In the parallelogram $ABCD$, $\angle BAD =60 ^ \circ$. Let $E $ be the intersection point of the diagonals. The circle circumscribed to the triangle $ACD$ intersects the line $AB$ at the point $K$ (different from $A$), the line $BD$ at the point $P$ (different from $D$), and to the line $BC$ in $L$ (different from $C$). The line $EP$ intersects the circumscribed circle of the triangle $CEL$ at the points $E$ and $M$. Show that the triangles $KLM$ and $CAP$ are congruent.

Alex dissects a paper triangle into two triangles. Each minute after this he dissects one of obtained triangles into two triangles. After some time (at least one hour) it appeared that all obtained triangles were congruent. Find all initial triangles for which this is possible.

Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially [b]bolded[/b]. What percent of the total area is partially bolded? [asy] import graph; size(7.01cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.42,xmax=14.59,ymin=-10.08,ymax=5.26; pair A=(0,0), B=(4,0), C=(0,4), D=(4,4), F=(2,0), G=(3,0), H=(1,4), I=(2,4), J=(3,4), K=(0,-2), L=(4,-2), M=(0,-6), O=(0,-4), P=(4,-4), Q=(2,-2), R=(2,-6), T=(6,4), U=(10,0), V=(10,4), Z=(10,2), A_1=(8,4), B_1=(8,0), C_1=(6,-2), D_1=(10,-2), E_1=(6,-6), F_1=(10,-6), G_1=(6,-4), H_1=(10,-4), I_1=(8,-2), J_1=(8,-6), K_1=(8,-4); draw(C--H--(1,0)--A--cycle,linewidth(1.6)); draw(M--O--Q--R--cycle,linewidth(1.6)); draw(A_1--V--Z--cycle,linewidth(1.6)); draw(G_1--K_1--J_1--E_1--cycle,linewidth(1.6)); draw(C--D); draw(D--B); draw(B--A); draw(A--C); draw(H--(1,0)); draw(I--F); draw(J--G); draw(C--H,linewidth(1.6)); draw(H--(1,0),linewidth(1.6)); draw((1,0)--A,linewidth(1.6)); draw(A--C,linewidth(1.6)); draw(K--L); draw((4,-6)--L); draw((4,-6)--M); draw(M--K); draw(O--P); draw(Q--R); draw(O--Q); draw(M--O,linewidth(1.6)); draw(O--Q,linewidth(1.6)); draw(Q--R,linewidth(1.6)); draw(R--M,linewidth(1.6)); draw(T--V); draw(V--U); draw(U--(6,0)); draw((6,0)--T); draw((6,2)--Z); draw(A_1--B_1); draw(A_1--Z); draw(A_1--V,linewidth(1.6)); draw(V--Z,linewidth(1.6)); draw(Z--A_1,linewidth(1.6)); draw(C_1--D_1); draw(D_1--F_1); draw(F_1--E_1); draw(E_1--C_1); draw(G_1--H_1); draw(I_1--J_1); draw(G_1--K_1,linewidth(1.6)); draw(K_1--J_1,linewidth(1.6)); draw(J_1--E_1,linewidth(1.6)); draw(E_1--G_1,linewidth(1.6)); dot(A,linewidth(1pt)+ds); dot(B,linewidth(1pt)+ds); dot(C,linewidth(1pt)+ds); dot(D,linewidth(1pt)+ds); dot((1,0),linewidth(1pt)+ds); dot(F,linewidth(1pt)+ds); dot(G,linewidth(1pt)+ds); dot(H,linewidth(1pt)+ds); dot(I,linewidth(1pt)+ds); dot(J,linewidth(1pt)+ds); dot(K,linewidth(1pt)+ds); dot(L,linewidth(1pt)+ds); dot(M,linewidth(1pt)+ds); dot((4,-6),linewidth(1pt)+ds); dot(O,linewidth(1pt)+ds); dot(P,linewidth(1pt)+ds); dot(Q,linewidth(1pt)+ds); dot(R,linewidth(1pt)+ds); dot((6,0),linewidth(1pt)+ds); dot(T,linewidth(1pt)+ds); dot(U,linewidth(1pt)+ds); dot(V,linewidth(1pt)+ds); dot((6,2),linewidth(1pt)+ds); dot(Z,linewidth(1pt)+ds); dot(A_1,linewidth(1pt)+ds); dot(B_1,linewidth(1pt)+ds); dot(C_1,linewidth(1pt)+ds); dot(D_1,linewidth(1pt)+ds); dot(E_1,linewidth(1pt)+ds); dot(F_1,linewidth(1pt)+ds); dot(G_1,linewidth(1pt)+ds); dot(H_1,linewidth(1pt)+ds); dot(I_1,linewidth(1pt)+ds); dot(J_1,linewidth(1pt)+ds); dot(K_1,linewidth(1pt)+ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy] $ \textbf{(A)}12\frac 12\qquad\textbf{(B)}20\qquad\textbf{(C)}25\qquad\textbf{(D)}33 \frac 13\qquad\textbf{(E)}37\frac 12 $

