In trapezoid $ABCD$, $AD \parallel BC$ and $\angle ABC + \angle CDA = 270^{\circ}$. Compute $AB^2$ given that $AB \cdot \tan(\angle BCD) = 20$ and $CD = 13$. [i]Proposed by Lewis Chen[/i]

In quadrilateral $ABCD$, angles $A, B, C, D$ form an increasing arithmetic sequence. Also, $\angle ACB = 90^o$ . If $CD = 14$ and the length of the altitude from $C$ to $AB$ is $9$, compute the area of $ABCD$.

Consider a $1$ by $2$ by $3$ rectangular prism. Find the length of the shortest path between opposite corners $A$ and $B$ that does not leave the surface of the prism. [img]https://cdn.artofproblemsolving.com/attachments/7/1/899b79b0f938e63fe7b093cecff63ed1255acb.png[/img]

Which of the following statements are true? (a) If a polygon can be divided into two congruent polygons by a broken line segment, it can be divided into two congruent polygons by a straight line segment. (b) If a convex polygon can be divided into two congruent polygons by a broken line segment, it can be so divided by a straight line segment. (c) If a convex polygon can be divided into two polygons by a broken line segment, one of which can be mapped onto the other by a combination of rotations and translations, it can be so divided by a straight line segment. (S Markelov,)

Let $ABCD$ be a cyclic quadrilateral in which internal angle bisectors $\angle ABC$ and $\angle ADC$ intersect on diagonal $AC$. Let $M$ be the midpoint of $AC$. Line parallel to $BC$ which passes through $D$ cuts $BM$ at $E$ and circle $ABCD$ in $F$ ($F \neq D$ ). Prove that $BCEF$ is parallelogram [i]Proposed by Steve Dinh[/i]

Let $f(n)$ be the least number of distinct points in the plane such that for each $k = 1, 2, \cdots, n$ there exists a straight line containing exactly $k$ of these points. Find an explicit expression for $f(n).$ [i]Simplified version.[/i] Show that $f(n)=\left[\frac{n+1}{2}\right]\left[\frac{n+2}{2}\right].$ Where $[x]$ denoting the greatest integer not exceeding $x.$

There were ten points $X_1, \ldots , X_{10}$ on a line in this particular order. Pete constructed an isosceles triangle on each segment $X_1X_2, X_2X_3,\ldots, X_9X_{10}$ as a base with the angle $\alpha{}$ at its apex. It so happened that all the apexes of those triangles lie on a common semicircle with diameter $X_1X_{10}$. Find $\alpha{}$. [i]Egor Bakaev[/i]

If $I$ is the incenter of a triangle $ABC$ and $R$ is the radius of its circumcircle then \[AI+BI+CI\leq 3R\]

Let $\Gamma$ be a semicircle with diameter $AB$. On this diameter is selected a point $C$, and on the semicircle are selected points $D$ and $E$ so that $E$ lies between $B$ and $D$. It turned out that $\angle ACD = \angle ECB$. The intersection point of the tangents to $\Gamma$ at points $D$ and $E$ is denoted by $F$. Prove that $\angle EFD=\angle ACD+ \angle ECB$.

In a convex hexagon $ABCDEF$ the triangles $BDF, ACE$ are equilateral and congruent. Prove that the three lines connecting the midpoints of opposite sides are concurrent.

Let $ABCDE$ be a convex pentagon such that $AE\parallel BC$ and $\angle ADE = \angle BDC$. The diagonals $AC$ and $BE$ intersect at point $F$. Prove that $\angle CBD= \angle ADF$.

Let $ABCD$ be a quadrilateral inscribed in a circle $k$. $AC$ and $BD$ meet at $E$. The rays $\overrightarrow{CB}, \overrightarrow{DA}$ meet at $F$. Prove that the line through the incenters of $\triangle ABE\,,\, \triangle ABF$ and the line through the incenters of $\triangle CDE\,,\, \triangle CDF$ meet at a point lying on the circle $k$. [i]Proposed by N. Beluhov[/i]

What condition should the angles of triangle $ ABC $ satisfy so that the bisector of angle $ A $, the median drawn from vertex $ B $ and the altitude drawn from vertex $ C $ intersect at one point?

Given an acute angle triangle $ABC$ with circumcircle $\Gamma$ and circumcenter $O$. A point $P$ is taken on the line $BC$ but not on $[BC]$. Let $K$ be the reflection of the second intersection of the line $AP$ and $\Gamma$ with respect to $OP$. If $M$ is the intersection of the lines $AK$ and $OP$, prove that $\angle OMB+\angle OMC=180^{\circ}$.

