A pyramid has a square base with sides of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube? $ \textbf{(A)}\ 5\sqrt{2}-7 \qquad \textbf{(B)}\ 7-4\sqrt{3} \qquad \textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{9} \qquad \textbf{(E)}\ \frac{\sqrt{3}}{9} $

The walls of a room are in the shape of a triangle $ABC$ with $\angle ABC = 90^\circ$, $\angle BAC = 60^\circ$, and $AB=6$. Chong stands at the midpoint of $BC$ and rolls a ball toward $AB$. Suppose that the ball bounces off $AB$, then $AC$, then returns exactly to Chong. Find the length of the path of the ball.

Let the vertices $A$ and $C$ of a right triangle $ABC$ be on the arc with center $B$ and radius $20$. A semicircle with diameter $[AB]$ is drawn to the inner region of the arc. The tangent from $C$ to the semicircle touches the semicircle at $D$ other than $B$. Let $CD$ intersect the arc at $F$. What is $|FD|$? $ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac 52 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5 $

In tetrahedron $ABCD$ let $h_a, h_b, h_c$ and $h_d$ be the lengths of the altitudes from each vertex to the opposite side of that vertex. Prove that \[\frac{1}{h_a} <\frac{1}{h_b}+\frac{1}{h_c}+\frac{1}{h_d}.\]

The circular base of a hemisphere of radius $2$ rests on the base of a square pyramid of height $6$. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid? $ \textbf{(A)}\ 3\sqrt{2} \qquad \textbf{(B)}\ \frac{13}{3} \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ \frac{13}{2} $

Let $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8$ be a regular octagon. Let $M_1$, $M_3$, $M_5$, and $M_7$ be the midpoints of sides $\overline{A_1 A_2}$, $\overline{A_3 A_4}$, $\overline{A_5 A_6}$, and $\overline{A_7 A_8}$, respectively. For $i = 1, 3, 5, 7$, ray $R_i$ is constructed from $M_i$ towards the interior of the octagon such that $R_1 \perp R_3$, $R_3 \perp R_5$, $R_5 \perp R_7$, and $R_7 \perp R_1$. Pairs of rays $R_1$ and $R_3$, $R_3$ and $R_5$, $R_5$ and $R_7$, and $R_7$ and $R_1$ meet at $B_1$, $B_3$, $B_5$, $B_7$ respectively. If $B_1 B_3 = A_1 A_2$, then $\cos 2 \angle A_3 M_3 B_1$ can be written in the form $m - \sqrt{n}$, where $m$ and $n$ are positive integers. Find $m + n$.

Let $ABC$ be an acute angled triangle and $CD$ be the altitude through $C$. If $AB = 8$ and $CD = 6$, find the distance between the midpoints of $AD$ and $BC$.

A circle of radius 2 is centered at $ O$. Square $ OABC$ has side length 1. Sides $ \overline{AB}$ and $ \overline{CB}$ are extended past $ b$ to meet the circle at $ D$ and $ E$, respectively. What is the area of the shaded region in the figure, which is bounded by $ \overline{BD}$, $ \overline{BE}$, and the minor arc connecting $ D$ and $ E$? [asy] defaultpen(linewidth(0.8)); pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B; fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray); clip(B--Arc(O, 2, 30, 60)--cycle); draw(Circle(origin, 2)); draw((-2,0)--(2,0)^^(0,-2)--(0,2)); draw(A--D^^C--E); label("$A$", A, dir(point--A)); label("$C$", C, dir(point--C)); label("$O$", O, dir(point--O)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$B$", B, SW);[/asy] $ \textbf{(A) } \frac {\pi}3 \plus{} 1 \minus{} \sqrt {3} \qquad \textbf{(B) } \frac {\pi}2\left( 2 \minus{} \sqrt {3}\right) \qquad \textbf{(C) } \pi\left(2 \minus{} \sqrt {3}\right) \qquad \textbf{(D) } \frac {\pi}{6} \plus{} \frac {\sqrt {3} \minus{} 1}{2} \\ \qquad \indent \textbf{(E) } \frac {\pi}{3} \minus{} 1 \plus{} \sqrt {3}$

The altitude drawn to the base of an isosceles triangle is $ 8$, and the perimeter $ 32$. The area of the triangle is: $ \textbf{(A)}\ 56\qquad \textbf{(B)}\ 48\qquad \textbf{(C)}\ 40\qquad \textbf{(D)}\ 32\qquad \textbf{(E)}\ 24$

A circle with center $ O$ has area $ 156\pi$. Triangle $ ABC$ is equilateral, $ \overline{BC}$ is a chord on the circle, $ OA \equal{} 4\sqrt3$, and point $ O$ is outside $ \triangle ABC$. What is the side length of $ \triangle ABC$? $ \textbf{(A)}\ 2\sqrt3 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 4\sqrt3 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$

