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 $

A square of area $40$ is inscribed in a semicircle as shown. What is the area of the semicircle? [asy] defaultpen(linewidth(0.8)); real r=sqrt(50), s=sqrt(10); draw(Arc(origin, r, 0, 180)); draw((r,0)--(-r,0), dashed); draw((s,0)--(s,2*s)--(-s,2*s)--(-s,0));[/asy] $ \textbf{(A) }20\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }30\pi\qquad\textbf{(D) }40\pi\qquad\textbf{(E) }50\pi $

A cubical cake with edge length $ 2$ inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where $ M$ is the midpoint of a top edge. The piece whose top is triangle $ B$ contains $ c$ cubic inches of cake and $ s$ square inches of icing. What is $ c\plus{}s$? [asy]unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); label("$M$",(-1,0),W); label("$C$",(-0.1,-0.3)); label("$A$",(-0.4,0.7)); label("$B$",(0.7,0.4));[/asy]$ \textbf{(A)}\ \frac{24}{5} \qquad \textbf{(B)}\ \frac{32}{5} \qquad \textbf{(C)}\ 8\plus{}\sqrt5 \qquad \textbf{(D)}\ 5\plus{}\frac{16\sqrt5}{5} \qquad \textbf{(E)}\ 10\plus{}5\sqrt5$

Square $ EFGH$ is inside the square $ ABCD$ so that each side of $ EFGH$ can be extended to pass through a vertex of $ ABCD$. Square $ ABCD$ has side length $ \sqrt {50}$ and $ BE \equal{} 1$. What is the area of the inner square $ EFGH$? [asy]unitsize(4cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair D=(0,0), C=(1,0), B=(1,1), A=(0,1); pair F=intersectionpoints(Circle(D,2/sqrt(5)),Circle(A,1))[0]; pair G=foot(A,D,F), H=foot(B,A,G), E=foot(C,B,H); draw(A--B--C--D--cycle); draw(D--F); draw(C--E); draw(B--H); draw(A--G); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,NNW); label("$F$",F,ENE); label("$G$",G,SSE); label("$H$",H,WSW);[/asy]$ \textbf{(A)}\ 25\qquad \textbf{(B)}\ 32\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 40\qquad \textbf{(E)}\ 42$

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}$

Circles of radii 5, 5, 8, and $m/n$ are mutually externally tangent, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

A square has sides of length $2$. Set $S$ is the set of all line segments that have length $2$ and whose endpoints are on adjacent sides of the square. The midpoints of the line segments in set $S$ enclose a region whose area to the nearest hundredth is $k$. Find $100k$.

A rectangle has sides of length $a$ and $36$. A hinge is installed at each vertex of the rectangle and at the midpoint of each side of length $36$. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of $24,$ the hexagon has the same area as the original rectangle. Find $a^2$. [asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (12*2.236, 36); E = (12*2.236, 18); F = (12*2.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236 [/asy]

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}$

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 $m + n$

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$

A square sheet of paper $ABCD$ is so folded that $B$ falls on the mid point of $M$ of $CD$. Prove that the crease will divide $BC$ in the ration $5 : 3$.

Two sequences of integers, $ a_1, a_2, a_3, \ldots$ and $ b_1, b_2, b_3, \ldots$, satisfy the equation \[ (a_n \minus{} a_{n \minus{} 1})(a_n \minus{} a_{n \minus{} 2}) \plus{} (b_n \minus{} b_{n \minus{} 1})(b_n \minus{} b_{n \minus{} 2}) \equal{} 0 \] for each integer $ n$ greater than $ 2$. Prove that there is a positive integer $ k$ such that $ a_k \equal{} a_{k \plus{} 2008}$.

On a sheet divided into squares, each square measuring $2cm$, two circles are drawn such that both circles are inscribed in a square as in the figure below. Determine the minimum distance between the two circles.

Two circles of equal radius can tightly fit inside right triangle $ABC$, which has $AB=13$, $BC=12$, and $CA=5$, in the three positions illustrated below. Determine the radii of the circles in each case. [asy] size(400); defaultpen(linewidth(0.7)+fontsize(12)); picture p = new picture; pair s1 = (20,0), s2 = (40,0); real r1 = 1.5, r2 = 10/9, r3 = 26/7; pair A=(12,5), B=(0,0), C=(12,0); draw(p,A--B--C--cycle); label(p,"$B$",B,SW); label(p,"$A$",A,NE); label(p,"$C$",C,SE); add(p); add(shift(s1)*p); add(shift(s2)*p); draw(circle(C+(-r1,r1),r1)); draw(circle(C+(-3*r1,r1),r1)); draw(circle(s1+C+(-r2,r2),r2)); draw(circle(s1+C+(-r2,3*r2),r2)); pair D=s2+156/17*(A-B)/abs(A-B), E=s2+(169/17,0), F=extension(D,E,s2+A,s2+C); draw(incircle(s2+B,D,E)); draw(incircle(s2+A,D,F)); label("Case (i)",(6,-3)); label("Case (ii)",s1+(6,-3)); label("Case (iii)",s2+(6,-3));[/asy]

If the sum of the lengths of the six edges of a trirectangular tetrahedron $ PABC$ (i.e., $ \angle APB \equal{} \angle BPC \equal{} \angle CPA \equal{} 90^\circ$) is $ S$, determine its maximum volume.