Determine all triangles that can be split into two congruent pieces by one cut. A cut consists of segments $P_1P_2$, $P_2P_3$, . . . , $P_{n-1}P_n$ where points $P_1, P_2, . . . , P_n$ are distinct, points $P_1$ and $P_n$ lie on the perimeter of the triangle and the rest of the points lie in the interior of the triangle such that the segments are disjoint except for the endpoints.

An isosceles right triangle with legs of length $8$ is partitioned into $16$ congruent triangles as shown. The shaded area is [asy] for (int a=0; a <= 3; ++a) { for (int b=0; b <= 3-a; ++b) { fill((a,b)--(a,b+1)--(a+1,b)--cycle,grey); } } for (int c=0; c <= 3; ++c) { draw((c,0)--(c,4-c),linewidth(1)); draw((0,c)--(4-c,c),linewidth(1)); draw((c+1,0)--(0,c+1),linewidth(1)); } label("$8$",(2,0),S); label("$8$",(0,2),W); [/asy] $\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 64$

Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron. $\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2$

The point $E$ is taken inside the square $ABCD$, the point $F$ is taken outside, so that the triangles $ABE$ and $BCF$ are congruent . Find the angles of the triangle $ABE$, if it is known that$EF$ is equal to the side of the square, and the angle $BFD$ is right.

The segments connecting the interior point of a convex non-sided $n$-gon with its vertices divide the $n$-gon into $n$ congruent triangles. For what is the smallest $n$ that is possible?

Given a quadrilateral, the midpoints $A, B, C, D$ of its consecutive sides, and the midpoints of its diagonals, $P$ and $Q$. Prove that $\vartriangle BCP = \vartriangle ADQ$.

Regular hexagon $ ABCDEF$ has vertices $ A$ and $ C$ at $ (0,0)$ and $ (7,1)$, respectively. What is its area? $ \textbf{(A) } 20\sqrt {3} \qquad \textbf{(B) } 22\sqrt {3} \qquad \textbf{(C) } 25\sqrt {3} \qquad \textbf{(D) } 27\sqrt {3} \qquad \textbf{(E) } 50$

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.

Let $ABC$ be a triangle with circumcircle $c$ and circumcenter $O$, and let $D$ be a point on the side $BC$ different from the vertices and the midpoint of $BC$. Let $K$ be the point where the circumcircle $c_1$ of the triangle $BOD$ intersects $c$ for the second time and let $Z$ be the point where $c_1$ meets the line $AB$. Let $M$ be the point where the circumcircle $c_2$ of the triangle $COD$ intersects $c$ for the second time and let $E$ be the point where $c_2$ meets the line $AC$. Finally let $N$ be the point where the circumcircle $c_3$ of the triangle $AEZ$ meets $c$ again. Prove that the triangles $ABC$ and $NKM$ are congruent.

Let $A_r,B_r, C_r$ be points on the circumference of a given circle $S$. From the triangle $A_rB_rC_r$, called $\Delta_r$, the triangle $\Delta_{r+1}$ is obtained by constructing the points $A_{r+1},B_{r+1}, C_{r+1} $on $S$ such that $A_{r+1}A_r$ is parallel to $B_rC_r$, $B_{r+1}B_r$ is parallel to $C_rA_r$, and $C_{r+1}C_r$ is parallel to $A_rB_r$. Each angle of $\Delta_1$ is an integer number of degrees and those integers are not multiples of $45$. Prove that at least two of the triangles $\Delta_1,\Delta_2, \ldots ,\Delta_{15}$ are congruent.

The diagonals of a convex quadrilateral dissect it into four similar triangles. Prove that this quadrilateral can also be dissected into two congruent triangles.

