We are given an acute triangle $ABC$. Let $D$ be the point on its circumcircle such that $AD$ is a diameter. Suppose that points $K$ and $L$ lie on segments $AB$ and $AC$, respectively, and that $DK$ and $DL$ are tangent to circle $AKL$. Show that line $KL$ passes through the orthocenter of triangle $ABC$.

Quadrilateral $ABCD$ is circumscribed around the circle with centre $I$. Let points $M$ and $N$ be the midpoints of sides $AB$ and $CD$ respectively and let $\frac{IM}{AB} = \frac{IN}{CD}$. Prove that $ABCD$ is either a trapezoid or a parallelogram.

In a given plane $\varrho$ consider a convex quadrilateral $ABCD$ and denote $E=AC\cap BD.$ Moreover, consider a point $V\notin\varrho$. On rays $VA,VB,VC,VD$ find points $A',B',C',D'$ respectively such that $E,A',B',C',D'$ are coplanar and $A'B'C'D'$ is a parallelogram. Discuss conditions of solvability.

What is the largest possible distance between two points, one on the sphere of radius 19 with center $(-2, -10, 5)$ and the other on the sphere of radius 87 with center $(12, 8, -16)$?

Let $\omega$ be the circumcircle of $ABC$, and $KL$ be the diameter of $\omega$ passing through $M$ midpoint of $AB$ ($K,C$ lies on different sides of $AB$). A circle passing through $L$ and $M$ meets $CK$ at points $P$ and $Q$ ($Q$ lies on $KP$). Let $LQ$ meet the circumcircle of $KMQ$ again at $R$. Prove that $APBR$ is cyclic.

Point D is from AC of triangle ABC so that 2AD=DC. Let DE be perpendicular to BC and AE intersects BD at F. It is known that triangle BEF is equilateral. Find <ADB?

Suppose there are $18$ light houses on the Mexican gulf. Each of the lighthouses lightens an angle with size $20$ degrees. Prove that we can choose the directions of the lighthouses such that the whole gulf is lightened.

In triangle $ABC$ we have $|AB| \ne |AC|$. The bisectors of $\angle ABC$ and $\angle ACB$ meet $AC$ and $AB$ at $E$ and $F$, respectively, and intersect at I. If $|EI| = |FI|$ find the measure of $\angle BAC$.

We are given sufficiently many stones of the forms of a rectangle $2\times 1$ and square $1\times 1$. Let $n > 3$ be a natural number. In how many ways can one tile a rectangle $3 \times n$ using these stones, so that no two $2 \times 1$ rectangles have a common point, and each of them has the longer side parallel to the shorter side of the big rectangle?

We are given a $\Delta ABC$. Point $D$ on the circumscribed circle k is such that $CD$ is a symmedian in $\Delta ABC$. Let $X$ and $Y$ be on the rays $\overrightarrow{CB}$ and $\overrightarrow{CA}$, so that $CX=2CA$ and $CY=2CB$. Prove that the circle, tangent externally to $k$ and to the lines $CA$ and $CB$, is tangent to the circumscribed circle of $\Delta XDY$.

$BC$ is a segment, $M$ is point on $BC$, $A$ is a point such that $A, B, C$ are non-collinear. (i) Prove that if $M$ is the midpoint of $BC$, then $AB^2 + AC^2 = 2(AM^2 + BM^2).$ (ii) If there exists another point (except $M$) on segment $BC$ satisfying (i), find the region of point $A$ might occupy.

In a quadrilateral $ABCD$, a circle is inscribed that is tangent to the sides $AB, BC, CD$ and $DA$ at points $M, N, P$ and $Q$, respectively. If $(AM) (CP) = (BN) (DQ)$, prove that $ABCD$ is an cyclic quadrilateral.

Let $ABCDEF$ be a convex hexagon with area $S$ such that $AB \parallel DE$, $BC \parallel EF$, $CD \parallel FA$ holds, and whose all angles are obtuse and opposite sides are not the same length. Prove that the following inequality holds: $$A_{ABC} + A_{BCD} + A_{CDE} + A_{DEF} + A_{EFA} + A_{FAB} < S$$ , where $A_{XYZ}$ is the area of triangle $XYZ$

The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to $ \textbf{(A)} \ 50 \qquad \textbf{(B)} \ 52 \qquad \textbf{(C)} \ 54 \qquad \textbf{(D)} \ 56 \qquad \textbf{(E)} \ 58 \qquad$ [asy]unitsize(3mm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; for(int i=0; i<3; ++i) { for(int j=0; j<3; ++j) { draw(shift(3*i,3*j)*p); } }[/asy]

Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles. [asy] unitsize(1.5 cm); pair A, B, C, D, G, H; A = (0,0); B = (2,0); D = (0.5,1.5); C = B + D - A; G = reflect(A,B)*(C) + C - B; H = reflect(B,C)*(H) + A - B; draw(H--A--D--C--G); draw(interp(A,G,-0.1)--interp(A,G,1.1)); draw(interp(C,H,-0.1)--interp(C,H,1.1)); draw(D--G--H--cycle, dashed); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, E); dot("$D$", D, NW); dot("$G$", G, NE); dot("$H$", H, SE); [/asy]

Let $ABC$ be a scalene triangle, $\Gamma$ its circumscribed circle and $H$ the point where the altitudes of triangle $ABC$ meet. The circumference with center at $H$ passing through $A$ cuts $\Gamma$ at a second point $D$. In the same way, the circles with center at $H$ and passing through $B$ and $C$ cut $\Gamma$ again at points $E$ and $F$, respectively. Prove that $H$ is also the point in which the altitudes of the triangle $DEF$ meet.

