Consider a pyramid $P-ABCD$ whose base $ABCD$ is a square and whose vertex $P$ is equidistant from $A$, $B$, $C$, and $D$. If $AB=1$ and $\angle APD=2\theta$ then the volume of the pyramid is $\text{(A)} \ \frac{\sin \theta}{6} \qquad \text{(B)} \ \frac{\cot \theta}{6} \qquad \text{(C)} \ \frac1{6\sin \theta} \qquad \text{(D)} \ \frac{1-\sin 2\theta}{6} \qquad \text{(E)} \ \frac{\sqrt{\cos 2\theta}}{6\sin \theta}$

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 $a+b+c.$

Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let $t$ be the amount of time, in seconds, before Jenny and Kenny can see each other again. If $t$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Diameter $AB$ of a circle has length a 2-digit integer (base ten). Reversing the digits gives the length of the perpendicular chord $CD$. The distance from their intersection point $H$ to the center $O$ is a positive rational number. Determine the length of $AB$.

In a plane, the six different points $A, B, C, A', B', C'$ are given such that triangles $ABC$ and $A'B'C'$ are congruent, i.e. $AB=A'B', BC=B'C', CA=C'A'.$ Let $G$ be the centroid of $ABC$ and $A_1$ be an intersection point of the circle with diameter $AA'$ and the circle with center $A'$ and passing through $G.$ Define $B_1$ and $C_1$ similarly. Prove that \[ AA_1^2+BB_1^2+CC_1^2 \leq AB^2+BC^2+CA^2 \]

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}$

In $ \triangle ABC$ , $ AB \equal{} 13$, $ AC \equal{} 5$, and $ BC \equal{} 12$. Points $ M$ and $ N$ lie on $ \overline{AC}$ and $ \overline{BC}$, respectively, with $ CM \equal{} CN \equal{} 4$. Points $ J$ and $ K$ are on $ \overline{AB}$ so that $ \overline{MJ}$ and $ \overline{NK}$ are perpendicular to $ \overline{AB}$. What is the area of pentagon $ CMJKN$? [asy]unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair C=(0,0), B=(12,0), A=(0,5), M=(0,4), Np=(4,0); pair K=foot(Np,A,B), J=foot(M,A,B); draw(A--B--C--cycle); draw(M--J); draw(Np--K); label("$C$",C,SW); label("$A$",A,NW); label("$B$",B,SE); label("$N$",Np,S); label("$M$",M,W); label("$J$",J,NE); label("$K$",K,NE);[/asy]$ \textbf{(A)}\ 15 \qquad \textbf{(B)}\ \frac{81}{5} \qquad \textbf{(C)}\ \frac{205}{12} \qquad \textbf{(D)}\ \frac{240}{13} \qquad \textbf{(E)}\ 20$

Let $a,b,c$ be positive numbers with $\sqrt{a}+\sqrt{b}+\sqrt{c}= \frac{\sqrt{3}}{2}$ Prove that the system of equations \[\sqrt{y-a}+\sqrt{z-a}=1\] \[\sqrt{z-b}+\sqrt{x-b}=1\] \[\sqrt{x-c}+\sqrt{y-c}=1\] has exactly one solution $(x,y,z)$ in real numbers. It was proposed by Poland. Have fun! :lol:

In triangle $ABC$, $AB=13$, $BC=14$ and $CA=15$. Segment $BC$ is split into $n+1$ congruent segments by $n$ points. Among these points are the feet of the altitude, median, and angle bisector from $A$. Find the smallest possible value of $n$. [i]Proposed by Evan Chen[/i]

Let $\overline{CH}$ be an altitude of $\triangle ABC$. Let $R$ and $S$ be the points where the circles inscribed in the triangles $ACH$ and $BCH$ are tangent to $\overline{CH}$. If $AB = 1995$, $AC = 1994$, and $BC = 1993$, then $RS$ can be expressed as $m/n$, where $m$ and $n$ are relatively prime integers. Find $m + n$

