A rectangular piece of paper $ABCD$ has sides of lengths $AB = 1$, $BC = 2$. The rectangle is folded in half such that $AD$ coincides with $BC$ and $EF$ is the folding line. Then fold the paper along a line $BM$ such that the corner $A$ falls on line $EF$. How large, in degrees, is $\angle ABM$? [asy] size(180); pathpen = rgb(0,0,0.6)+linewidth(1); pointpen = black+linewidth(3); pointfontpen = fontsize(10); pen dd = rgb(0,0,0.6) + linewidth(0.7) + linetype("4 4"), dr = rgb(0.8,0,0), dg = rgb(0,0.6,0), db = rgb(0,0,0.6)+linewidth(1); pair A=(0,1), B=(0,0), C=(2,0), D=(2,1), E=A/2, F=(2,.5), M=(1/3^.5,1), N=reflect(B,M)*A; D(B--M--D("N",N,NE)--B--D("C",C,SE)--D("D",D,NE)--M); D(D("M",M,plain.N)--D("A",A,NW)--D("B",B,SW),dd); D(D("E",E,W)--D("F",F,plain.E),dd); [/asy]

$20.$ The circle $\omega$ touches the circle $\Omega$ internally at point $P.$ The centre $O$ of $\Omega$ is outside $\omega.$ Let $XY$ be a diameter of $\Omega$ which is also tangent to $\omega.$ Assume $PY>PX.$ Let $PY$ intersect $\omega$ at $z.$ If $YZ=2PZ,$ what is the magnitude of $\angle{PYX}$ in degrees $?$

A quadrilateral is given, in which the lengths of some two sides are equal to $1$ and $4$. Also, the diagonal of length $2$ divides it into two isosceles triangles. Find the perimeter of this quadrilateral.

Let $A_1A_2A_3A_4A_5$ be a regular pentagon with side length 1. The sides of the pentagon are extended to form the 10-sided polygon shown in bold at right. Find the ratio of the area of quadrilateral $A_2A_5B_2B_5$ (shaded in the picture to the right) to the area of the entire 10-sided polygon. [asy] size(8cm); defaultpen(fontsize(10pt)); pair A_2=(-0.4382971011,5.15554989475), B_4=(-2.1182971011,-0.0149584477027), B_5=(-4.8365942022,8.3510997895), A_3=(0.6,8.3510997895), B_1=(2.28,13.521608132), A_4=(3.96,8.3510997895), B_2=(9.3965942022,8.3510997895), A_5=(4.9982971011,5.15554989475), B_3=(6.6782971011,-0.0149584477027), A_1=(2.28,3.18059144705); filldraw(A_2--A_5--B_2--B_5--cycle,rgb(.8,.8,.8)); draw(B_1--A_4^^A_4--B_2^^B_2--A_5^^A_5--B_3^^B_3--A_1^^A_1--B_4^^B_4--A_2^^A_2--B_5^^B_5--A_3^^A_3--B_1,linewidth(1.2)); draw(A_1--A_2--A_3--A_4--A_5--cycle); pair O = (A_1+A_2+A_3+A_4+A_5)/5; label("$A_1$",A_1, 2dir(A_1-O)); label("$A_2$",A_2, 2dir(A_2-O)); label("$A_3$",A_3, 2dir(A_3-O)); label("$A_4$",A_4, 2dir(A_4-O)); label("$A_5$",A_5, 2dir(A_5-O)); label("$B_1$",B_1, 2dir(B_1-O)); label("$B_2$",B_2, 2dir(B_2-O)); label("$B_3$",B_3, 2dir(B_3-O)); label("$B_4$",B_4, 2dir(B_4-O)); label("$B_5$",B_5, 2dir(B_5-O)); [/asy]

Let $M$ be the midpoint of side $AC$ of the triangle $ABC$. Let $P$ be a point on the side $BC$. If $O$ is the point of intersection of $AP$ and $BM$ and $BO = BP$, determine the ratio $\frac{OM}{PC}$ .

