In $\triangle PQR$, $PQ=8$, $QR=13$, and $RP=15$. Prove that there is a point $S$ on line segment $\overline{PR}$, but not at its endpoints, such that $PS$ and $QS$ are also integers. [asy] size(200); defaultpen(linewidth(0.8)); pair P=origin,Q=(8,0),R=(7,10),S=(3/2,15/7); draw(P--Q--R--cycle); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,NE); draw(Q--S,linetype("4 4")); label("$S$",S,NW); [/asy]

Points $A$ and $B$ lie on circle $\omega$. Point $P$ lies on the extension of segment $AB$ past $B$. Line $\ell$ passes through $P$ and is tangent to $\omega$. The tangents to $\omega$ at points $A$ and $B$ intersect $\ell$ at points $D$ and $C$ respectively. Given that $AB=7$, $BC=2$, and $AD=3$, compute $BP$.

The diagram below shows a regular hexagon with an inscribed square where two sides of the square are parallel to two sides of the hexagon. There are positive integers $m$, $n$, and $p$ such that the ratio of the area of the hexagon to the area of the square can be written as $\tfrac{m+\sqrt{n}}{p}$ where $m$ and $p$ are relatively prime. Find $m + n + p$. [asy] import graph; size(4cm); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((0,1)--(1,1)--(1.5,1.87)--(1,2.73)--(0,2.73)--(-0.5,1.87)--cycle); filldraw((1.13,2.5)--(-0.13,2.5)--(-0.13,1.23)--(1.13,1.23)--cycle,grey); draw((0,1)--(1,1)); draw((1,1)--(1.5,1.87)); draw((1.5,1.87)--(1,2.73)); draw((1,2.73)--(0,2.73)); draw((0,2.73)--(-0.5,1.87)); draw((-0.5,1.87)--(0,1)); draw((1.13,2.5)--(-0.13,2.5)); draw((-0.13,2.5)--(-0.13,1.23)); draw((-0.13,1.23)--(1.13,1.23)); draw((1.13,1.23)--(1.13,2.5)); [/asy]

Given a triangle with $120^\circ$. Let $x,\ y,\ z$ be the side lengths of the triangle such that $x<y<z$. (1) Find all triplets $(x,\ y,\ z)$ of positive integers $x,\ y,\ z$ such that $x+y-z=2$. (2) Find all triplets $(x,\ y,\ z)$ of positive integers $x,\ y,\ z$ such that $x+y-z=3$. (3) Let $a,\ b$ be non-negative integers. Express the number of $(x,\ y,\ z)$ such that $x+y-z=2^a3^b$ in terms of $a,\ b$. 2012 Hitotsubashi University entrance exam, problem 1

Let $ABC$ be equilateral, and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a+b\sqrt{c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime. What is $a^{2}+b^{2}+c^{2}$?

For a prime $q$, let $\Phi_q(x)=x^{q-1}+x^{q-2}+\cdots+x+1$. Find the sum of all primes $p$ such that $3 \le p \le 100$ and there exists an odd prime $q$ and a positive integer $N$ satisfying \[\dbinom{N}{\Phi_q(p)}\equiv \dbinom{2\Phi_q(p)}{N} \not \equiv 0 \pmod p. \][i]Proposed by Sammy Luo[/i]

Circles $ C_1,C_2$ intersect at $ P$. A line $ \Delta$ is drawn arbitrarily from $ P$ and intersects with $ C_1,C_2$ at $ B,C$. What is locus of $ A$ such that the median of $ AM$ of triangle $ ABC$ has fixed length $ k$.

