Let $T_1$ and $T_2$ be the points of tangency of the excircles of a triangle $ABC$ with its sides $BC$ and $AC$ respectively. It is known that the reflection of the incenter of $ABC$ across the midpoint of $AB$ lies on the circumcircle of triangle $CT_1T_2$. Find $\angle BCA$.

Let equilateral triangle $\vartriangle ABC$ be inscribed in a circle $\omega_1$ with radius $4$. Consider another circle $\omega_2$ with radius $2$ internally tangent to $\omega_1$ at $A$. Let $\omega_2$ intersect sides $\overline{AB}$ and $\overline{AC}$ at $D$ and $E$, respectively, as shown in the diagram. Compute the area of the shaded region. [img]https://cdn.artofproblemsolving.com/attachments/a/5/8255cdb8b041719d735607da8139aa4016375d.png[/img]

Altitudes $AA_1$ and $BB_1$ of triangle ABC meet in point $H$. Line $CH$ meets the semicircle with diameter $AB$, passing through $A_1, B_1$, in point $D$. Segments $AD$ and $BB_1$ meet in point $M$, segments $BD$ and $AA_1$ meet in point $N$. Prove that the circumcircles of triangles $B_1DM$ and $A_1DN$ touch.

On one of the sides of the $60$ degree angle with vertex $O$ a fixed point $F$ is marked. On the other side of the angle a point $A$ is chosen, and on the ray $OF$, but not the segment $OF$, a point $B$ such that $OA=FB$. On the segment $AB$ equilateral triangle $ABC$ and $ABD$ are built such that points $O$ and $C$ lie in the same half-plane with respect to $AB$, and $D$ in the other. a) Prove that the point $C$ does not depend on $A$. b) Prove that all points $D$ lie on a line.

The points $D, E$ on the side $AB$ of the triangle $ABC$ are such that $\frac{AD}{DB}\frac{AE}{EB} = \left(\frac{AC}{CB}\right)^2$. Show that $\angle ACD = \angle BCE$.

In an acute triangle $\triangle{ABC}$, let $AE$ and $BF$ be highs of it, and $H$ its orthocenter. The symmetric line of $AE$ with respect to the angle bisector of $\sphericalangle{A}$ and the symmetric line of $BF$ with respect to the angle bisector of $\sphericalangle{B}$ intersect each other on the point $O$. The lines $AE$ and $AO$ intersect again the circuncircle to $\triangle{ABC}$ on the points $M$ and $N$ respectively. Let $P$ be the intersection of $BC$ with $HN$; $R$ the intersection of $BC$ with $OM$; and $S$ the intersection of $HR$ with $OP$. Show that $AHSO$ is a paralelogram.

A trapezoid $ABCD$ is bicentral. The vertex $A$, the incenter $I$, the circumcircle $\omega$ and its center $O$ are given and the trapezoid is erased. Restore it using only a ruler.

We say two vertices of a simple polygon are [i]visible[/i] from each other if either they are adjacent, or the segment joining them is completely inside the polygon (except two endpoints that lie on the boundary). Find all positive integers $n$ such that there exists a simple polygon with $n$ vertices in which every vertex is visible from exactly $4$ other vertices. (A simple polygon is a polygon without hole that does not intersect itself.) [i]Proposed by Morteza Saghafian[/i]

A triangle $ ABC$ is given with $ \left|AB\right| > \left|AC\right|$. Line $ l$ tangents in a point $ A$ the circumcirle of $ ABC$. A circle centered in $ A$ with radius $ \left|AC\right|$ cuts $ AB$ in the point $ D$ and the line $ l$ in points $ E, F$ (such that $ C$ and $ E$ are in the same halfplane with respect to $ AB$). Prove that the line $ DE$ passes through the incenter of $ ABC$.

A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below, so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of the resulting trapezoid. [asy] size(150); defaultpen(linewidth(0.8)); draw(origin--(8,0)--(8,5)--(0,5)--cycle,linewidth(1)); draw(origin--(8/3,5)^^(16/3,5)--(8,0),linetype("4 4")); draw(origin--(4,3)--(8,0)^^(8/3,5)--(4,3)--(16/3,5),linetype("0 4")); label("$5$",(0,5/2),W); label("$8$",(4,0),S); [/asy]

Given convex $ n$-gon $ A_1\ldots A_n$. Let $ P_i$ ($ i \equal{} 1,\ldots , n$) be such points on its boundary that $ A_iP_i$ bisects the area of polygon. All points $ P_i$ don't coincide with any vertex and lie on $ k$ sides of $ n$-gon. What is the maximal and the minimal value of $ k$ for each given $ n$?

The inscribed circle $\omega$ of the triangle $ABC$ touches its sides $BC, CA$, and $AB$ at the points $D, E$, and $F$, respectively. Let the points $X, Y$, and $Z$ of the circle $\omega$ be diametrically opposite to the points $D, E$, and $F$, respectively. Line $AX, BY$ and $CZ$ intersect the sides $BC, CA$ and $AB$ at the points $D', E'$ and $F'$, respectively. On the segments $AD', BE'$ and $CF'$ noted the points $X', Y'$ and $Z'$, respectively, so that $D'X'= AX$, $E'Y' = BY$, $F'Z' = CZ$. Prove that the points $X', Y'$ and $Z'$ coincide.

There are given points with integer coordinate $(m,n)$ such that $1\leq m,n\leq 4$. Two players, Ana and Ben, are playing a game: First Ana color one of the coordinates with red one, then she pass the turn to Ben who color one of the remaining coordinates with yellow one, then this process they repeate again one after other. The game win the first player who can create a rectangle with same color of vertices and the length of sides are positive integer numbers, otherwise the game is a tie. Does there exist a strategy for any of the player to win the game?

