Prove that \[\text{max}_{|z|=1}\{|P(z)|\}\ge\sqrt{2|a_ma_n|+\sum_{k=m}^{n} |a_k|^2}\]

Determine the greatest positive integer $k$ that satisfies the following property: The set of positive integers can be partitioned into $k$ subsets $A_1, A_2, \ldots, A_k$ such that for all integers $n \geq 15$ and all $i \in \{1, 2, \ldots, k\}$ there exist two distinct elements of $A_i$ whose sum is $n.$ Proposed by Igor Voronovich, Belarus

Determine all sequences $(x_1,x_2,\ldots,x_{2011})$ of positive integers, such that for every positive integer $n$ there exists an integer $a$ with \[\sum^{2011}_{j=1} j x^n_j = a^{n+1} + 1\] Proposed by Warut Suksompong, Thailand

Show that there exist infinitely many pairs $(a, b)$ of positive integers with the property that $a+b$ divides $ab+1$, $a-b$ divides $ab-1$, $b>1$ and $a>b\sqrt{3}-1$

Suppose that $1000$ students are standing in a circle. Prove that there exists an integer $k$ with $100 \leq k \leq 300$ such that in this circle there exists a contiguous group of $2k$ students, for which the first half contains the same number of girls as the second half. Proposed by Gerhard Wöginger, Austria

Let $ABC$ be a triangle with $AB=AC$ and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D,B$ and $C$ at the point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A,E$ and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incentre of triangle $KAB$. Proposed by Jan Vonk, Belgium and Hojoo Lee, South Korea

Let $S$ be a nonempty set of primes satisfying the property that for each proper subset $P$ of $S$, all the prime factors of the number $\left(\prod_{p\in P}p\right)-1$ are also in $S$. Determine all possible such sets $S$.

In a $2\times n$ array we have positive reals s.t. the sum of the numbers in each of the $n$ columns is $1$. Show that we can select a number in each column s.t. the sum of the selected numbers in each row is at most $\frac{n+1}4$.