Let $ ABC$ be a triangle and $ PQRS$ a square with $ P$ on $ AB$, $ Q$ on $ AC$, and $ R$ and $ S$ on $ BC$. Let $ H$ on $ BC$ such that $ AH$ is the altitude of the triangle from $ A$ to base $ BC$. Prove that: (a) $ \frac{1}{AH} \plus{}\frac{1}{BC}\equal{}\frac{1}{PQ}$ (b) the area of $ ABC$ is twice the area of $ PQRS$ iff $ AH\equal{}BC$

In the diagram below, square $ABCD$ with side length 23 is cut into nine rectangles by two lines parallel to $\overline{AB}$ and two lines parallel to $\overline{BC}$. The areas of four of these rectangles are indicated in the diagram. Compute the largest possible value for the area of the central rectangle. [asy] size(250); defaultpen (linewidth (0.7) + fontsize (10)); draw ((0,0)--(23,0)--(23,23)--(0,23)--cycle); label("$A$", (0,23), NW); label("$B$", (23, 23), NE); label("$C$", (23,0), SE); label("$D$", (0,0), SW); draw((0,6)--(23,6)); draw((0,19)--(23,19)); draw((5,0)--(5,23)); draw((12,0)--(12,23)); label("13", (17/2, 21)); label("111",(35/2,25/2)); label("37",(17/2,3)); label("123",(2.5,12.5));[/asy] [i]Proposed by Lewis Chen[/i]

[asy] size(120); real t = 2/sqrt(3); real x = 1 + sqrt(3); pair A = t*dir(90), D = x*A; pair B = t*dir(210), E = x*B; pair C = t*dir(330), F = x*C; draw(D--E--F--cycle); draw(Circle(A, 1)); draw(Circle(B, 1)); draw(Circle(C, 1)); //Credit to MSTang for the diagram[/asy] Each of the three circles in the adjoining figure is externally tangent to the other two, and each side of the triangle is tangent to two of the circles. If each circle has radius three, then the perimeter of the triangle is $\textbf{(A) }36+9\sqrt{2}\qquad\textbf{(B) }36+6\sqrt{3}\qquad\textbf{(C) }36+9\sqrt{3}\qquad\textbf{(D) }18+18\sqrt{3}\qquad \textbf{(E) }45$

We are given a combination lock consisting of $6$ rotating discs. Each disc consists of digits $0, 1, 2,\ldots , 9$ in that order (after digit $9$ comes $0$). Lock is opened by exactly one combination. A move consists of turning one of the discs one digit in any direction and the lock opens instantly if the current combination is correct. Discs are initially put in the position $000000$, and we know that this combination is not correct. [list] a) What is the least number of moves necessary to ensure that we have found the correct combination? b) What is the least number of moves necessary to ensure that we have found the correct combination, if we know that none of the combinations $000000, 111111, 222222, \ldots , 999999$ is correct?[/list] [i]Proposed by Ognjen Stipetić and Grgur Valentić[/i]

Let $ABC$ be a triangle, and let $P$ be a distinct point on the plane. Moreover, let $A'B'C'$ be a homothety of $ABC$ with ratio $2$ and center $P$, and let $O$ and $O'$ be the circumcenters of $ABC$ and $A'B'C'$, respectively. The circumcircles of $AB'C'$, $A'BC'$, and $A'B'C$ meet at points $X$, $Y$, and $Z$, different from $A'$, $B'$, and $C'$. In a similar way, the circumcircles of $A'BC$, $AB'C$, and $ABC'$ meet at $X'$, $Y'$, and $Z'$, different from $A$, $B$, $C$. Let $W$ and $W'$ be the circumcenters of $XYZ$ and $X'Y'Z'$, respectively. Prove that $OW$ is parallel to $O'W'$. [i]Proposed by Mateus Thimóteo, Brazil.[/i]

In the tetrahedron $ABCD$ , the dihedral angles at the $BC, CD$, and $DA$ edges are equal to $\alpha$, and for the remaining edges equal to $\beta$. Find the ratio $AB / CD$.

In triangle \( ABC \), let \( I \) be the \( A \)-excenter. Points \( X \) and \( Y \) are placed on line \( BC \) such that \( B \) is between \( X \) and \( C \), and \( C \) is between \( Y \) and \( B \). Moreover, \( B \) and \( C \) are the contact points of \( BC \) with the \( A \)-excircle of triangles \( BAY \) and \( AXC \), respectively. Let \( J \) be the \( A \)-excenter of triangle \( AXY \), and let \( H' \) be the reflection of the orthocenter of triangle \( ABC \) with respect to its circumcenter. Prove that \( I \), \( J \), and \( H' \) are collinear. Proposed by Ali Nazarboland

From point $M$ in triangle $ABC$ perpendiculars are dropped to each altitude. It can be shown that each of the line segments of altitudes, measured between the vertex and the foot of the perpendicular drawn to it, are of equal length. Prove that these lengths are each equal to the diameter of the circle inscribed in the triangle.

Let $ABC$ a triangle and $C$ it´s circuncircle. Let $D$ a point in arc $AB$ that not contain $A$, diferent of $B$ and $C$ such that $CD$ and $AB$ are not parallel. Let $E$ the intersection of $CD$ and $AB$ and $O$ the circumcircle of triangle $DBE$. Prove that the measure of $\angle OBE$ does not depend of the choice of $D$.