Two squares and an isosceles triangle are positioned as shown in the figure (the up left vertex of the large square lies on the side of the triangle). Prove that points $A, B$ and $C$ are collinear. [img]https://cdn.artofproblemsolving.com/attachments/d/c/03515e40f74ced1f8243c11b3e610ef92137ac.png[/img]

Let $ABC$ be a triangle with circumcentre $O$ and incentre $I$ such that $O$ is different from $I$. Let $AK, BL, CM$ be the altitudes of $ABC$, let $U, V , W$ be the mid-points of $AK, BL, CM$ respectively. Let $D, E, F$ be the points at which the in-circle of $ABC$ respectively touches the sides $BC, CA, AB$. Prove that the lines $UD, VE, WF$ and $OI$ are concurrent.

Given a finite set $B$ of points in space, any two distances between the points of this set are different. Each point of the set $B$ is connected by a line segment to the closest point of the set $B$. This way we will get a set of sections, one of which (any chosen one) we paint red, all the remaining sections we paint green. Prove that there are two points of the set $B$ that cannot be connected by a line composed of green segments.

Find out whether there exist two congruent cubes with a common center such that each face of one cube has a common point with each face of the other.

Let $A_1A_2A_3\ldots A_n$ be a regular $n$-gon. Let $B_1$ and $B_{n-1}$ be the midpoints of its sides $A_1A_2$ and $A_{n-1}A_n$. Also, for every $i\in\left\{2,3,4,\ldots ,n-2\right\}$. Let $S$ be the point of intersection of the lines $A_1A_{i+1}$ and $A_nA_i$, and let $B_i$ be the point of intersection of the angle bisector bisector of the angle $\measuredangle A_iSA_{i+1}$ with the segment $A_iA_{i+1}$. Prove that $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=180^{\circ}$. [i]Proposed by Dusan Dukic, Serbia and Montenegro[/i]

In an acute triangle $ABC$, the circle with diameter $AC$ intersects $AB$ and $AC$ at $K$ and $L$ different from $A$ and $C$ respectively. The circumcircle of $ABC$ intersects the line $CK$ at the point $F$ different from $C$ and the line $AL$ at the point $D$ different from $A$. A point $E$ is choosen on the smaller arc of $AC$ of the circumcircle of $ABC$ . Let $N$ be the intersection of the lines $BE$ and $AC$ . If $AF^{2}+BD^{2}+CE^{2}=AE^{2}+CD^{2}+BF^{2}$ prove that $\angle KNB= \angle BNL$ .

The bisectors $AA_1$ and $CC_1$ of triangle $ABC$ intersect at point $I$. The circumscribed circles of triangles $AIC_1$ and $CIA_1$ intersect the arcs $AC$ and $BC$ (not containing points $B$ and $A$ respectively) of the circumscribed circle of triangle $ABC$ at points $C_2$ and $A_2$, respectively. Prove that lines $A_1A_2$ and $C_1C_2$ intersect on the circumscribed circle of triangle $ABC$.

Point $X$ is chosen inside the non trapezoid quadrilateral $ABCD$ such that $\angle AXD +\angle BXC=180$. Suppose the angle bisector of $\angle ABX$ meets the $D$-altitude of triangle $ADX$ in $K$, and the angle bisector of $\angle DCX$ meets the $A$-altitude of triangle $ADX$ in $L$.We know $BK \perp CX$ and $CL \perp BX$. If the circumcenter of $ADX$ is on the line $KL$ prove that $KL \perp AD$. Proposed by [i]Alireza Dadgarnia[/i]

The area and the perimeter of the triangle with sides $10,8,6$ are equal. Find all the triangles with integral sides whose area and perimeter are equal.

Let $ABCDEF$ be a convex hexagon satisfying $AC=DF$, $CE=FB$ and $EA=BD$. Prove that the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ intersect in one point.

Consider an acute triangle $ABC$ with $|AB| > |CA| > |BC|$. The vertices $D, E$, and $F$ are the base points of the altitudes from $A, B$, and $C$, respectively. The line through F parallel to $DE$ intersects $BC$ in $M$. The angular bisector of $\angle MF E$ intersects $DE$ in $N$. Prove that $F$ is the circumcentre of $\vartriangle DMN$ if and only if $B$ is the circumcentre of $\vartriangle FMN$.

In right triangle $ \triangle ACE$, we have $ AC \equal{} 12$, $ CE \equal{} 16$, and $ EA \equal{} 20$. Points $ B$, $ D$, and $ F$ are located on $ \overline{AC}$, $ \overline{CE}$, and $ \overline{EA}$, respectively, so that $ AB \equal{} 3$, $ CD \equal{} 4$, and $ EF \equal{} 5$. What is the ratio of the area of $ \triangle DBF$ to that of $ \triangle ACE$? [asy] size(200);defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair C = (0,0); pair E = (16,0); pair A = (0,12); pair F = waypoint(E--A,0.25); pair B = waypoint(A--C,0.25); pair D = waypoint(C--E,0.25); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("$A$",A,NW);label("$B$",B,W);label("$C$",C,SW);label("$D$",D,S);label("$E$",E,SE);label("$F$",F,NE); label("$3$",midpoint(A--B),W); label("$9$",midpoint(B--C),W); label("$4$",midpoint(C--D),S); label("$12$",midpoint(D--E),S); label("$5$",midpoint(E--F),NE); label("$15$",midpoint(F--A),NE); draw(A--C--E--cycle); draw(B--F--D--cycle);[/asy]$ \textbf{(A)}\ \frac {1}{4}\qquad \textbf{(B)}\ \frac {9}{25}\qquad \textbf{(C)}\ \frac {3}{8}\qquad \textbf{(D)}\ \frac {11}{25}\qquad \textbf{(E)}\ \frac {7}{16}$