$ABCD$ is a rectangle $\overline{AB} = 5\sqrt{3}$, $\overline{AD} = 30$. Extend $\overline{BC}$ past $C$ and construct point $P$ on this extension such that $\angle APD = 60^{\circ}$. Point $H$ is on $\overline{AP}$ such that $\overline{DH} \perp \overline{AP}$. Find the length of $\overline{DH}$. [i]Proposed by Kevin Wu[/i]

Let $m$ and $n$ be positive integers. Mr. Fat has a set $S$ containing every rectangular tile with integer side lengths and area of a power of $2$. Mr. Fat also has a rectangle $R$ with dimensions $2^m \times 2^n$ and a $1 \times 1$ square removed from one of the corners. Mr. Fat wants to choose $m + n$ rectangles from $S$, with respective areas $2^0, 2^1, \ldots, 2^{m + n - 1}$, and then tile $R$ with the chosen rectangles. Prove that this can be done in at most $(m + n)!$ ways. [i]Palmer Mebane.[/i]

A satellite is considered accessible from the point $A{}$ of the planet's surface if it is located relative to the tangent plane drawn at point $A{}$, strictly on the other side than the planet. What is the smallest number of satellites that need to be launched over a spherical planet so that at some point the signals of at least two satellites are available from each point on the planet's surface? [i]Proposed by S. Volchenkov[/i]

Akash's birthday cake is in the form of a $4 \times 4 \times 4$ inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into $64$ smaller cubes, each measuring $1 \times 1 \times 1$ inch, as shown below. How many of the small pieces will have icing on exactly two sides? [asy] /* Created by SirCalcsALot and sonone Code modfied from https://artofproblemsolving.com/community/c3114h2152994_the_old__aops_logo_with_asymptote */ import three; currentprojection=orthographic(1.75,7,2); //++++ edit colors, names are self-explainatory ++++ //pen top=rgb(27/255, 135/255, 212/255); //pen right=rgb(254/255,245/255,182/255); //pen left=rgb(153/255,200/255,99/255); pen top = rgb(170/255, 170/255, 170/255); pen left = rgb(81/255, 81/255, 81/255); pen right = rgb(165/255, 165/255, 165/255); pen edges=black; int max_side = 4; //+++++++++++++++++++++++++++++++++++++++ path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle; path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle; path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle; for(int i=0; i<max_side; ++i){ for(int j=0; j<max_side; ++j){ draw(shift(i,j,-1)*surface(topface),top); draw(shift(i,j,-1)*topface,edges); draw(shift(i,-1,j)*surface(rightface),right); draw(shift(i,-1,j)*rightface,edges); draw(shift(-1,j,i)*surface(leftface),left); draw(shift(-1,j,i)*leftface,edges); } } picture CUBE; draw(CUBE,surface(leftface),left,nolight); draw(CUBE,surface(rightface),right,nolight); draw(CUBE,surface(topface),top,nolight); draw(CUBE,topface,edges); draw(CUBE,leftface,edges); draw(CUBE,rightface,edges); // begin made by SirCalcsALot int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}}; for (int i = 0; i < max_side; ++i) { for (int j = 0; j < max_side; ++j) { for (int k = 0; k < min(heights[i][j], max_side); ++k) { add(shift(i,j,k)*CUBE); } } } [/asy] $\textbf{(A)}\ 12\qquad~~\textbf{(B)}\ 16\qquad~~\textbf{(C)}\ 18\qquad~~\textbf{(D)}\ 20\qquad~~\textbf{(E)}\ 24$

An ice cream cone has radius $1$ and height $4$ inches. What is the number of inches in the radius of a sphere of ice cream which has the same volume of the cone? $\text{(A) }\frac{1}{2}\qquad\text{(B) }1\qquad\text{(C) }\frac{3}{2}\qquad\text{(D) }2\qquad\text{(E) }\frac{5}{2}$