Let $P$ be a point outside a circle $\omega$. The tangents from $P$ to $\omega$ are drawn touching $\omega$ at points $A$ and $B$. Let $M$ and $N$ be the midpoints of $AP$ and $AB$, respectively. Line $MN$ is extended to cut $\omega$ at $C$ so that $N$ lies between $M$ and $C$. Line $PC$ intersects $\omega$ again at $D$, and lines $ND$ and $PB$ intersect at $O$. Prove that $MNOP$ is a rhombus.

Let $ABC$ be a triangle with orthocenter $H$ and circumcircle $\Gamma$. Let $D$ be the reflection of $A$ in $B$ and let $E$ the reflection of $A$ in $C$. Let $M$ be the midpoint of segment $DE$. Show that the tangent to $\Gamma$ in $A$ is perpendicular to $HM$.

In an acute triangle $ABC$ with $\overline{AB} < \overline{AC}$, angle bisector of $A$ and perpendicular bisector of $\overline{BC}$ intersect at $D$. Let $P$ be an interior point of triangle $ABC$. Line $CP$ meets the circumcircle of triangle $ABP$ again at $K$. Prove that $B, D, K$ are collinear if and only if $AD$ and $BC$ meet on the circumcircle of triangle $APC$.

Let $ABCD$ be a given convex quadrilateral in a plane. Prove that there exist a line with four different points $P,Q,R,S$ on it and a square $A’B’C’D’$ such that $P$ lies on both line $AB$ and $A’B’,$ $Q$ lies on both line $BC$ and $B’C’,$ $R$ lies on both line $CD$ and $C’D’,$ $S$ lies on both line $DA$ and $D’A’.$

In convex pentagon $ABCDE$ points $A_1$, $B_1$, $C_1$, $D_1$, $E_1$ are intersections of pairs of diagonals $(BD, CE)$, $(CE, DA)$, $(DA, EB)$, $(EB, AC)$ and $(AC, BD)$ respectively. Prove that if four of quadrilaterals $AB_{1}A_{1}B$, $BC_{1}B_{1}C$, $CD_{1}C_{1}D$, $DE_{1}D_{1}E$ and $EA_{1}E_{1}A$ are cyclic then the fifth one is also cyclic.

Consider a triangle $ABC$ . The points $D,E$ are on sides $AB,AC$ such that $BDEC$ is a cyclic quadrilateral. Let $P$ be the intersection of $BE$ and $CD$. $H$ is a point on $AC$ such that $\angle PHA = 90^{\circ}$. Let $M,N$ be the midpoints of $AP,BC$. Prove that: $ ACD \sim MNH $.

The figure below shows a $90 \times90$ square with each side divided into three equal segments. Some of the endpoints of these segments are connected by straight lines. Find the area of the shaded region. [asy] import graph; size(6cm); real labelscalefactor = 0.5; pen dotstyle = black; draw((-4,6)--(86,6)--(86,96)--(-4,96)--cycle); filldraw((16,76)--(-4,36)--(32,60)--(56,96)--cycle,grey); filldraw((32,60)--(-4,6)--(50,42)--(86,96)--cycle,grey); filldraw((50,42)--(26,6)--(66,26)--(86,66)--cycle,grey); draw((-4,6)--(26,6)); draw((26,6)--(56,6)); draw((56,6)--(86,6)); draw((-4,6)--(86,6)); draw((86,6)--(86,96)); draw((86,96)--(-4,96)); draw((-4,96)--(-4,6)); draw((26,96)--(-4,36)); draw((56,96)--(-4,6)); draw((86,96)--(26,6)); draw((86,66)--(56,6)); draw((-4,66)--(56,96)); draw((-4,36)--(86,96)); draw((-4,6)--(86,66)); draw((26,6)--(86,36)); draw((16,76)--(-4,36)); draw((-4,36)--(32,60)); draw((32,60)--(56,96)); draw((56,96)--(16,76)); draw((32,60)--(-4,6)); draw((-4,6)--(50,42)); draw((50,42)--(86,96)); draw((86,96)--(32,60)); draw((50,42)--(26,6)); draw((26,6)--(66,26)); draw((66,26)--(86,66)); draw((86,66)--(50,42)); dot((-4,96),dotstyle); dot((26,96),dotstyle); dot((56,96),dotstyle); dot((86,96),dotstyle); dot((-4,6),dotstyle); dot((-4,36),dotstyle); dot((-4,66),dotstyle); dot((27.09,6),dotstyle); dot((56,6),dotstyle); dot((86,36),dotstyle); dot((86,66),dotstyle); dot((86,6),dotstyle); [/asy]