Square $ ABCD$ has side length $ 2$. A semicircle with diameter $ \overline{AB}$ is constructed inside the square, and the tangent to the semicricle from $ C$ intersects side $ \overline{AD}$ at $ E$. What is the length of $ \overline{CE}$? [asy] defaultpen(linewidth(0.8)); pair A=origin, B=(1,0), C=(1,1), D=(0,1), X=tangent(C, (0.5,0), 0.5, 1), F=C+2*dir(C--X), E=intersectionpoint(C--F, A--D); draw(C--D--A--B--C--E); draw(Arc((0.5,0), 0.5, 0, 180)); pair point=(0.5,0.5); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E));[/asy] $ \textbf{(A)}\ \frac {2 \plus{} \sqrt5}{2} \qquad \textbf{(B)}\ \sqrt 5 \qquad \textbf{(C)}\ \sqrt 6 \qquad \textbf{(D)}\ \frac52 \qquad \textbf{(E)}\ 5 \minus{} \sqrt5$

A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below, so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of the resulting trapezoid. [asy] size(150); defaultpen(linewidth(0.8)); draw(origin--(8,0)--(8,5)--(0,5)--cycle,linewidth(1)); draw(origin--(8/3,5)^^(16/3,5)--(8,0),linetype("4 4")); draw(origin--(4,3)--(8,0)^^(8/3,5)--(4,3)--(16/3,5),linetype("0 4")); label("$5$",(0,5/2),W); label("$8$",(4,0),S); [/asy]

In the figure, $\overline{AB}$ and $\overline{CD}$ are diameters of the circle with center $O$, $\overline{AB} \perp \overline{CD}$, and chord $\overline{DF}$ intersects $\overline{AB}$ at $E$. If $DE = 6$ and $EF = 2$, then the area of the circle is [asy] size(120); defaultpen(linewidth(0.7)); pair O=origin, A=(-5,0), B=(5,0), C=(0,5), D=(0,-5), F=5*dir(40), E=intersectionpoint(A--B, F--D); draw(Circle(O, 5)); draw(A--B^^C--D--F); dot(O^^A^^B^^C^^D^^E^^F); markscalefactor=0.05; draw(rightanglemark(B, O, D)); label("$A$", A, dir(O--A)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$D$", D, dir(O--D)); label("$F$", F, dir(O--F)); label("$O$", O, NW); label("$E$", E, SE);[/asy] $\textbf{(A)}\ 23\pi \qquad \textbf{(B)}\ \dfrac{47}{2}\pi \qquad \textbf{(C)}\ 24\pi \qquad \textbf{(D)}\ \dfrac{49}{2}\pi \qquad \textbf{(E)}\ 25\pi$

Sides $AB,BC,CD$ and $DA$ of convex polygon $ABCD$ have lengths 3,4,12, and 13, respectively, and $\measuredangle CBA$ is a right angle. The area of the quadrilateral is [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); real r=degrees((12,5)), s=degrees((3,4)); pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s)); draw(A--B--C--D--cycle); markscalefactor=0.05; draw(rightanglemark(A,B,C)); pair point=incenter(A,C,D); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$3$", A--B, dir(A--B)*dir(-90)); label("$4$", B--C, dir(B--C)*dir(-90)); label("$12$", C--D, dir(C--D)*dir(-90)); label("$13$", D--A, dir(D--A)*dir(-90));[/asy] $\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48$

On side $\overline{AB}$ of square $ABCD$, point $E$ is selected. Points $F$ and $G$ are located on sides $\overline{AB}$ and $\overline{AD}$, respectively, such that $\overline{FG} \perp \overline{CE}$. Let $P$ be the intersection point of segments $\overline{FG}$ and $\overline{CE}$. Given that $[EPF] = 1$, $[EPGA] = 8$, and $[CPFB] = 15$, compute $[PGDC]$. (Here $[\mathcal P]$ denotes the area of the polygon $\mathcal P$.) [i]Proposed by Aaron Lin[/i]

In a rectangular plot of land, a man walks in a very peculiar fashion. Labeling the corners $ABCD$, he starts at $A$ and walks to $C$. Then, he walks to the midpoint of side $AD$, say $A_1$. Then, he walks to the midpoint of side $CD$ say $C_1$, and then the midpoint of $A_1D$ which is $A_2$. He continues in this fashion, indefinitely. The total length of his path if $AB=5$ and $BC=12$ is of the form $a + b\sqrt{c}$. Find $\displaystyle\frac{abc}{4}$.