Given circle $O$ with radius $R$, the inscribed triangle $ABC$ is an acute scalene triangle, where $AB$ is the largest side. $AH_A, BH_B,CH_C$ are heights on $BC,CA,AB$. Let $D$ be the symmetric point of $H_A$ with respect to $H_BH_C$, $E$ be the symmetric point of $H_B$ with respect to $H_AH_C$. $P$ is the intersection of $AD,BE$, $H$ is the orthocentre of $\triangle ABC$. Prove: $OP\cdot OH$ is fixed, and find this value in terms of $R$. (Edited)

Let $I$ be the center of incircle of $\triangle ABC$, and $J$ be the center of excircle tangent to $[BC]$. If $m(\widehat B) = 45^\circ$, $m(\widehat A) = 120^\circ$, and $|IJ|=\sqrt 3$, then what is $|BC|$? $ \textbf{(A)}\ \frac 32 \qquad\textbf{(B)}\ \frac {\sqrt 3}2 \qquad\textbf{(C)}\ \frac 34 \qquad\textbf{(D)}\ \frac {\sqrt 6}2 \qquad\textbf{(E)}\ \sqrt3 - 1 $

Let three circles $ \Gamma_1, \Gamma_2, \Gamma_3$, which are non-overlapping and mutually external, be given in the plane. For each point $ P$ in the plane, outside the three circles, construct six points $ A_1, B_1, A_2, B_2, A_3, B_3$ as follows: For each $ i \equal{} 1, 2, 3$, $ A_i, B_i$ are distinct points on the circle $ \Gamma_i$ such that the lines $ PA_i$ and $ PB_i$ are both tangents to $ \Gamma_i$. Call the point $ P$ exceptional if, from the construction, three lines $ A_1B_1, A_2 B_2, A_3 B_3$ are concurrent. Show that every exceptional point of the plane, if exists, lies on the same circle.

$O$ is a point in triangle $ABC$. We draw perpendicular from $O$ to $BC,AC,AB$ which intersect $BC,AC,AB$ at $A_{1},B_{1},C_{1}$. Prove that $O$ is circumcenter of triangle $ABC$ iff perimeter of $ABC$ is not less than perimeter of triangles $AB_{1}C_{1},BC_{1}A_{1},CB_{1}A_{1}$.

Can it happen that $6$ parallelepipeds, no two of which have common points, are placed in space so that there is a point outside of them from which no vertex of a parallelepiped is visible? (The parallelepipeds are not transparent.) (V Proizvolov)

Given quadrilateral $ABCD$ inscribed into a circle with diagonal $AC$ as diameter. Let $E$ be a point on segment $BC$ s.t. $\sphericalangle DAC=\sphericalangle EAB$. Point $M$ is midpoint of $CE$. Prove that $BM=DM$.

We choose a point on each side of a parallelogram $ABCD$, let these four points be $P, Q, R$ and $S$. Then we divide the parallelogram into several regions using line segments as shown in the figure. The areas of the grey regions are given, except for one (see the figure). Find the area of the region marked with a question mark. [img]https://cdn.artofproblemsolving.com/attachments/4/7/dbd009042dabdb2eafc8fc74960e9011038dae.png[/img]

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 $\triangle ABC$ be an arbitary triangle, and construct $P,Q,R$ so that each of the angles marked is $30^\circ$. Prove that $\triangle PQR$ is an equilateral triangle. [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair ext30(pair pt1, pair pt2) { pair r1 = pt1+rotate(-30)*(pt2-pt1), r2 = pt2+rotate(30)*(pt1-pt2); draw(anglemark(r1,pt1,pt2,25)); draw(anglemark(pt1,pt2,r2,25)); return intersectionpoints(pt1--r1, pt2--r2)[0]; } pair A = (0,0), B=(10,0), C=(3,7), P=ext30(B,C), Q=ext30(C,A), R=ext30(A,B); draw(A--B--C--A--R--B--P--C--Q--A); draw(P--Q--R--cycle, linetype("8 8")); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$P$", P, NE); label("$Q$", Q, NW); label("$R$", R, S);[/asy]