Let $D$ be a point inside the equilateral triangle $\triangle ABC$ such that $|AD|=\sqrt{2}, |BD|=3, |CD|=\sqrt{5}$. $m(\widehat{ADB}) = ?$ $\textbf{(A)}\ 120^{\circ} \qquad\textbf{(B)}\ 105^{\circ} \qquad\textbf{(C)}\ 100^{\circ} \qquad\textbf{(D)}\ 95^{\circ} \qquad\textbf{(E)}\ 90^{\circ}$

A right triangle $ABC$ with hypotenuse $AB$ has side $AC = 15$. Altitude $CH$ divides $AB$ into segments $AH$ And $HB$, with $HB = 16$. The area of $\triangle ABC$ is: [asy] size(200); defaultpen(linewidth(0.8)+fontsize(11pt)); pair A = origin, H = (5,0), B = (13,0), C = (5,6.5); draw(C--A--B--C--H^^rightanglemark(C,H,B,16)); label("$A$",A,W); label("$B$",B,E); label("$C$",C,N); label("$H$",H,S); label("$15$",C/2,NW); label("$16$",(H+B)/2,S); [/asy] $\textbf{(A) }120\qquad \textbf{(B) }144\qquad \textbf{(C) }150\qquad \textbf{(D) }216\qquad \textbf{(E) }144\sqrt5$

Each side of triangle $ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC$, and $E$ is the midpoint of $AD$. The length of $BE$, in the same unit, is: ${{ \textbf{(A)}\ \sqrt{18} \qquad\textbf{(B)}\ \sqrt{28} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ \sqrt{63} }\qquad\textbf{(E)}\ \sqrt{98} } $

Cyclic pentagon $ ABCDE$ has a right angle $ \angle{ABC} \equal{} 90^{\circ}$ and side lengths $ AB \equal{} 15$ and $ BC \equal{} 20$. Supposing that $ AB \equal{} DE \equal{} EA$, find $ CD$.

If circular arcs $ AC$ and $ BC$ have centers at $ B$ and $ A$, respectively, then there exists a circle tangent to both $ \stackrel{\frown}{AC}$ and $ \stackrel{\frown}{BC}$, and to $ \overline{AB}$. If the length of $ \stackrel{\frown}{BC}$ is $ 12$, then the circumference of the circle is [asy]unitsize(4cm); defaultpen(fontsize(8pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,.375); pair A=(-.5,0); pair B=(.5,0); pair C=shift(-.5,0)*dir(60); draw(Arc(A,1,0,60)); draw(Arc(B,1,120,180)); draw(A--B); draw(Circle(O,.375)); dot(A); dot(B); dot(C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N);[/asy]$ \textbf{(A)}\ 24 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 28$

Let $d$ be one of the common tangent lines of externally tangent circles $k_1$ and $k_2$. $d$ touches $k_1$ at $A$. Let $[AB]$ be a diameter of $k_1$. The tangent from $B$ to $k_2$ touches $k_2$ at $C$. If $|AB|=8$ and the diameter of $k_2$ is $7$, then what is $|BC|$? $ \textbf{(A)}\ 7 \qquad\textbf{(B)}\ 6\sqrt 2 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 5\sqrt 3 $

Let $ABCD$ be a nondegenerate isosceles trapezoid with integer side lengths such that $BC \parallel AD$ and $AB=BC=CD$. Given that the distance between the incenters of triangles $ABD$ and $ACD$ is $8!$, determine the number of possible lengths of segment $AD$. [i]Ray Li[/i]

In $ \triangle ABC $, $ AB = 3 $, $ BC = 4 $, and $ CA = 5 $. Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$, $\overline{BC}$ at $B$ and $D$, and $\overline{AC}$ at $F$ and $G$. Given that $EF=DF$ and $\tfrac{DG}{EG} = \tfrac{3}{4}$, length $DE=\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$.