In the adjoining figure of a rectangular solid, $\angle DHG=45^\circ$ and $\angle FHB=60^\circ$. Find the cosine of $\angle BHD$. [asy] size(200); import three;defaultpen(linewidth(0.7)+fontsize(10)); currentprojection=orthographic(1/3+1/10,1-1/10,1/3); real r=sqrt(3); triple A=(0,0,r), B=(0,r,r), C=(1,r,r), D=(1,0,r), E=O, F=(0,r,0), G=(1,0,0), H=(1,r,0); draw(D--G--H--D--A--B--C--D--B--F--H--B^^C--H); draw(A--E^^G--E^^F--E, linetype("4 4")); label("$A$", A, N); label("$B$", B, dir(0)); label("$C$", C, N); label("$D$", D, W); label("$E$", E, NW); label("$F$", F, S); label("$G$", G, W); label("$H$", H, S); triple H45=(1,r-0.15,0.1), H60=(1-0.05, r, 0.07); label("$45^\circ$", H45, dir(125), fontsize(8)); label("$60^\circ$", H60, dir(25), fontsize(8));[/asy] $\textbf {(A) } \frac{\sqrt{3}}{6} \qquad \textbf {(B) } \frac{\sqrt{2}}{6} \qquad \textbf {(C) } \frac{\sqrt{6}}{3} \qquad \textbf {(D) } \frac{\sqrt{6}}{4} \qquad \textbf {(E) } \frac{\sqrt{6}-\sqrt{2}}{4}$

Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon. A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a+b\sqrt{c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c.$

In quadrilateral $ABCD$, $BC=8$, $CD=12$, $AD=10$, and $m\angle A= m\angle B = 60^\circ$. Given that $AB=p + \sqrt{q}$, where $p$ and $q$ are positive integers, find $p+q$.

Square $ ABCD$ has side length $ s$, a circle centered at $ E$ has radius $ r$, and $ r$ and $ s$ are both rational. The circle passes through $ D$, and $ D$ lies on $ \overline{BE}$. Point $ F$ lies on the circle, on the same side of $ \overline{BE}$ as $ A$. Segment $ AF$ is tangent to the circle, and $ AF \equal{} \sqrt {9 \plus{} 5\sqrt {2}}$. What is $ r/s$? [asy]unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B=(0,0), C=(3,0), D=(3,3), A=(0,3); pair Ep=(3+5*sqrt(2)/6,3+5*sqrt(2)/6); pair F=intersectionpoints(Circle(A,sqrt(9+5*sqrt(2))),Circle(Ep,5/3))[0]; pair[] dots={A,B,C,D,Ep,F}; draw(A--F); draw(Circle(Ep,5/3)); draw(A--B--C--D--cycle); dot(dots); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,SW); label("$E$",Ep,E); label("$F$",F,NW);[/asy]$ \textbf{(A) } \frac {1}{2}\qquad \textbf{(B) } \frac {5}{9}\qquad \textbf{(C) } \frac {3}{5}\qquad \textbf{(D) } \frac {5}{3}\qquad \textbf{(E) } \frac {9}{5}$

A circle of radius $r$ has chords $\overline{AB}$ of length $10$ and $\overline{CD}$ of length $7$. When $\overline{AB}$ and $\overline{CD}$ are extended through $B$ and $C$, respectively, they intersect at $P$, which is outside the circle. If $\angle APD = 60^{\circ}$ and $BP = 8$, then $r^{2} =$ $ \textbf{(A)}\ 70\qquad\textbf{(B)}\ 71\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 73\qquad\textbf{(E)}\ 74 $

Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ$. Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ$. Find the number of degrees in $\angle CMB$.

Equilateral $\triangle ABC$ has side length $\sqrt{111}$. There are four distinct triangles $AD_1E_1$, $AD_1E_2$, $AD_2E_3$, and $AD_2E_4$, each congruent to $\triangle ABC$, with $BD_1 = BD_2=\sqrt{11}$. Find $\sum^4_{k=1}(CE_k)^2$.

An equilateral triangle is given. A point lies on the incircle of this triangle. If the smallest two distances from the point to the sides of the triangle is $1$ and $4$, the sidelength of this equilateral triangle can be expressed as $\tfrac{a\sqrt b}c$ where $(a,c)=1$ and $b$ is not divisible by the square of an integer greater than $1$. Find $a+b+c$.