How many non-congruent triangles with perimeter $ 7$ have integer side lengths? $ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

An acute triangle $ABC$ is inscribed in a circle of radius 1 with centre $O;$ all the angles of $ABC$ are greater than $45^\circ.$ $B_{1}$ is the foot of perpendicular from $B$ to $CO,$ $B_{2}$ is the foot of perpendicular from $B_{1}$ to $AC.$ Similarly, $C_{1}$ is the foot of perpendicular from $C$ to $BO,$ $C_{2}$ is the foot of perpendicular from $C_{1}$ to $AB.$ The lines $B_{1}B_{2}$ and $C_{1}C_{2}$ intersect at $A_{3}.$ The points $B_{3}$ and $C_{3}$ are defined in the same way. Find the circumradius of triangle $A_{3}B_{3}C_{3}.$ [i]Proposed by F.Bakharev, F.Petrov[/i]

Given an acute triangle $ PBC$ with $ PB\neq PC.$ Points $ A,D$ lie on $ PB,PC,$ respectively. $ AC$ intersects $ BD$ at point $ O.$ Let $ E,F$ be the feet of perpendiculars from $ O$ to $ AB,CD,$ respectively. Denote by $ M,N$ the midpoints of $ BC,AD.$ $ (1)$: If four points $ A,B,C,D$ lie on one circle, then $ EM\cdot FN \equal{} EN\cdot FM.$ $ (2)$: Determine whether the converse of $ (1)$ is true or not, justify your answer.

Inside the convex pentagon $ABCDE$, a point $O$ was chosen, and it turned out that all five triangles $AOB$, $BOC$, $COD$, $DOE$ and $EOA$ are congrunet to each other. Prove that these triangles are isosceles or right-angled.

Let $ABCD$ be a quadrilateral with $AC=20$, $AD=16$. We take point $P$ on segment $CD$ so that triangle $ABP$ and $ACD$ are congruent. If the area of triangle $APD$ is $28$, find the area of triangle $BCP$. Note that $XY$ expresses the length of segment $XY$.

Find all natural numbers $k$ for which there exists a set $M$ of ten real numbers such that there are exactly $k$ pairwise non-congruent triangles whose side lengths are three (not necessarily distinct) elements of $M$.

Four points lie on a plane such that no three of them are collinear. Consider the four triangles formed by taking any three points at a time. If the inradii of these four triangles are all equal, prove that the four triangles are congruent.

Let $ABC$ be a triangle such that $AB <AC$, $\omega$ its inscribed circle and $\Gamma$ its circumscribed circle. Let also $\omega_b$ be the excircle relative to vertex $B$, then $B'$ is the point of tangency between $\omega_b$ and $(AC)$. Similarly, let the circle $\omega_c$ be the excircle exinscribed relative to vertex $C$, then $C'$ is the point of tangency between $\omega_c$ and $(AB)$. Finally, let $I$ be the center of $\omega$ and $X$ the point of $\Gamma$ such that $\angle XAI$ is a right angle. Prove that the triangles $XBC'$ and $XCB'$ are congruent.

In rectangle $ ADEH$, points $ B$ and $ C$ trisect $ \overline{AD}$, and points $ G$ and $ F$ trisect $ \overline{HE}$. In addition, $ AH \equal{} AC \equal{} 2.$ What is the area of quadrilateral $ WXYZ$ shown in the figure? [asy]defaultpen(linewidth(0.7));pointpen=black; pathpen=black; size(7cm); pair A,B,C,D,E,F,G,H,W,X,Y,Z; A=(0,2); B=(1,2); C=(2,2); D=(3,2); H=(0,0); G=(1,0); F=(2,0); E=(3,0); D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,NE); D('F',F,NE); D('G',G,NW); D('H',H,NW); D(A--F); D(B--E); D(D--G); D(C--H); Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); D('W',W,N); D('X',X,plain.E); D('Y',Y,S); D('Z',Z,plain.W); D(A--D--E--H--cycle);[/asy] $ \textbf{(A) } \frac 12 \qquad \textbf{(B) } \frac {\sqrt {2}}2\qquad \textbf{(C) } \frac {\sqrt {3}}2 \qquad \textbf{(D) } \frac {2\sqrt {2}}3 \qquad \textbf{(E) } \frac {2\sqrt {3}}3$

Prove that the sum of the plane angles at each of the vertices of a given tetrahedron is $ 180^{\circ} $ if and only if all its faces are congruent.

A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^\circ$ in such a way that $BD=BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incenter of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.