Four circles, no two of which are congruent, have centers at $A$, $B$, $C$, and $D$, and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\frac{5}{8}$ times the radius of circle $B$, and the radius of circle $C$ is $\frac{5}{8}$ times the radius of circle $D$. Furthermore, $AB = CD = 39$ and $PQ = 48$. Let $R$ be the midpoint of $\overline{PQ}$. What is $AR+BR+CR+DR$? $ \textbf{(A)}\ 180 \qquad\textbf{(B)}\ 184 \qquad\textbf{(C)}\ 188 \qquad\textbf{(D)}\ 192\qquad\textbf{(E)}\ 196 $

In $ \triangle ABC$, we have $ AC \equal{} BC \equal{} 7$ and $ AB \equal{} 2$. Suppose that $ D$ is a point on line $ AB$ such that $ B$ lies between $ A$ and $ D$ and $ CD \equal{} 8$. What is $ BD$? $ \textbf{(A)}\ 3\qquad \textbf{(B)}\ 2 \sqrt {3}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 4 \sqrt {2}$

In the accompanying figure, the outer square has side length 40. A second square S' of side length 15 is constructed inside S with the same center as S and with sides parallel to those of S. From each midpoint of a side of S, segments are drawn to the two closest vertices of S'. The result is a four-pointed starlike figure inscribed in S. The star figure is cut out and then folded to form a pyramid with base S'. Find the volume of this pyramid. [asy] draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0)); draw((2.5,2.5)--(4,0)--(5.5,2.5)--(8,4)--(5.5,5.5)--(4,8)--(2.5,5.5)--(0,4)--(2.5,2.5)--(5.5,2.5)--(5.5,5.5)--(2.5,5.5)--(2.5,2.5)); [/asy]

In a rectangle $ABCD$ two not intersecting circles $\omega_1$ and $\omega_2$ are drawn such that $\omega_1$ is tangent to $AB$ and $AD$ at points $P$ and $S$ respectively, and $\omega_2$ is tangent to $CB$ and $CD$ at $T$ and $Q$ respectively. It is known that $PQ=11, ST=10, BD=14$. Find the distance between centers of circles $\omega_1$ and $\omega_2$. [i]I. Voronovich[/i]

A flagpole is originally $ 5$ meters tall. A hurricane snaps the flagpole at a point $ x$ meters above the ground so that the upper part, still attached to the stump, touches the ground $ 1$ meter away from the base. What is $ x$? $ \textbf{(A)}\ 2.0 \qquad \textbf{(B)}\ 2.1 \qquad \textbf{(C)}\ 2.2 \qquad \textbf{(D)}\ 2.3 \qquad \textbf{(E)}\ 2.4$

Circles with centers $ A$ and $ B$ have radii 3 and 8, respectively. A common internal tangent intersects the circles at $ C$ and $ D$, respectively. Lines $ AB$ and $ CD$ intersect at $ E$, and $ AE \equal{} 5$. What is $ CD$? [asy]unitsize(2.5mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair A=(0,0), Ep=(5,0), B=(5+40/3,0); pair M=midpoint(A--Ep); pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1]; pair D=B+8*dir(180+degrees(C)); dot(A); dot(C); dot(B); dot(D); draw(C--D); draw(A--B); draw(Circle(A,3)); draw(Circle(B,8)); label("$A$",A,W); label("$B$",B,E); label("$C$",C,SE); label("$E$",Ep,SSE); label("$D$",D,NW);[/asy]$ \textbf{(A) } 13\qquad \textbf{(B) } \frac {44}{3}\qquad \textbf{(C) } \sqrt {221}\qquad \textbf{(D) } \sqrt {255}\qquad \textbf{(E) } \frac {55}{3}$

Point $P$ is taken interior to a square with side-length $a$ and such that is it equally distant from two consecutive vertices and from the side opposite these vertices. If $d$ represents the common distance, then $d$ equals: $\textbf{(A)}\ \dfrac{3a}{5} \qquad \textbf{(B)}\ \dfrac{5a}{8} \qquad \textbf{(C)}\ \dfrac{3a}{8} \qquad \textbf{(D)}\ \dfrac{a\sqrt{2}}{2} \qquad \textbf{(E)}\ \dfrac{a}{2}$