In triangle $ABC$ medians of triangle $BE$ and $AD$ are perpendicular to each other. Find the length of $\overline{AB}$, if $\overline{BC}=6$ and $\overline{AC}=8$

In trapezoid $ ABCD$, $ \overline{AB}$ and $ \overline{CD}$ are perpendicular to $ \overline{AD}$, with $ AB\plus{}CD\equal{}BC$, $ AB<CD$, and $ AD\equal{}7$. What is $ AB\cdot CD$? $ \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$

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$

In circle $ O$, the midpoint of radius $ OX$ is $ Q$; at $ Q$, $ \overline{AB} \perp \overline{XY}$. The semi-circle with $ \overline{AB}$ as diameter intersects $ \overline{XY}$ in $ M$. Line $ \overline{AM}$ intersects circle $ O$ in $ C$, and line $ \overline{BM}$ intersects circle $ O$ in $ D$. Line $ \overline{AD}$ is drawn. Then, if the radius of circle $ O$ is $ r$, $ AD$ is: [asy]defaultpen(linewidth(.8pt)); unitsize(2.5cm); real m = 0; real b = 0; pair O = origin; pair X = (-1,0); pair Y = (1,0); pair Q = midpoint(O--X); pair A = (Q.x, -1*sqrt(3)/2); pair B = (Q.x, -1*A.y); pair M = (Q.x + sqrt(3)/2,0); m = (B.y - M.y)/(B.x - M.x); b = (B.y - m*B.x); pair D = intersectionpoint(Circle(O,1),M--(1.5,1.5*m + b)); m = (A.y - M.y)/(A.x - M.x); b = (A.y - m*A.x); pair C = intersectionpoint(Circle(O,1),M--(1.5,1.5*m + b)); draw(Circle(O,1)); draw(Arc(Q,sqrt(3)/2,-90,90)); draw(A--B); draw(X--Y); draw(B--D); draw(A--C); draw(A--D); dot(O);dot(M); label("$B$",B,NW); label("$C$",C,NE); label("$Y$",Y,E); label("$D$",D,SE); label("$A$",A,SW); label("$X$",X,W); label("$Q$",Q,SW); label("$O$",O,SW); label("$M$",M,NE+2N);[/asy]$ \textbf{(A)}\ r\sqrt {2} \qquad \textbf{(B)}\ r\qquad \textbf{(C)}\ \text{not a side of an inscribed regular polygon}\qquad \textbf{(D)}\ \frac {r\sqrt {3}}{2}\qquad \textbf{(E)}\ r\sqrt {3}$

In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3 \cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$.

A circle of radius $ 1$ is surrounded by $ 4$ circles of radius $ r$ as shown. What is $ r$? [asy]defaultpen(linewidth(.9pt)); real r = 1 + sqrt(2); pair A = dir(45)*(r + 1); pair B = dir(135)*(r + 1); pair C = dir(-135)*(r + 1); pair D = dir(-45)*(r + 1); draw(Circle(origin,1)); draw(Circle(A,r));draw(Circle(B,r));draw(Circle(C,r));draw(Circle(D,r)); draw(A--(dir(45)*r + A)); draw(B--(dir(45)*r + B)); draw(C--(dir(45)*r + C)); draw(D--(dir(45)*r + D)); draw(origin--(dir(25))); label("$r$",midpoint(A--(dir(45)*r + A)), SE); label("$r$",midpoint(B--(dir(45)*r + B)), SE); label("$r$",midpoint(C--(dir(45)*r + C)), SE); label("$r$",midpoint(D--(dir(45)*r + D)), SE); label("$1$",origin,W);[/asy]$ \textbf{(A)}\ \sqrt {2}\qquad \textbf{(B)}\ 1 \plus{} \sqrt {2}\qquad \textbf{(C)}\ \sqrt {6}\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 2 \plus{} \sqrt {2}$

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} } $

In rectangle $ ABCD$, $ AB\equal{}100$. Let $ E$ be the midpoint of $ \overline{AD}$. Given that line $ AC$ and line $ BE$ are perpendicular, find the greatest integer less than $ AD$.

A triangle with sides of $ 5$, $ 12$, and $ 13$ has both an inscibed and a circumscribed circle. What is the distance between the centers of those circles? $ \textbf{(A)}\ \frac{3\sqrt{5}}{2}\qquad \textbf{(B)}\ \frac{7}{2}\qquad \textbf{(C)}\ \sqrt{15}\qquad \textbf{(D)}\ \frac{\sqrt{65}}{2}\qquad \textbf{(E)}\ \frac{9}{2}$