A rectangular piece of paper with side lengths 5 by 8 is folded along the dashed lines shown below, so that the folded flaps just touch at the corners as shown by the dotted lines. Find the area of the resulting trapezoid. [asy] size(150); defaultpen(linewidth(0.8)); draw(origin--(8,0)--(8,5)--(0,5)--cycle,linewidth(1)); draw(origin--(8/3,5)^^(16/3,5)--(8,0),linetype("4 4")); draw(origin--(4,3)--(8,0)^^(8/3,5)--(4,3)--(16/3,5),linetype("0 4")); label("$5$",(0,5/2),W); label("$8$",(4,0),S); [/asy]

Circles with centers $A ,~ B,$ and $C$ each have radius $r$, where $1 < r < 2$. The distance between each pair of centers is $2$. If $B'$ is the point of intersection of circle $A$ and circle $C$ which is outside circle $B$, and if $C'$ is the point of intersection of circle $A$ and circle $B$ which is outside circle $C$, then length $B'C'$ equals $\textbf{(A) }3r-2\qquad\textbf{(B) }r^2\qquad\textbf{(C) }r+\sqrt{3(r-1)}\qquad$ $\textbf{(D) }1+\sqrt{3(r^2-1)}\qquad\textbf{(E) }\text{none of these}$ [asy] //Holy crap, CSE5 is freaking amazing! import cse5; pathpen=black; pointpen=black; dotfactor=3; size(200); pair A=(1,2),B=(2,0),C=(0,0); D(CR(A,1.5)); D(CR(B,1.5)); D(CR(C,1.5)); D(MP("$A$",A)); D(MP("$B$",B)); D(MP("$C$",C)); pair[] BB,CC; CC=IPs(CR(A,1.5),CR(B,1.5)); BB=IPs(CR(A,1.5),CR(C,1.5)); D(BB[0]--CC[1]); MP("$B'$",BB[0],NW);MP("$C'$",CC[1],NE); //Credit to TheMaskedMagician for the diagram [/asy]

On a $4 \times 4 \times 3$ rectangular parallelepiped, vertices $A$, $B$, and $C$ are adjacent to vertex $D$. The perpendicular distance from $D$ to the plane containing $A$, $B$, and $C$ is closest to $\text{(A)}\ 1.6 \qquad \text{(B)}\ 1.9 \qquad \text{(C)}\ 2.1 \qquad \text{(D)}\ 2.7 \qquad \text{(E)}\ 2.9$

Given the areas of the three squares in the figure, what is the area of the interior triangle? [asy] real r=22.61986495; pair A=origin, B=(12,0), C=(12,5); draw(A--B--C--cycle); markscalefactor=0.1; draw(rightanglemark(C, B, A)); draw(scale(12)*shift(0,-1)*unitsquare); draw(scale(5)*shift(12/5,0)*unitsquare); draw(scale(13)*rotate(r)*unitsquare); pair P=shift(0,-1)*(13/sqrt(2) * dir(r+45)), Q=(14.5,1.2), R=(6, -7); label("169", P, N); label("25", Q, N); label("144", R, N); [/asy] $ \textbf{(A)}\ 13\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 1800$

The area of trapezoid $ ABCD$ is $ 164 \text{cm}^2$. The altitude is $ 8 \text{cm}$, $ AB$ is $ 10 \text{cm}$, and $ CD$ is $ 17 \text{cm}$. What is $ BC$, in centimeters? [asy]/* AMC8 2003 #21 Problem */ size(4inch,2inch); draw((0,0)--(31,0)--(16,8)--(6,8)--cycle); draw((11,8)--(11,0), linetype("8 4")); draw((11,1)--(12,1)--(12,0)); label("$A$", (0,0), SW); label("$D$", (31,0), SE); label("$B$", (6,8), NW); label("$C$", (16,8), NE); label("10", (3,5), W); label("8", (11,4), E); label("17", (22.5,5), E);[/asy] $ \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$

In the plane, consider a point $ X$ and a polygon $ \mathcal{F}$ (which is not necessarily convex). Let $ p$ denote the perimeter of $ \mathcal{F}$, let $ d$ be the sum of the distances from the point $ X$ to the vertices of $ \mathcal{F}$, and let $ h$ be the sum of the distances from the point $ X$ to the sidelines of $ \mathcal{F}$. Prove that $ d^2 \minus{} h^2\geq\frac {p^2}{4}.$

Let triangle $ABC$ be a right triangle in the xy-plane with a right angle at $C$. Given that the length of the hypotenuse $AB$ is 60, and that the medians through $A$ and $B$ lie along the lines $y=x+3$ and $y=2x+4$ respectively, find the area of triangle $ABC$.

In triangle $ ABC$, angle $ C$ is a right angle and $ CB > CA$. Point $ D$ is located on $ \overline{BC}$ so that angle $ CAD$ is twice angle $ DAB$. If $ AC/AD \equal{} 2/3$, then $ CD/BD \equal{} m/n$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m \plus{} n$. $ \textbf{(A)}\ 10\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 22\qquad \textbf{(E)}\ 26$

The $8$ eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of $50$ mm and a length of $80$ mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectrangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least $200$ mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters. [asy] size(200); defaultpen(linewidth(0.7)); path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin; path laceR=reflect((75,0),(75,-240))*laceL; draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray); for(int i=0;i<=3;i=i+1) { path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5); unfill(circ1); draw(circ1); unfill(circ2); draw(circ2); } draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));[/asy]