Points $A$ and $B$ are on a circle $\omega$ with center $O$ such that $\tfrac{\pi}{3}< \angle AOB <\tfrac{2\pi}{3}$. Let $C$ be the circumcenter of the triangle $AOB$. Let $l$ be a line passing through $C$ such that the angle between $l$ and the segment $OC$ is $\tfrac{\pi}{3}$. $l$ cuts tangents in $A$ and $B$ to $\omega$ in $M$ and $N$ respectively. Suppose circumcircles of triangles $CAM$ and $CBN$, cut $\omega$ again in $Q$ and $R$ respectively and theirselves in $P$ (other than $C$). Prove that $OP\perp QR$. [i]Proposed by Mehdi E'tesami Fard, Ali Khezeli[/i]

Let the $ABCD$ be a quadrilateral without parallel sides, inscribed in a circle. Let $P$ and $Q$ be the intersection points between the lines containing the quadrilateral opposite sides. Show that the bisectors to the angles at $P$ and $Q$ are parallel to the bisectors of the angles at the intersection point of the diagonals of the quadrilateral.

In acute triangle $ABC$ let $H$ be its orthocenter and $I$ be its incenter. Let $D$ be the projection of point $I$ onto the line $BC$ and $E$ be the reflection of point $A$ in point $I$. Further, let $F$ be the projection of point $H$ onto the line $ED$. Prove that points $B, H, F$ and $C$ lie on circle.

Three squares $ABB_1B_2,BCC_1C_2,CAA_1A_2$ are constructed in the exterior of a triangle $ABC$. In the exterior of these squares, another three squares $A_1B_2B_3B_4,B_1C_2C_3C_4,C_1A_2A_3A_4$ are constructed. Prove that the area of a triangle with sides $C_3A_4,A_3B_4,B_3C_4$ is $16$ times the area of $\triangle ABC$.

In the rectangle $ABCD$ with $AB> BC$, the perpendicular bisecotr of $AC$ intersects the side $CD$ at point $E$. The circle with the center at point $E$ and the radius $AE$ intersects again the side $AB$ at point $F$. If point $O$ is the orthogonal projection of point $C$ on the line $EF$, prove that points $B, O$ and $D$ are collinear.

A rectangular prism with positive integer side lengths formed by stacking unit cubes is called [i]bipartisan[/i] if the same number of unit cubes can be seen on the surface as those which cannot be seen on the surface. How many non-congruent bipartisan rectangular prisms are there? [i]2018 CCA Math Bonanza Team Round #8[/i]

Let $n$ be an integer greater than $2$. Prove that the $n$th power of the length of the hypotenuse of a right triangle is greater than the sum of the $n$th powers of the lengths of the legs.

The polygon $M{}$ is bicentric. The polygon $P{}$ has vertices at the points of contact of the sides of $M{}$ with the inscribed circle. The polygon $Q{}$ is formed by the external bisectors of the angles of $M{}.$ Prove that $P{}$ and $Q{}$ are homothetic.

Determine all positive integers $m$ satisfying the condition that there exists a unique positive integer $n$ such that there exists a rectangle which can be decomposed into $n$ congruent squares and can also be decomposed into $m+n$ congruent squares.

Let us consider a triangle $\Delta{PQR}$ in the co-ordinate plane. Show for every function $f: \mathbb{R}^2\to \mathbb{R}\;,f(X)=ax+by+c$ where $X\equiv (x,y) \text{ and } a,b,c\in\mathbb{R}$ and every point $A$ on $\Delta PQR$ or inside the triangle we have the inequality: \begin{align*} & f(A)\le \text{max}\{f(P),f(Q),f(R)\} \end{align*}

Given four distinct parallel planes, prove that there exists a regular tetrahedron with a vertex on each plane.

Several cells are marked in a $100 \times 100$ table. Vasya wants to split the square into several rectangles such that each rectangle does not contain more than two marked cells and there are at most $k$ rectangles containing less than two cells. What is the smallest $k$ such that Vasya will certainly be able to do this?