Let $\triangle ABC$ be an acute triangle, and let $D,E,F$ respectively be three points on sides $BC,CA,AB$ such that $AEDF$ is a cyclic quadrilateral. Let $O_B$ and $O_C$ be the circumcenters of $\triangle BDF$ and $\triangle CDE$, respectively. Finally, let $D'$ be a point on segment $BC$ such that $BD'=CD$. Prove that $\triangle BD'O_B$ and $\triangle CD'O_C$ have the same surface.

[b]p1.[/b] On the island of Nevermind some people are liars; they always lie. The remaining habitants of the island are truthlovers; they tell only the truth. Three habitants of the island, $A, B$, and $C$ met this morning. $A$ said: “All of us are liars”. $B$ said: “Only one of us is a truthlover”. Who of them is a liar and who of them is a truthlover? [b]p2.[/b] Pinocchio has $9$ pieces of paper. He is allowed to take a piece of paper and cut it in $5$ pieces or $7$ pieces which increases the number of his pieces. Then he can take again one of his pieces of paper and cut it in $5$ pieces or $7$ pieces. He can do this again and again as many times as he wishes. Can he get $2004$ pieces of paper? [b]p3.[/b] In Dragonland there are coins of $1$ cent, $2$ cents, $10$ cents, $20$ cents, and $50$ cents. What is the largest amount of money one can have in coins, yet still not be able to make exactly $1$ dollar? [b]p4.[/b] Find all solutions $a, b, c, d, e$ if it is known that they represent distinct digits and satisfy the following: $\begin{tabular}{ccccc} & a & b & c & d \\ + & a & c & a & c \\ \hline c & d & e & b & c \\ \end{tabular}$ [b]p5.[/b] Two players play the following game. On the lowest left square of an $8\times 8$ chessboard there is a rook. The first player is allowed to move the rook up or to the right by an arbitrary number of squares. The second player 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? PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

In an acute triangle $\triangle ABC$, let there be point $D$ on segment $AC, E$ on segment $AB$ such that $\angle ADE = \angle ABC$. Let the bisector of $\angle A$ hit $BC$ at $K$. Let the foot of the perpendicular from $K$ to $DE$ be $P$, and the foot of the perpendicular from $A$ to $DE$ be $L$. Let $Q$ be the midpoint of $AL$. If the incenter of $\triangle ABC$ lies on the circumcircle of $\triangle ADE$, prove that $P,Q$ and the incenter of $\triangle ADE$ are collinear.

Show that the radian measure of an acute angle is less than the arithmetic mean of its sine and its tangent.

In a plane with a square grid, where the side length of the base square is $1$, lies a right triangle. All its vertices are lattice points and all side lengths are integer. Prove that the center of the incircle is also a lattice point.

Let $C$ and $D$ be two distinct points on a semicircle of diameter $AB$. Let $E$ be the intersection of $AC$ and $BD$, $F$ be the intersection of $AD$ and $BC$ and $X, Y$, and $Z$ are the midpoints of $AB, CD$, and $EF$, respectively. Prove that the points $X, Y,$ and $Z$ are collinear.

Given is a triangle $ABC$ with circumcircle $\gamma$. Points $E, F$ lie on $AB, AC$ such that $BE=CF$. Let $(AEF)$ meet $\gamma$ at $D$. The perpendicular from $D$ to $EF$ meets $\gamma$ at $G$ and $AD$ meets $EF$ at $P$. If $PG$ meets $\gamma$ at $J$, prove that $\frac {JE} {JF}=\frac{AE} {AF}$.

In a scalene triangle $ ABC$ the altitudes $ AA_{1}$ and $ CC_{1}$ intersect at $ H, O$ is the circumcenter, and $ B_{0}$ the midpoint of side $ AC$. The line $ BO$ intersects side $ AC$ at $ P$, while the lines $ BH$ and $ A_{1}C_{1}$ meet at $ Q$. Prove that the lines $ HB_{0}$ and $ PQ$ are parallel.

Consider all right triangles with integer sides such that the length of the hypotenuse and one of the two sides are consecutive. How many such triangles exist?

In triangle $ABC$, $\angle B=2\angle C$ and $\angle A>90^\circ$. Let $D$ be the point on the line $AB$ such that $CD$ is perpendicular to $AC$, and let $M$ be the midpoint of $BC$. Prove that $\angle AMB=\angle DMC$.

Given $2009 \times 4018$ rectangular board. Frame is a rectangle $n \times n$ or $n \times(n + 2)$ for $ ( n \geq 3 )$ without all cells which don’t have any common points with boundary of rectangle. Rectangles $1\times1,1\times 2,1\times 3$ and $ 2\times 4$ are also frames. Two players by turn paint all cells of some frame that has no painted cells yet. Player that can't make such move loses. Who has a winning strategy?