Triangle $ ABC$ has side lengths $ AB \equal{} 5$, $ BC \equal{} 6$, and $ AC \equal{} 7$. Two bugs start simultaneously from $ A$ and crawl along the sides of the triangle in opposite directions at the same speed. They meet at point $ D$. What is $ BD$? $ \textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Let $ ABC$ be a triangle with $ \angle A \equal{} 45^\circ$. Let $ P$ be a point on side $ BC$ with $ PB \equal{} 3$ and $ PC \equal{} 5$. Let $ O$ be the circumcenter of $ ABC$. Determine the length $ OP$.

Equilateral triangle $ABC$ has been creased and folded so that vertex $A$ now rests at $A'$ on $\overline{BC}$ as shown. If $BA' = 1$ and $A'C = 2$ then the length of crease $\overline{PQ}$ is [asy] size(170); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=(1.5,3*sqrt(3)/2), C=(3,0), D=(1,0), P=B+1.6*dir(B--A), Q=C+1.2*dir(C--A); draw(B--P--D--B^^P--Q--D--C--Q); draw(Q--A--P, linetype("4 4")); label("$A$", A, N); label("$B$", B, W); label("$C$", C, E); label("$A'$", D, S); label("$P$", P, W); label("$Q$", Q, E); [/asy] $ \textbf{(A)}\ \frac{8}{5}\qquad\textbf{(B)}\ \frac{7}{20}\sqrt{21}\qquad\textbf{(C)}\ \frac{1+\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{13}{8}\qquad\textbf{(E)}\ \sqrt{3} $

Given a quadrilateral $ABCD$ such that $AB^2 + CD^2 = AD^2 + BC^2$, prove that $AC \perp BD$.

An equilateral triangle with side length $6$ has a square of side length $6$ attached to each of its edges as shown. The distance between the two farthest vertices of this figure (marked $A$ and $B$ in the figure) can be written as $m + \sqrt{n}$ where $m$ and $n$ are positive integers. Find $m + n$. [asy] draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle); draw((1,0)--(1+sqrt(3)/2,1/2)--(1/2+sqrt(3)/2,1/2+sqrt(3)/2)--(1/2,sqrt(3)/2)); draw((0,0)--(-sqrt(3)/2,1/2)--(-sqrt(3)/2+1/2,1/2+sqrt(3)/2)--(1/2,sqrt(3)/2)); dot((-sqrt(3)/2+1/2,1/2+sqrt(3)/2)); label("A", (-sqrt(3)/2+1/2,1/2+sqrt(3)/2), N); draw((1,0)--(1,-1)--(0,-1)--(0,0)); dot((1,-1)); label("B", (1,-1), SE); [/asy]

Let $ABCD$ be a cyclic quadrilateral, and $E$ be the intersection of its diagonals. If $m(\widehat{ADB}) = 22.5^\circ$, $|BD|=6$, and $|AD|\cdot|CE|=|DC|\cdot|AE|$, find the area of the quadrilateral $ABCD$.

In convex quadrilateral $ABCD$, $\angle A \cong \angle C$, $AB = CD = 180$, and $AD \neq BC$. The perimeter of $ABCD$ is 640. Find $\lfloor 1000 \cos A \rfloor$. (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x$.)

Consider a triangle $ABC$ with $AB=AC$, and $D$ the foot of the altitude from the vertex $A$. The point $E$ lies on the side $AB$ such that $\angle ACE= \angle ECB=18^{\circ}$. If $AD=3$, find the length of the segment $CE$.

In quadrilateral $ABCD$, we have $AB = 5$, $BC = 6$, $CD = 5$, $DA = 4$, and $\angle ABC = 90^\circ$. Let $AC$ and $BD$ meet at $E$. Compute $\dfrac{BE}{ED}$.

A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is 6 cm, and that of $BC$ is 2 cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle. [asy] size(150); defaultpen(linewidth(0.65)+fontsize(11)); real r=10; pair O=(0,0),A=r*dir(45),B=(A.x,A.y-r),C; path P=circle(O,r); C=intersectionpoint(B--(B.x+r,B.y),P); draw(Arc(O, r, 45, 360-17.0312)); draw(A--B--C);dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,SE); [/asy]