Points $A$, $B$, $C$, and $D$ lie on a circle such that chords $\overline{AC}$ and $\overline{BD}$ intersect at a point $E$ inside the circle. Suppose that $\angle ADE =\angle CBE = 75^\circ$, $BE=4$, and $DE=8$. The value of $AB^2$ can be written in the form $a+b\sqrt{c}$ for positive integers $a$, $b$, and $c$ such that $c$ is not divisible by the square of any prime. Find $a+b+c$. [i]Proposed by Tony Kim[/i]

Let $A_1, A_2, \ldots, A_n$ be an $n$ -sided regular polygon. If $\frac{1}{A_1 A_2} = \frac{1}{A_1 A_3} + \frac{1}{A_1A_4}$, find $n$.

If $I$ is the incenter of a triangle $ABC$ and $R$ is the radius of its circumcircle then \[AI+BI+CI\leq 3R\]

Rectangle $ABCD$ has $AB=6$ and $BC=3$. Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$. What is the degree measure of $\angle AMD$? $ \textbf{(A)}\ 15 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 75 $

Let $ABCD$ a convex quadrilateral with $AB=BC=CD$, with $AC$ not equal to $BD$ and $E$ be the intersection point of it's diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120$.

In a circle, parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$, $\beta$, and $\alpha + \beta$ radians, respectively, where $\alpha + \beta < \pi$. If $\cos \alpha$, which is a positive rational number, is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?

In the figure, $\triangle ABC$ has $\angle A =45^{\circ}$ and $\angle B =30^{\circ}$. A line $DE$, with $D$ on $AB$ and $\angle ADE =60^{\circ}$, divides $\triangle ABC$ into two pieces of equal area. (Note: the figure may not be accurate; perhaps $E$ is on $CB$ instead of $AC$.) The ratio $\frac{AD}{AB}$ is [asy] size((220)); draw((0,0)--(20,0)--(7,6)--cycle); draw((6,6)--(10,-1)); label("A", (0,0), W); label("B", (20,0), E); label("C", (7,6), NE); label("D", (9.5,-1), W); label("E", (5.9, 6.1), SW); label("$45^{\circ}$", (2.5,.5)); label("$60^{\circ}$", (7.8,.5)); label("$30^{\circ}$", (16.5,.5)); [/asy] $ \textbf{(A)}\ \frac{1}{\sqrt{2}} \qquad\textbf{(B)}\ \frac{2}{2+\sqrt{2}} \qquad\textbf{(C)}\ \frac{1}{\sqrt{3}} \qquad\textbf{(D)}\ \frac{1}{\sqrt[3]{6}} \qquad\textbf{(E)}\ \frac{1}{\sqrt[4]{12}} $

Prove that $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}$

$M$ is an arbitrary point inside $\triangle ABC$. $AM$ intersects the circumcircle of the triangle again at $A_1$. Find the points $M$ that minimise $\frac{MB\cdot MC}{MA_1}$.

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]

Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area? $ \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}$

Consider a regular heptagon ( polygon of $7$ equal sides and angles) $ABCDEFG$ as in the figure below:- $(a).$ Prove $\frac{1}{\sin\frac{\pi}{7}}=\frac{1}{\sin\frac{2\pi}{7}}+\frac{1}{\sin\frac{3\pi}{7}}$ $(b).$ Using $(a)$ or otherwise, show that $\frac{1}{AG}=\frac{1}{AF}+\frac{1}{AE}$ [asy] draw(dir(360/7)..dir(2*360/7),blue); draw(dir(2*360/7)..dir(3*360/7),blue); draw(dir(3*360/7)..dir(4*360/7),blue); draw(dir(4*360/7)..dir(5*360/7),blue); draw(dir(5*360/7)..dir(6*360/7),blue); draw(dir(6*360/7)..dir(7*360/7),blue); draw(dir(7*360/7)..dir(360/7),blue); draw(dir(2*360/7)..dir(4*360/7),blue); draw(dir(4*360/7)..dir(1*360/7),blue); label("$A$",dir(4*360/7),W); label("$B$",dir(5*360/7),S); label("$C$",dir(6*360/7),S); label("$D$",dir(7*360/7),E); label("$E$",dir(1*360/7),E); label("$F$",dir(2*360/7),N); label("$G$",dir(3*360/7),W); [/asy]

Points $A$ and $B$ lie on circle $\omega$. Point $P$ lies on the extension of segment $AB$ past $B$. Line $\ell$ passes through $P$ and is tangent to $\omega$. The tangents to $\omega$ at points $A$ and $B$ intersect $\ell$ at points $D$ and $C$ respectively. Given that $AB=7$, $BC=2$, and $AD=3$, compute $BP$.