Prove that the intersection of a plane and a regular tetrahedron can be an obtuse-angled triangle and that the obtuse angle in any such triangle is always smaller than $ 120^{\circ}.$

In triangle $ABC$, points $K ,P$ are chosen on the side $AB$ so that $AK = BL$, and points $M,N$ are chosen on the side $BC$ so that $CN = BM$. Prove that $KN + LM \ge AC$. (I. Bogdanov)

Find the largest value of the area of the projection of the cylinder onto the plane if its radius is $r$ and its height is $h$ (orthogonal projection).

Point $O$ is the origin of a space. Points $A_1, A_2,\dots, A_n$ have nonnegative coordinates. Prove the following inequality: $$|\overrightarrow{OA_1}|+|\overrightarrow {OA_2}|+\dots+|\overrightarrow {OA_n}|\leq \sqrt{3}|\overrightarrow {OA_1}+\overrightarrow{OA_2}+\dots+\overrightarrow{OA_n}|$$ [I]Proposed by A. Khrabrov[/i]

The angle $\angle XOY =\alpha $ and the points $A$ and $B$ on OY are given such that $OA = a$ and $OB = b$ with $a > b$. A circle passes through the points $A$ and $B$ and is tangent to $OX$. a) Calculate the radius of that circle in terms of $a, b$ and $\alpha $. b) If $a$ and $b$ are constants and $\alpha $ varies, show that the minimum value of the radius of the circle is $\frac{a-b}{2}$.

Let $ABC$ be a triangle and $\Gamma$ the $A$- exscribed circle whose center is $J$ . Let $D$ and $E$ be the touchpoints of $\Gamma$ with the lines $AB$ and $AC$, respectively. Let $S$ be the area of the quadrilateral $ADJE$, Find the maximum value that $\frac{S}{AJ^2}$ has and when equality holds.

Mummy-trolley huts are located on a straight line at points with coordinates $x_1, x_2,...., x_n$. In this village are going to build $3$ stores $A, B$ and $C$, of which will be brought every day to all Moomin-trolls chocolates, bread and water. For the delivery of chocolate, the store takes the distance from the store to the hut, raised to the square; for bread delivery , take the distance from the store to the hut; for water delivery take distance $1$, if the distance is greater than $1$ km, but do not take anything otherwise. a) Where to build each of the stores so that the total cost of all Moomin-trolls for delivery wasthe smallest? b) Where to place the TV tower, if the fee for each Moomin-troll is the maximum distance from the TV tower to the farthest hut from it? c) How will the answer change if the Moomin-troll huts are not located in a straight line, and on the plane? [hide=original wording] На прямiй розташованi хатинки Мумi-тролей в точках з координатами x1, x2, . . . , xn. В цьому селi бираються побудувати 3 магазина A, B та C, з яких будуть кожен день привозити всiм Мумi-тролям шоколадки, хлiб та воду. За доставку шоколадки мага- зин бере вiдстань вiд магазину до хатинки, пiднесену до квадрату; за доставку хлiба – вiдстань вiд магазину до хатинки; за доставку води беруть 1, якщо вiдстань бiльша 1 км, та нiчого не беруть в супротивному випадку. 1. Де побудувати кожний з магазинiв, щоб загальнi витрати всiх Мумi-тролей на доставку були найменшими? 2. Де розташувати телевежу, якщо плата для кожного Мумi-троля – максимальна вiдстань вiд телевежi до самої вiддаленої вiд неї хатинки? 3. Як змiниться вiдповiдь, якщо хатинки Мумi-тролей розташованi не на прямiй, а на площинi?[/hide]

Let $M$ be an arbitrary interior point of a tetrahedron $ABCD$, and let $S_A,S_B,S_C,S_D$ be the areas of the faces $BCD,ACD,ABD,ABC$, respectively. Prove that $$S_A\cdot MA+S_B\cdot MB+S_C\cdot MC+S_D\cdot MD\ge9V,$$where $V$ is the volume of $ABCD$. When does equality hold?

Let $P$ be an interior point of triangle $ABC$. Let $a,b,c$ be the sidelengths of triangle $ABC$ and let $p$ be it's semiperimeter. Find the maximum possible value of $$ \min\left(\frac{PA}{p-a},\frac{PB}{p-b},\frac{PC}{p-c}\right)$$ taking into consideration all possible choices of triangle $ABC$ and of point $P$. by Elton Bojaxhiu, Albania

There are several triangles. From them a new triangle is obtained according to the following rule. The largest side of the new triangle is equal to the sum of the large sides of the data, the middle one is equal to the sum of the middle sides, and the smallest one is the sum of the smaller ones. Prove that if all the angles of these triangles were less than $a$, and $\phi$, where $\phi$ is the largest angle of the resulting triangle, then $\cos \phi \ge 1-\sin (a/2)$.

