Point $D$ is on side $CB$ of triangle $ABC$. If \[ \angle{CAD} = \angle{DAB} = 60^\circ,\quad AC = 3\quad\mbox{ and }\quad AB = 6, \] then the length of $AD$ is $\text{(A)} \ 2 \qquad \text{(B)} \ 2.5 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 3.5 \qquad \text{(E)} \ 4$

Let ABCD be a convex quadrilateral such that $AB + CD = \sqrt{2}AC$ and $BC + DA = \sqrt{2}BD$. Prove that ABCD is a parallelogram.

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]

In $\triangle A B C$, let $D, E$, and $F$ be the midpoints of the sides of the triangle, and let $P, Q,$ and $R$ be the midpoints of the corresponding medians, $AD ,B E,$ and $C F$, respectively, as shown in the figure at the right. Prove that the value of \[\frac{AQ^2 + A R^2 + B P^2 + B R^2 + C P^2+ C Q^2 }{A B^2 + B C^2 + C A^2}\] does not depend on the shape of $\triangle A B C$ and find that value. [asy] defaultpen(linewidth(0.7)+fontsize(10));size(200); pair A=origin, B=(14,0), C=(9,12), D=midpoint(C--B), E=midpoint(C--A), F=midpoint(A--B), R=midpoint(C--F), P=midpoint(D--A), Q=midpoint(E--B); draw(A--B--C--A, linewidth(1)); draw(A--D^^B--E^^C--F); draw(B--R--A--Q--C--P--cycle, dashed); pair point=centroid(A,B,C); 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)); label("$F$", F, dir(point--F)); label("$P$", P, dir(40)*dir(point--P)); label("$Q$", Q, dir(40)*dir(point--Q)); label("$R$", R, dir(40)*dir(point--R)); dot(P^^Q^^R);[/asy]

Let $ABC$ be an acute triangle. Denote by $B_0$ and $C_0$ the feet of the altitudes from vertices $B$ and $C$, respectively. Let $X$ be a point inside the triangle $ABC$ such that the line $BX$ is tangent to the circumcircle of the triangle $AXC_0$ and the line $CX$ is tangent to the circumcircle of the triangle $AXB_0$. Show that the line $AX$ is perpendicular to $BC$.

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]$

The cyclic octagon $ABCDEFGH$ has sides $a,a,a,a,b,b,b,b$ respectively. Find the radius of the circle that circumscribes $ABCDEFGH.$

Let $\triangle ABC$ be an arbitary triangle, and construct $P,Q,R$ so that each of the angles marked is $30^\circ$. Prove that $\triangle PQR$ is an equilateral triangle. [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair ext30(pair pt1, pair pt2) { pair r1 = pt1+rotate(-30)*(pt2-pt1), r2 = pt2+rotate(30)*(pt1-pt2); draw(anglemark(r1,pt1,pt2,25)); draw(anglemark(pt1,pt2,r2,25)); return intersectionpoints(pt1--r1, pt2--r2)[0]; } pair A = (0,0), B=(10,0), C=(3,7), P=ext30(B,C), Q=ext30(C,A), R=ext30(A,B); draw(A--B--C--A--R--B--P--C--Q--A); draw(P--Q--R--cycle, linetype("8 8")); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$P$", P, NE); label("$Q$", Q, NW); label("$R$", R, S);[/asy]

The sides of a triangle are $ 30$, $ 70$, and $ 80$ units. If an altitude is dropped upon the side of length $ 80$, the larger segment cut off on this side is: $ \textbf{(A)}\ 62\qquad \textbf{(B)}\ 63\qquad \textbf{(C)}\ 64\qquad \textbf{(D)}\ 65\qquad \textbf{(E)}\ 66$

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 $a^{2}+b^{2}+c^{2}$?

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$

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]

The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon? [asy] size(130); pair A, B, C, D, E, F, G, H, I, J, K, L; A = dir(120); B = dir(60); C = dir(0); D = dir(-60); E = dir(-120); F = dir(180); draw(A--B--C--D--E--F--cycle); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); G = midpoint(A--B); H = midpoint(B--C); I = midpoint(C--D); J = midpoint(D--E); K = midpoint(E--F); L = midpoint(F--A); draw(G--H--I--J--K--L--cycle); label("$A$", A, dir(120)); label("$B$", B, dir(60)); label("$C$", C, dir(0)); label("$D$", D, dir(-60)); label("$E$", E, dir(-120)); label("$F$", F, dir(180)); [/asy] $\textbf{(A)}\ \displaystyle \frac{1}{2} \qquad \textbf{(B)}\ \displaystyle \frac{\sqrt 3}{3} \qquad \textbf{(C)}\ \displaystyle \frac{2}{3} \qquad \textbf{(D)}\ \displaystyle \frac{3}{4} \qquad \textbf{(E)}\ \displaystyle \frac{\sqrt 3}{2}$

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} $

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 $

Two congruent $ 30^{\circ}$-$ 60^{\circ}$-$ 90^{\circ}$ are placed so that they overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the area common to both triangles is $ \textbf{(A)}\ 6\sqrt3 \qquad \textbf{(B)}\ 8\sqrt3 \qquad \textbf{(C)}\ 9\sqrt3 \qquad \textbf{(D)}\ 12\sqrt3 \qquad \textbf{(E)}\ 24$

Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles.

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 $CE$.

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 $\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \dfrac{1+\sqrt{3}}{\sqrt{2}} \qquad \textbf{(C)}\ \sqrt{3} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \dfrac{4}{\sqrt{3}}$

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 $a+b+c$.

Let $ABC$ be a triangle with $AB = 42$, $AC = 39$, $BC = 45$. Let $E$, $F$ be on the sides $\overline{AC}$ and $\overline{AB}$ such that $AF = 21, AE = 13$. Let $\overline{CF}$ and $\overline{BE}$ intersect at $P$, and let ray $AP$ meet $\overline{BC}$ at $D$. Let $O$ denote the circumcenter of $\triangle DEF$, and $R$ its circumradius. Compute $CO^2-R^2$. [i]Proposed by Yang Liu[/i]

Given a square cardboard of area $\frac{1}{4}$, and a paper triangle of area $\frac{1}{2}$ such that the square of its sidelength is a positive integer. Prove that the triangle can be folded in some ways such that the squace can be placed inside the folded figure so that both of its faces are completely covered with paper. [i]Proposed by N.Beluhov, Bulgaria[/i]

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 $\angle CMB$.