Consider an acute-angled triangle $ABC$. Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$, and let $O$ be the circumcenter of triangle $ABC$. Assume that $\angle C \geq \angle B+30^{\circ}$. Prove that $\angle A+\angle COP < 90^{\circ}$.

Prove that for all positive real numbers $a,b,c$, \[ \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \geq 1. \]

Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.

Let $n$ be an odd integer greater than 1 and let $c_1, c_2, \ldots, c_n$ be integers. For each permutation $a = (a_1, a_2, \ldots, a_n)$ of $\{1,2,\ldots,n\}$, define $S(a) = \sum_{i=1}^n c_i a_i$. Prove that there exist permutations $a \neq b$ of $\{1,2,\ldots,n\}$ such that $n!$ is a divisor of $S(a)-S(b)$.

Let $ABC$ be a triangle with $\angle BAC = 60^{\circ}$. Let $AP$ bisect $\angle BAC$ and let $BQ$ bisect $\angle ABC$, with $P$ on $BC$ and $Q$ on $AC$. If $AB + BP = AQ + QB$, what are the angles of the triangle?

Let $a > b > c > d$ be positive integers and suppose that \[ ac + bd = (b+d+a-c)(b+d-a+c). \] Prove that $ab + cd$ is not prime.

Click for solution Solution by Iura: Lemma. If $ab=cd$ then there exist $m,n,p,q$ with $a=mn, b=pq, c=mp, d=nq$. Proof is obvious. Opening the brackets we get $a^2+c^2-ac=b^2+bd+d^2$ Multiplying by $4$ we get $(2a-c)^2+3c^2=(2b+d)^2+3d^2$ thus $3(c+d)(c-d)=(2b+d-2a+c)(2b+d+2a+c)$. Now using this Lemma we get either \[c+d=mn, c-d=pq, 2a+c-2b+d=3mp, 2b+d+2a+c=nq,\] or \[c+d=mn, c-d=pq, 2a+c-2b+d=mp, 2b+d+2a+c=3nq.\] For the first case, notice that $16(ab+cd)=(3p^2+n^2)(3m-q)(m+q)$, and it's easy to handle it now. The second case is analogous.