Given two acute triangles $\vartriangle ABC, \vartriangle DEF$. If $AB \ge DE, BC \ge EF$ and $CA \ge FD$, show that the area of $\vartriangle ABC$ is not less than the area of $\vartriangle DEF$

Let $ABCD$ be a tetrahedron that is not degenerate and not necessarily regular, where sides $AD$ and $BC$ have the same length $a$, sides $BD$ and $AC$ have the same length $b$, side $AB$ has length $c_1$ and the side $CD$ has length $c_2$. There is a point $P$ for which the sum of the distances to the vertices of the tetrahedron is minimal. Determine this sum depending on the quantities $a, b, c_1$ and $c_2$.

Let $ \,ABC\,$ be a triangle and $ \,P\,$ an interior point of $ \,ABC\,$. Show that at least one of the angles $ \,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is less than or equal to $ 30^{\circ }$.

There are three polygons and the area of each one is $3$. They are drawn inside a square of area $6$. Find the greatest value of $m$ such that among those three polygons, we can always find two polygons so that the area of their overlap is not less than $m$.

The parallelogram $ABCD$ has $AB=a,AD=1,$ $\angle BAD=A$, and the triangle $ABD$ has all angles acute. Prove that circles radius $1$ and center $A,B,C,D$ cover the parallelogram if and only \[a\le\cos A+\sqrt3\sin A.\]

Points $M,N$ and $P$ on the sides $BC, CA$ and $AB$ of $\vartriangle ABC$ are such that $\vartriangle MNP$ is acute. Denote by $h$ and $H$ the lengths of the shortest altitude of $\vartriangle ABC$ and the longest altitude of $\vartriangle MNP$. Prove that $h \le 2H$.

Let $\alpha,\beta,\gamma$ be positive real numbers such that $\alpha+\beta+\gamma < \pi$, $\alpha+\beta > \gamma$,$ \beta+\gamma > \alpha$, $\gamma + \alpha > \beta.$ Prove that with the segments of lengths $\sin \alpha, \sin \beta, \sin \gamma $ we can construct a triangle and that its area is not greater than \[A=\dfrac 18\left( \sin 2\alpha+\sin 2\beta+ \sin 2\gamma \right).\] [i]Proposed by Soviet Union[/i]

In an acute-angled triangle $ABC$, in which $AB<AC$, the point $M$ is the midpoint of the side $BC, K$ is the midpoint of the broken line segment $BAC$ . Prove that $\sqrt2 KM > AB$. (George Naumenko)

On the $BE$ side of a regular $ABE$ triangle, a $BCDE$ rhombus is built outside it. The segments $AC$ and $BD$ intersect at point $F$. Prove that $AF <BD$.

Determine the greatest possible value of the constant $c$ that satisfies the following condition: for any convex heptagon the sum of the lengthes of all it’s diagonals is greater than $cP$, where $P$ is the perimeter of the heptagon. (I. Zhuk)

Let ABC be a triangle and $M$ be an interior point. Prove that \[ \min\{MA,MB,MC\}+MA+MB+MC<AB+AC+BC.\]

Let $ d$ be the sum of the lengths of all the diagonals of a plane convex polygon with $ n$ vertices (where $ n>3$). Let $ p$ be its perimeter. Prove that: \[ n\minus{}3<{2d\over p}<\Bigl[{n\over2}\Bigr]\cdot\Bigl[{n\plus{}1\over 2}\Bigr]\minus{}2,\] where $ [x]$ denotes the greatest integer not exceeding $ x$.

One convex quadrilateral is inside another. Can it turn out that the sum of the lengths of the diagonals of the outer quadrilateral is less than the sum of the lengths of the diagonals of the inner?

The triangle $ ABC$ is inscribed in a circle. The interior bisectors of the angles $ A,B$ and $ C$ meet the circle again at $ A', B'$ and $ C'$ respectively. Prove that the area of triangle $ A'B'C'$ is greater than or equal to the area of triangle $ ABC.$

The insphere of a tetrahedron $ABCD$ touches the faces $ABC$, $BCD$, $CDA$, $DAB$ at $D^{\prime}$, $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ respectively. Denote by $S_{AB}$ the area of the triangle $AC^{\prime}B^{\prime}$. Define similarly $S_{AC}$, $S_{BC},$ $S_{AD}$, $S_{BD}$, $S_{CD}$. Prove that there exists a triangle with sidelengths $\sqrt{S_{AB}S_{CD}}$, $\sqrt{S_{AC}S_{BD}}$ , $\sqrt{S_{AD}S_{BC}}$. Proposed by: S.Arutyunyan