Let $ABC$ and $PQR$ be two triangles such that [list] [b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$. [b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$ [/list] Prove that $AB+AC=PQ+PR$.

Let $\alpha, \beta$ and $\gamma$ denote the angles of the triangle $ABC$. The perpendicular bisector of $AB$ intersects $BC$ at the point $X$, the perpendicular bisector of $AC$ intersects it at $Y$. Prove that $\tan(\beta) \cdot \tan(\gamma) = 3$ implies $BC= XY$ (or in other words: Prove that a sufficient condition for $BC = XY$ is $\tan(\beta) \cdot \tan(\gamma) = 3$). Show that this condition is not necessary, and give a necessary and sufficient condition for $BC = XY$.

Let $ABC$ be an acute triangle with $AB<AC$. Point $M{}$ from the side $(BC)$ is the foot of the bisector from the vertex $A{}$. The perpendicular bisector of the segment $[AM]$ intersects the side $(AC)$ in $E{}$, the side $(AB)$ in $D$ and the line $(BC)$ in $F{}$. Prove that $\frac{DB}{CE}=\frac{FB}{FC}=\left(\frac{AB}{AC}\right)^2$.

Let $ABC$ be a scalene (i.e. non-isosceles) triangle. Let $U$ be the center of the circumcircle of this triangle and $I$ the center of the incircle. Assume that the second point of intersection different from $C$ of the angle bisector of $\gamma = \angle ACB$ with the circumcircle of $ABC$ lies on the perpendicular bisector of $UI$. Show that $\gamma$ is the second-largest angle in the triangle $ABC$.

Janabel has a device that, when given two distinct points $U$ and $V$ in the plane, draws the perpendicular bisector of $UV$. Show that if three lines forming a triangle are drawn, Janabel can mark the orthocenter of the triangle using this device, a pencil, and no other tools. [i]Proposed by Fedir Yudin.[/i]

Let $M$ be the intersection of two diagonal, $AC$ and $BD$, of a rhombus $ABCD$, where angle $A<90^\circ$. Construct $O$ on segment $MC$ so that $OB<OC$ and let $t=\frac{MA}{MO}$, provided that $O \neq M$. Construct a circle that has $O$ as centre and goes through $B$ and $D$. Let the intersections between the circle and $AB$ be $B$ and $X$. Let the intersections between the circle and $BC$ be $B$ and $Y$. Let the intersections of $AC$ with $DX$ and $DY$ be $P$ and $Q$, respectively. Express $\frac{OQ}{OP}$ in terms of $t$.

Square $ABCD$ is shown in the diagram below. Points $E$, $F$, and $G$ are on sides $\overline{AB}$, $\overline{BC}$ and $\overline{DA}$, respectively, so that lengths $\overline{BE}$, $\overline{BF}$, and $\overline{DG}$ are equal. Points $H$ and $I$ are the midpoints of segments $\overline{EF}$ and $\overline{CG}$, respectively. Segment $\overline{GJ}$ is the perpendicular bisector of segment $\overline{HI}$. The ratio of the areas of pentagon $AEHJG$ and quadrilateral $CIHF$ can be written as $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] draw((0,0)--(50,0)--(50,50)--(0,50)--cycle); label("$A$",(0,50),NW); label("$B$",(50,50),NE); label("$C$",(50,0),SE); label("$D$",(0,0),SW); label("$E$",(0,100/3-1),W); label("$F$",(100/3-1,0),S); label("$G$",(20,50),N); label("$H$",((100/3-1)/2,(100/3-1)/2),SW); label("$I$",(35,25),NE); label("$J$",(((100/3-1)/2+35)/2,((100/3-1)/2+25)/2),S); draw((0,100/3-1)--(100/3-1,0)); draw((20,50)--(50,0)); draw((100/6-1/2,100/6-1/2)--(35,25)); draw((((100/3-1)/2+35)/2,((100/3-1)/2+25)/2)--(20,50)); [/asy]

Given Triangle $ABC$, $\angle B= 2 \angle C$, and $\angle A>90^\circ$. Let $M$ be midpoint of $BC$. Perpendicular of $AC$ at $C$ intersects $AB$ at $D$. Show $\angle AMB = \angle DMC$ [hide]If possible, don't use projective geometry[/hide]

A girls' camp is located $ 300$ rods from a straight road. On this road, a boys' camp is located $ 500$ rods from the girls' camp. It is desired to build a canteen on the road which shall be exactly the same distance from each camp. The distance of the canteen from each of the camps is: $ \textbf{(A)}\ 400\text{ rods} \qquad\textbf{(B)}\ 250\text{ rods} \qquad\textbf{(C)}\ 87.5\text{ rods} \qquad\textbf{(D)}\ 200\text{ rods}\\ \textbf{(E)}\ \text{none of these}$

Let $S_1$ be a semicircle with centre $O$ and diameter $AB$.A circle $C_1$ with centre $P$ is drawn, tangent to $S_1$, and tangent to $AB$ at $O$. A semicircle $S_2$ is drawn, with centre $Q$ on $AB$, tangent to $S_1$ and to $C_1$. A circle $C_2$ with centre $R$ is drawn, internally tangent to $S_1$ and externally tangent to $S_2$ and $C_1$. Prove that $OPRQ$ is a rectangle.

Let $K$ and $L$ be the points on the semicircle with diameter $AB$. Denote intersection of $AK$ and $AL$ as $T$ and let $N$ be the point such that $N$ is on segment $AB$ and line $TN$ is perpendicular to $AB$. If $U$ is the intersection of perpendicular bisector of $AB$ an $KL$ and $V$ is a point on $KL$ such that angles $UAV$ and $UBV$ are equal. Prove that $NV$ is perpendicular to $KL$.

A $30$-gon $A_1A_2\cdots A_{30}$ is inscribed in a circle of radius $2$. Prove that one can choose a point $B_k$ on the arc $A_kA_{k+1}$ for $1 \leq k \leq 29$ and a point $B_{30}$ on the arc $A_{30}A_1$, such that the numerical value of the area of the $60$-gon $A_1B_1A_2B_2 \dots A_{30}B_{30}$ is equal to the numerical value of the perimeter of the original $30$-gon.

In a plane a circle with radius $R$ and center $w$ and a line $\Lambda$ are given. The distance between $w$ and $\Lambda$ is $d, d > R$. The points $M$ and $N$ are chosen on $\Lambda$ in such a way that the circle with diameter $MN$ is externally tangent to the given circle. Show that there exists a point $A$ in the plane such that all the segments $MN$ are seen in a constant angle from $A.$

Let $D$, $E$ and $F$ be points in which incircle of triangle $ABC$ touches sides $BC$, $CA$ and $AB$, respectively, and let $I$ be a center of that circle.Furthermore, let $P$ be a foot of perpendicular from point $I$ to line $AD$, and let $M$ be midpoint of $DE$. If $\{N\}=PM\cap{AC}$, prove that $DN \parallel EF$

In triangle $ABC$, the perpendicular bisectors of sides $AB$ and $BC$ intersect side $AC$ at points $P$ and $Q$, respectively, with point $P$ lying on the segment $AQ$. Prove that the circumscribed circles of the triangles $PBC$ and $QBA$ intersect on the bisector of the angle $PBQ$.