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$.

In quadrilateral $ ABCD$, it is given that $ \angle A \equal{} 120^\circ$, angles $ B$ and $ D$ are right angles, $ AB \equal{} 13$, and $ AD \equal{} 46$. Then $ AC \equal{}$ $ \textbf{(A)}\ 60 \qquad \textbf{(B)}\ 62 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 65 \qquad \textbf{(E)}\ 72$

Let $ AB$ be the diameter of semicircle $O$ , $C, D $ be points on the arc $AB$, $P, Q$ be respectively the circumcenter of $\triangle OAC $ and $\triangle OBD $ . Prove that:$CP\cdot CQ=DP \cdot DQ$.[asy] import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black; real h=sqrt(55/64); pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B); D(arc(O,1,0,180),darkgreen); D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue); D(O); [/asy]

A ship sails $ 10$ miles in a straight line from $ A$ to $ B$, turns through an angle between $ 45^{\circ}$ and $ 60^{\circ}$, and then sails another $ 20$ miles to $ C$. Let $ AC$ be measured in miles. Which of the following intervals contains $ AC^2$? [asy]unitsize(2mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C=20*dir(50); draw(A--B--C); draw(A--C,linetype("4 4")); dot(A); dot(B); dot(C); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C),SE); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE);[/asy]$ \textbf{(A)}\ [400,500] \qquad \textbf{(B)}\ [500,600] \qquad \textbf{(C)}\ [600,700] \qquad \textbf{(D)}\ [700,800]$ $ \textbf{(E)}\ [800,900]$

In the adjoining figure triangle $ ABC$ is such that $ AB \equal{} 4$ and $ AC \equal{} 8$. If $ M$ is the midpoint of $ BC$ and $ AM \equal{} 3$, what is the length of $ BC$? $ \textbf{(A)}\ 2\sqrt{26} \qquad \textbf{(B)}\ 2\sqrt{31} \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 4\plus{}2\sqrt{13} \qquad$ $ \textbf{(E)}\ \text{not enough information given to solve the problem}$ [asy]draw((0,0)--(2.8284,2)--(8,0)--cycle); draw((2.8284,2)--(4,0)); label("A",(2.8284,2),N); label("B",(0,0),S); label("C",(8,0),S); label("M",(4,0),S);[/asy]

Two circles intersect at points $A$ and $B$. The minor arcs $AB$ measure $30^\circ$ on one circle and $60^\circ$ on the other circle. What is the ratio of the area of the larger circle to the area of the smaller circle? $\textbf{(A) }2\qquad \textbf{(B) }1+\sqrt3\qquad \textbf{(C) }3\qquad \textbf{(D) }2+\sqrt3\qquad \textbf{(E) }4\qquad$

The angles in a particular triangle are in arithmetic progression, and the side lengths are $4,5,x$. The sum of the possible values of $x$ equals $a+\sqrt{b}+\sqrt{c}$ where $a, b$, and $c$ are positive integers. What is $a+b+c$? $ \textbf{(A)}\ 36\qquad\textbf{(B)}\ 38\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 44$

Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is $\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \dfrac{1+\sqrt{3}}{\sqrt{2}} \qquad \textbf{(C)}\ \sqrt{3} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \dfrac{4}{\sqrt{3}}$

Let $ABC$ be a triangle where $M$ is the midpoint of $\overline{AC}$, and $\overline{CN}$ is the angle bisector of $\angle ACB$ with $N$ on $\overline{AB}$. Let $X$ be the intersection of the median $\overline{BM}$ and the bisector $\overline{CN}$. In addition $\bigtriangleup BXN$ is equilateral and $AC=2$. What is $BN^2$? $\textbf{(A)}\ \frac{10-6\sqrt{2}}{7} \qquad\textbf{(B)}\ \frac{2}{9} \qquad\textbf{(C)}\ \frac{5\sqrt{2} - 3\sqrt{3}}{8} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{6} \qquad\textbf{(E)}\ \frac{3\sqrt{3} - 4}{5}$.

A regular 12-gon is inscribed in a circle of radius 12. The sum of the lengths of all sides and diagonals of the 12-gon can be written in the form \[ a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}, \] where $a$, $b$, $c$, and $d$ are positive integers. Find $a + b + c + d$.

Sides $AB,~ BC,$ and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5,$ and $20$, respectively. If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5} $, then side $AD$ has length $\textbf{(A) }24\qquad\textbf{(B) }24.5\qquad\textbf{(C) }24.6\qquad\textbf{(D) }24.8\qquad\textbf{(E) }25$ [size=70]*A polygon is called “simple” if it is not self intersecting.[/size]

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 $a+b+c$.

A right circular cone has a base with radius 600 and height $200\sqrt{7}$. A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is 125, and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is $375\sqrt{2}$. Find the least distance that the fly could have crawled.

Let $ AB$ be the diameter of semicircle $O$ , $C, D $ be points on the arc $AB$, $P, Q$ be respectively the circumcenter of $\triangle OAC $ and $\triangle OBD $ . Prove that:$CP\cdot CQ=DP \cdot DQ$.[asy] import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black; real h=sqrt(55/64); pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B); D(arc(O,1,0,180),darkgreen); D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue); D(O); [/asy]

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 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]

In triangle ABC, side $a = \sqrt{3}$, side $b = \sqrt{3}$, and side $c > 3$. Let $x$ be the largest number such that the magnitude, in degrees, of the angle opposite side $c$ exceeds $x$. Then $x$ equals: $\textbf{(A)}\ 150 \qquad \textbf{(B)}\ 120\qquad \textbf{(C)}\ 105 \qquad \textbf{(D)}\ 90 \qquad \textbf{(E)}\ 60$

The lengths of the sides of a triangle are consescutive integers, and the largest angle is twice the smallest angle. The cosine of the smallest angle is $\textbf {(A) } \frac 34 \qquad \textbf {(B) } \frac{7}{10} \qquad \textbf {(C) } \frac 23 \qquad \textbf {(D) } \frac{9}{14} \qquad \textbf {(E) } \text{none of these}$

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 $p+q$.

Triangle $ABC$ is a right triangle with $AC=7,$ $BC=24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD=BD=15.$ Given that the area of triangle $CDM$ may be expressed as $\frac{m\sqrt{n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p.$

In triangle $ABC$, $\angle ABC = 120^{\circ}$, $AB = 3$ and $BC = 4$. If perpendiculars constructed to $\overline{AB}$ at $A$ and to $\overline{BC}$ at $C$ meet at $D$, then $CD = $ $ \textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{8}{\sqrt{3}}\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ \frac{11}{2}\qquad\textbf{(E)}\ \frac{10}{\sqrt{3}} $

To the nearest degree, find the measure of the largest angle in a triangle with side lengths $3$, $5$, and $7$.