Given a triangle $ABC$ and a point $P$ on the side $AB$ . Let $Q$ be the intersection of the straight line $CP$ (different from $C$) with the circumcicle of the triangle. Prove the inequality $$\frac{\overline{PQ}}{\overline{CQ}} \le \left(\frac{\overline{AB}}{\overline{AC}+\overline{CB}}\right)^2$$ and that equality holds if and only if the $CP$ is bisector of the angle $ACB$. [img]https://cdn.artofproblemsolving.com/attachments/b/1/068fafd5564e77930160115a1cd409c4fdbf61.png[/img]

Let $O, H$ be the orthocenter and circumcenter of of an acute-angled triangke $ABC$ with $AB<AC$.Let $K$ be the midpoint of $AH$.The line through $K$ perpendicular to $OK$ meet $AB$ and the tangent to the circumcircle at $A$ at $X$ and $Y$ respectively. Prove that $\angle XOY=\angle AOB$

Higher Secondary P1 A polygon is called degenerate if one of its vertices falls on a line that joins its neighboring two vertices. In a pentagon $ABCDE$, $AB=AE$, $BC=DE$, $P$ and $Q$ are midpoints of $AE$ and $AB$ respectively. $PQ||CD$, $BD$ is perpendicular to both $AB$ and $DE$. Prove that $ABCDE$ is a degenerate pentagon.

In square $ABCD$ with side length $2$, let $M$ be the midpoint of $AB$. Let $N$ be a point on $AD$ such that $AN = 2ND$. Let point $P$ be the intersection of segment $MN$ and diagonal $AC$. Find the area of triangle $BPM$. [i]Proposed by Jacob Xu[/i]

On a circle $\Gamma$, points $A, B, N, C, D, M$ are chosen in a clockwise order in such a way that $N$ and $M$ are the midpoints of clockwise arcs $BC$ and $AD$ respectively. Let $P$ be the intersection of $AC$ and $BD$, and let $Q$ be a point on line $MB$ such that $PQ$ is perpendicular to $MN$. Point $R$ is chosen on segment $MC$ such that $QB = RC$, prove that the midpoint of $QR$ lies on $AC$.

The perpendicular bisector of the bisector $BL$ of the triangle $ABC$ intersects the bisectors of its external angles $A$ and $C$ at points $P$ and $Q$, respectively. Prove that the circle circumscribed around the triangle $PBQ$ is tangent to the circle circumscribed around the triangle $ABC$.

Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[ \angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.

$ABCD$ is a cyclic quadrilateral. The lines $AD$ and $BC$ meet at X, and the lines $AB$ and $CD$ meet at $Y$ . The line joining the midpoints $M$ and $N$ of the diagonals $AC$ and $BD$, respectively, meets the internal bisector of angle $AXB$ at $P$ and the external bisector of angle $BYC$ at $Q$. Prove that $PXQY$ is a rectangle

Let $\Gamma$ be the circumcircle of an acute triangle $ABC$. The perpendicular line to $AB$ passing through $C$ cuts $AB$ in $D$ and $\Gamma$ again in $E$. The bisector of the angle $C$ cuts $AB$ in $F$ and $\Gamma$ again in $G$. The line $GD$ meets $\Gamma$ again at $H$ and the line $HF$ meets $\Gamma$ again at $I$. Prove that $AI = EB$.

Consider all tetrahedra $ABCD$ in which the sum of the areas of the faces $ABD, ACD, BCD$ does not exceed $1$. Among such tetrahedra, find those with the maximum volume.

Suppose $ \vartriangle ABC$ is similar to $\vartriangle DEF$, with $ A$, $ B$, and $C$ corresponding to $D, E$, and $F$ respectively. If $\overline{AB} = \overline{EF}$, $\overline{BC} = \overline{FD}$, and $\overline{CA} = \overline{DE} = 2$, determine the area of $ \vartriangle ABC$.