What is the smallest number of $5\times 1$ tiles which must be placed on a $31\times 5$ rectangle (each covering exactly $5$ unit squares) so that no further tiles can be placed? How many different ways are there of placing the minimal number (so that further tiles are blocked)? What are the answers for a $52\times 5$ rectangle?

Let $ABCD$ be a parallelogram. The line $\Delta$ meets the lines $AB, BC, CD$ and $DA$ at $M, N, P$ and $Q,$ respectively. Let $R$ be the intersection point of the lines $AB,DN$ and let $S$ be intersection point of the lines $AD, BP.$ Prove that $RS \parallel \Delta.$ [asy] import graph; size(400); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); pen qqzzcc = rgb(0,0.6,0.8); pen wwwwff = rgb(0.4,0.4,1); draw((2,2)--(6,2),qqzzcc+linewidth(1.6pt)); draw((6,2)--(4,0),qqzzcc+linewidth(1.6pt)); draw((-1.95,(+12-2*-1.95)/2)--(12.24,(+12-2*12.24)/2),qqzzcc+linewidth(1.6pt)); draw((-1.95,(-0+3*-1.95)/3)--(12.24,(-0+3*12.24)/3),qqzzcc+linewidth(1.6pt)); draw((-1.95,(-0-0*-1.95)/6)--(12.24,(-0-0*12.24)/6),qqzzcc+linewidth(1.6pt)); draw((4,0)--(4,4),wwwwff+linewidth(1.2pt)+linetype("3pt 3pt")); draw((2,2)--(8.14,0),wwwwff+linewidth(1.2pt)+linetype("3pt 3pt")); draw((-1.95,(+32.56-4*-1.95)/4.14)--(12.24,(+32.56-4*12.24)/4.14),qqzzcc+linewidth(1.6pt)); dot((0,0),ds); label("$A$", (0,-0.3),NE*lsf); dot((4,0),ds); label("$B$", (4.02,-0.33),NE*lsf); dot((2,2),ds); label("$D$", (1.81,2.07),NE*lsf); dot((6,2),ds); label("$C$", (6.16,2.08),NE*lsf); dot((3,3),ds); label("$Q$", (2.97,3.22),NE*lsf); dot((5,1),ds); label("$N$", (4.99,1.19),NE*lsf); label("$\Delta$", (1.7,3.76),NE*lsf); dot((6,0),ds); label("$M$", (5.9,-0.33),NE*lsf); dot((4,2),ds); label("$P$", (4.02,2.08),NE*lsf); dot((4,4),ds); label("$S$", (3.94,4.12),NE*lsf); dot((8.14,0),ds); label("$E$", (8.2,0.09),NE*lsf); clip((-1.95,-6.96)--(-1.95,4.99)--(12.24,4.99)--(12.24,-6.96)--cycle); [/asy]

Let $O$ be a point inside a triangle $ABC$ and let the lines $AO,BO$, and $CO$ meet sides $BC,CA$, and $AB$ at points $A_1,B_1$, and $C_1$, respectively. If $AA_1$ is the longest among the segments $AA_1,BB_1,CC_1$, prove that $$OA_1+OB_1+OC_1\le AA_1.$$

Mark six points in a plane so that any three of them are vertices of a nondegenerate isosceles triangle.

In triangle $ABC$, $M$ is the intersection point of the medians, $O$ is the center of the inscribed circle. Prove that if the line $OM$ is parallel to the side $BC$, then the point $O$ is equidistant from the sides $AB$ and $AC$.