Let $ABC$ be an equilateral triangle. Denote by $D$ the midpoint of $\overline{BC}$, and denote the circle with diameter $\overline{AD}$ by $\Omega$. If the region inside $\Omega$ and outside $\triangle ABC$ has area $800\pi-600\sqrt3$, find the length of $AB$. [i]Proposed by Eugene Chen[/i]

Let $ABC$ be a triangle with $\angle BAC=40^\circ $, $O$ be the center of its circumscribed circle and $G$ is its centroid. Point $D$ of line $BC$ is such that $CD=AC$ and $C$ is between $B$ and $D$. If $AD\parallel OG$, find $\angle ACB$.

On the circumcircle of triangle $ABC$, point $P$ is taken in such a way that the perpendicular drawn by the point $P$ to the line $AC$ cuts the circle also at the point $Q$, the perpendicular drawn by the point $Q$ to the line $AB$ cuts the circle also at point R and the perpendicular drawn by point $R$ to the line BC cuts the circle also at the point $P$. Let $O$ be the center of this circle. Prove that $\angle POC = 90^o$ .

Is it possible to find a set of $100$ (or $200$) points on the boundary of a cube such that this set remains fixed under all rotations which leave the cube fixed ?

Facundo and Luca have been given a cake that is shaped like the quadrilateral in the figure. [img]https://cdn.artofproblemsolving.com/attachments/3/2/630286edc1935e1a8dd9e704ed4c813c900381.png[/img] They are going to make two straight cuts on the cake, thus obtaining $4$ portions in the shape of a quadrilateral. Then Facundo will be left with two portions that do not share any side, the other two will be for Luca. Show how they can cut the cuts so that both children get the same amount of cake. Justify why cutting in this way achieves the objective.

Let AOB and COD be angles which can be identified by a rotation of the plane (such that rays OA and OC are identified). A circle is inscribed in each of these angles; these circles intersect at points E and F. Show that angles AOE and DOF are equal.

Two medians of a triangle with unequal sides are $ 3$ inches and $ 6$ inches. Its area is $ 3 \sqrt{15}$ square inches. The length of the third median in inches, is: $ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \sqrt{3} \qquad \textbf{(C)}\ 3 \sqrt{6} \qquad \textbf{(D)}\ 6 \sqrt{3} \qquad \textbf{(E)}\ 6 \sqrt{6}$

Let $ABC$ be an acute triangle with $AB<AC$, and let $D$ and $E$ be points on side $BC$ such that $BD=CE$ and $D$ lies between $B$ and $E$. Suppose there exists a point $P$ inside $ABC$ such that $PD\parallel AE$ and $\angle PAB=\angle EAC$. Prove that $\angle PBA=\angle PCA$. [i]Calvin Deng.[/i]

In a non-convex hexagon, each angle is either $90$ or $270$ degrees. Is it true that for some lengths of the sides it can be cut into two hexagons similar to it and unequal to each other?

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

Eight points in a general position are given in the plane. The areas of all $56$ triangles with vertices at these points are written in a row. Prove that it is possible to insert the symbols "$+$" and "$-$" between them in such a way that the obtained sum is equal to zero.

Given a circle and a fixed point $P$ not lying on it. Find the geometrical locus of the orthocenters of the triangles $ABP$, where $AB$ is the diameter of the circle.

Let $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, $\Gamma_4$ be distinct circles such that $\Gamma_1$, $\Gamma_3$ are externally tangent at $P$, and $\Gamma_2$, $\Gamma_4$ are externally tangent at the same point $P$. Suppose that $\Gamma_1$ and $\Gamma_2$; $\Gamma_2$ and $\Gamma_3$; $\Gamma_3$ and $\Gamma_4$; $\Gamma_4$ and $\Gamma_1$ meet at $A$, $B$, $C$, $D$, respectively, and that all these points are different from $P$. Prove that \[ \frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}. \]