Points $ A, B, C$ and $ D$ are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square? [asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", (1,0), S); label("$D$", (0,1), W);[/asy] $ \textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 40$

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]

Find all right triangles with integer side lengths in which two congruent circles with prime radius can be inscribed such that they are externally tangent, both touch the hypotenuse, and each is tangent to another leg of the right triangle.

Four faces of tetrahedron $ABCD$ are congruent triangles whose angles form an arithmetic progression. If the lengths of the sides of the triangles are $a < b < c$, determine the radius of the sphere circumscribed about the tetrahedron as a function on $a, b$, and $c$. What is the ratio $c/a$ if $R = a \ ?$

How many non-congruent triangles have vertices at three of the eight points in the array shown below? [asy]dot((0,0)); dot((0,.5)); dot((.5,0)); dot((.5,.5)); dot((1,0)); dot((1,.5)); dot((1.5,0)); dot((1.5,.5));[/asy] $ \textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

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 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.

Let $ ABC$ is a triangle, let $ H$ is orthocenter of $ \triangle ABC$, let $ M$ is midpoint of $ BC$. Let $ (d)$ is a line perpendicular with $ HM$ at point $ H$. Let $ (d)$ meet $ AB, AC$ at $ E, F$ respectively. Prove that $ HE \equal{}HF$.

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$

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.

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?

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.

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.

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] $\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 40 \qquad \text{(E)}\ 64$

Prove that the sum of the face angles at each vertex of a tetrahedron is a straight angle if and only if the faces are congruent triangles.

Can we say that two triangles are congruent if the radii of the inscribed circles, the radii of the circumscribed circles, and the areas of these triangles are equal?

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.

Suppose $A_1, A_2, A_3, \ldots, A_{20}$is a 20 sides regular polygon. How many non-isosceles (scalene) triangles can be formed whose vertices are among the vertices of the polygon but the sides are not the sides of the polygon?

Two congruent equilateral triangles $ABC$ and $A'B'C'$ in the plane are given. Show that the midpoints of the segments $AA',BB', CC'$ either are collinear or form an equilateral triangle.

In triangle $\triangle ABC$, $AB<AC$, $M,D,E$ are the midpoints of $BC$, the arcs $BAC$ and $BC$ of the circumcircle of $\triangle ABC$ respectively. The incircle of $\triangle ABC$ touches $AB$ at $F$, $AE$ meets $BC$ at $G$, and the perpendicular to $AB$ at $B$ meets segment $EF$ at $N$. If $BN=EM$, prove that $DF$ is perpendicular to $FG$.

Given triangle $ABC$ . From the $P$ point three lines $(PA),(PB),(PC)$ are drawn. They cross the circumscribed circle at $A_1, B_1,C_1$ points respectively. It comes out that the $A_1B_1C_1$ triangle equals to the initial one. Prove that there are not more than eight such a points $P$ in a plane.

Two parallel sides of a quadrilateral have integer lengths. Prove that this quadrilateral can be cut into congruent triangles. (A Shapovalov)

Is it true that when all the faces of a tetrahedron have the same area, they are congruent triangles?

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?

We are given a triangle $ABC$ such that $\angle BAC < 90^{\circ}$. The point $D$ is on the opposite side of the line $AB$ to $C$ such that $|AD| = |BD|$ and $\angle ADB = 90^{\circ}$. Similarly, the point $E$ is on the opposite side of $AC$ to $B$ such that $|AE| = |CE|$ and $\angle AEC = 90^{\circ}$. The point $X$ is such that $ADXE$ is a parallelogram. Prove that $|BX| = |CX|$.