[b]p1.[/b] Two players play the following game. On the lowest left square of an $8 \times 8$ chessboard there is a rook (castle). The first player is allowed to move the rook up or to the right by an arbitrary number of squares. The second layer is also allowed to move the rook up or to the right by an arbitrary number of squares. Then the first player is allowed to do this again, and so on. The one who moves the rook to the upper right square wins. Who has a winning strategy? [b]p2.[/b] Find the smallest positive whole number that ends with $17$, is divisible by $17$, and the sum of its digits is $17$. [b]p3.[/b] Three consecutive $2$-digit numbers are written next to each other. It turns out that the resulting $6$-digit number is divisible by $17$. Find all such numbers. [b]p4.[/b] Let $ABCD$ be a convex quadrilateral (a quadrilateral $ABCD$ is called convex if the diagonals $AC$ and $BD$ intersect). Suppose that $\angle CBD = \angle CAB$ and $\angle ACD = \angle BDA$ . Prove that $\angle ABC = \angle ADC$. [b]p5.[/b] A circle of radius $1$ is cut into four equal arcs, which are then arranged to make the shape shown on the picture. What is its area? [img]https://cdn.artofproblemsolving.com/attachments/f/3/49c3fe8b218ab0a5378ecc635b797a912723f9.png[/img] PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

Let $ABC$ be a right isosceles triangle with right angle at $A$. Let $E$ and $F$ be points on A$B$ and $AC$ respectively such that $\angle ECB = 30^o$ and $\angle FBC = 15^o$. Lines $CE$ and $BF$ intersect at $P$ and line $AP$ intersects side $BC$ at $D$. Calculate the measure of angle $\angle FDC$.

Let $D$ be a point inside of equilateral $\triangle ABC$, and $E$ be a point outside of equilateral $\triangle ABC$ such that $m(\widehat{BAD})=m(\widehat{ABD})=m(\widehat{CAE})=m(\widehat{ACE})=5^\circ$. What is $m(\widehat{EDC})$ ? $ \textbf{(A)}\ 45^\circ \qquad\textbf{(B)}\ 40^\circ \qquad\textbf{(C)}\ 35^\circ \qquad\textbf{(D)}\ 30^\circ \qquad\textbf{(E)}\ 25^\circ $

The median $AM$ is drawn in the triangle $ABC$ ($AB \ne AC$). The point $P$ is the foot of the perpendicular drawn on the segment $AM$ from the point $B$. On the segment $AM$ we chose such a point $Q$ that $AQ = 2PM$. Prove that $\angle CQM = \angle BAM$.

Two particles move along the edges of equilateral triangle $ \triangle ABC$ in the direction \[ A\rightarrow B\rightarrow C\rightarrow A \]starting simultaneously and moving at the same speed. One starts at $ A$, and the other starts at the midpoint of $ \overline{BC}$. The midpoint of the line segment joining the two particles traces out a path that encloses a region $ R$. What is the ratio of the area of $ R$ to the area of $ \triangle ABC$? $ \textbf{(A)}\ \frac {1}{16}\qquad \textbf{(B)}\ \frac {1}{12}\qquad \textbf{(C)}\ \frac {1}{9}\qquad \textbf{(D)}\ \frac {1}{6}\qquad \textbf{(E)}\ \frac {1}{4}$

A non-equilateral triangle $\triangle ABC$ of perimeter $12$ is inscribed in circle $\omega$ .Points $P$ and $Q$ are arc midpoints of arcs $ABC$ and $ACB$ , respectively. Tangent to $\omega$ at $A$ intersects line $PQ$ at $R$. It turns out that the midpoint of segment $AR$ lies on line $BC$ . Find the length of the segment $BC$. [i] (А. Кузнецов)[/i]

Let $ABC$ be a scalene triangle with $\angle BAC>90^\circ$. Let $D$ and $E$ be two points on the side $BC$ such that $\angle BAD=\angle ACB$ and $\angle CAE=\angle ABC$. The angle-bisector of $\angle ACB$ meets $AD$ at $N$, If $MN\parallel BC$, determine $\angle (BM, CN)$. [i]Petru Braica[/i]

Two circles $ \left(\Omega_1\right),\left(\Omega_2\right) $ touch internally on the point $ T $. Let $ M,N $ be two points on the circle $ \left(\Omega_1\right) $ which are different from $ T $ and $ A,B,C,D $ be four points on $ \left(\Omega_2\right) $ such that the chords $ AB, CD $ pass through $ M,N $, respectively. Prove that if $ AC,BD,MN $ have a common point $ K $, then $ TK $ is the angle bisector of $ \angle MTN $. * $ \left(\Omega_2\right) $ is bigger than $ \left(\Omega_1\right) $