Click for solution grobber wrote: In a $2\times n$ array we have positive reals s.t. the sum of the numbers in each of the $n$ columns is $1$. Show that we can select a number in each column s.t. the sum of the selected numbers in each row is at most $\frac{n+1}4$. Here is my solution: We denote the numbers from the first row by $a_1$, $a_2$, ..., $a_n$ in increasing order: $a_1\leq a_2\leq ...\leq a_n$. Then, the corresponding numbers from the second row are obviously $1-a_1$, $1-a_2$, ..., $1-a_n$. Now, let $k$ be the largest index satisfying $a_1+a_2+...+a_k\leq\frac{n+1}{4}$. We WLOG assume that $k \neq n$ (since otherwise, we can simply select all numbers from the first row and no numbers from the second row, and we are done). Thus, $k+1 \leq n$, so that $a_{k+1}$ is well-defined. Moreover, $a_1+a_2+...+a_{k+1}>\frac{n+1}{4}$ (since $k$ was chosen to be the largest index satisfying $a_1+a_2+...+a_k\leq\frac{n+1}{4}$). Now, we are going to prove that $\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)\leq\frac{n+1}{4}$. In fact, recall that $a_1\leq a_2\leq ...\leq a_n$. Hence, the arithmetic mean of the numbers $a_{k+1}$, $a_{k+2}$, ..., $a_n$ is greater or equal than the number $a_{k+1}$ (the smallest of the numbers $a_{k+1}$, $a_{k+2}$, ..., $a_n$). In other words, $\frac{a_{k+1}+a_{k+2}+...+a_n}{n-k}\geq a_{k+1}$. On the other hand, recall again that $a_1\leq a_2\leq ...\leq a_n$. Hence, the arithmetic mean of the numbers $a_1$, $a_2$, ..., $a_{k+1}$ is smaller or equal than the number $a_{k+1}$ (the greatest of the numbers $a_1$, $a_2$, ..., $a_{k+1}$). In other words, $\frac{a_1+a_2+...+a_{k+1}}{k+1}\leq a_{k+1}$. Thus, $\frac{a_{k+1}+a_{k+2}+...+a_n}{n-k}\geq a_{k+1} \geq \frac{a_1+a_2+...+a_{k+1}}{k+1}$, and thus $a_{k+1}+a_{k+2}+...+a_n\geq\left(n-k\right)\cdot\frac{a_1+a_2+...+a_{k+1}}{k+1} \geq \left(n-k\right)\cdot\frac{\left(\frac{n+1}{4}\right)}{k+1}$ (since $n-k\geq 0$ and $a_1+a_2+...+a_{k+1}>\frac{n+1}{4}$). In other words, $a_{k+1}+a_{k+2}+...+a_n\geq \left(n-k\right)\cdot\frac{\left(\frac{n+1}{4}\right)}{k+1} = \frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}$. Hence, $\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)$ $=\left(n-k\right)-\left(a_{k+1}+a_{k+2}+...+a_n\right)\leq\left(n-k\right)-\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}$. Thus, in order to show that $\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)\leq\frac{n+1}{4}$, it will be enough to prove that $\left(n-k\right)-\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}\leq\frac{n+1}{4}$. This, however, is straightforward: $\left(n-k\right)-\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}\leq\frac{n+1}{4}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ n-k\leq\frac{n+1}{4}+\frac{\left(n+1\right)\left(n-k\right)}{4\left(k+1\right)}$ $\Longleftrightarrow\ \ \ \ \ n-k\leq\frac{n+1}{4}\left(1+\frac{n-k}{k+1}\right)\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ n-k\leq\frac{n+1}{4}\cdot\frac{n+1}{k+1}$ $\Longleftrightarrow\ \ \ \ \ n-k\leq\left(\frac{n+1}{2}\right)^2\cdot\frac{1}{k+1}\ \ \ \ \ \Longleftrightarrow\ \ \ \ \ \left(n-k\right)\left(k+1\right)\leq\left(\frac{n+1}{2}\right)^2$. But this is clear from AM-GM: $\left(n-k\right)\left(k+1\right)\leq\left(\frac{\left(n-k\right)+\left(k+1\right)}{2}\right)^2 = \left(\frac{n+1}{2}\right)^2$. So we have proved the inequality $\left(1-a_{k+1}\right)+\left(1-a_{k+2}\right)+...+\left(1-a_n\right)\leq\frac{n+1}{4}$. Together with $a_1+a_2+...+a_k\leq\frac{n+1}{4}$, this shows that if we choose the numbers $a_1$, $a_2$, ..., $a_k$ from the first row and the numbers $1-a_{k+1}$, $1-a_{k+2}$, ..., $1-a_n$ from the second row, then the sum of the chosen numbers in each row is $\leq\frac{n+1}{4}$. And the problem is solved. Unrelated to my solution above, here is a question on seshadri's solution: In the case $k\geq m$, how do we make sure that $k\geq n/2$ (we need this to speak of $a_{2k-n+1}$)? Maybe $m\leq n/2$ should be replaced by $m\geq n/2$ ? In this case, seshadri's solution seems to work (one minor mistake, though: the proof of $2k\leq n+m$ does not work in the special case when $n=m$, because we get a $\leq$ sign instead of the $<$ sign in $k-1/2\leq \sum a_i = \sum_{i=1}^m a_i+\sum_{i=m+1}^n a_i < m+(n-m)/2=(n+m)/2$; but fortunately there is an easy reason why $2k\leq n+m$ holds in the case when $n=m$). Darij

A quadrilateral $ABCD$ without parallel sides is circumscribed around a circle with centre $O$. Prove that $O$ is a point of intersection of middle lines of quadrilateral $ABCD$ (i.e. barycentre of points $A,\,B,\,C,\,D$) iff $OA\cdot OC=OB\cdot OD$.

Find the least positive integer that cannot be represented as $\frac{2^a-2^b}{2^c-2^d}$ for some positive integers $a, b, c, d$.

Let $\mathbb{R}^{+}$ denote the set of all positive real numbers. Find all functions $f:\mathbb{R}^{+}\longrightarrow \mathbb{R}$ satisfying \[f(x)+f(y)\le \frac{f(x+y)}{2}, \frac{f(x)}{x}+\frac{f(y)}{y}\ge \frac{f(x+y)}{x+y},\] for all $x, y\in \mathbb{R}^{+}$.