The ratio of $w$ to $x$ is $4 : 3$, the ratio of $y$ to $z$ is $3 : 2$, and the ratio of $z$ to $x$ is $1 : 6$. What is the ratio of $w$ to $y$? $\textbf{(A) }4:3 \qquad \textbf{(B) }3:2 \qquad \textbf{(C) } 8:3 \qquad \textbf{(D) } 4:1 \qquad \textbf{(E) } 16:3 $

Find $A^2$, where $A$ is the sum of the absolute values of all roots of the following equation: \begin{eqnarray*}x &=& \sqrt{19} + \frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{x}}}}}}}}}\end{eqnarray*}

A decorative window is made up of a rectangle with semicircles at either end. The ratio of $AD$ to $AB$ is $3:2$. And $AB$ is 30 inches. What is the ratio of the area of the rectangle to the combined area of the semicircle. [asy] import graph; size(5cm); real lsf=0; pen dps=linewidth(0.7)+fontsize(8); defaultpen(dps); pen ds=black; real xmin=-4.27,xmax=14.73,ymin=-3.22,ymax=6.8; draw((0,4)--(0,0)); draw((0,0)--(2.5,0)); draw((2.5,0)--(2.5,4)); draw((2.5,4)--(0,4)); draw(shift((1.25,4))*xscale(1.25)*yscale(1.25)*arc((0,0),1,0,180)); draw(shift((1.25,0))*xscale(1.25)*yscale(1.25)*arc((0,0),1,-180,0)); dot((0,0),ds); label("$A$",(-0.26,-0.23),NE*lsf); dot((2.5,0),ds); label("$B$",(2.61,-0.26),NE*lsf); dot((0,4),ds); label("$D$",(-0.26,4.02),NE*lsf); dot((2.5,4),ds); label("$C$",(2.64,3.98),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy] $ \textbf{(A)}\ 2:3 \qquad\textbf{(B)}\ 3:2\qquad\textbf{(C)}\ 6:\pi \qquad\textbf{(D)}\ 9: \pi \qquad\textbf{(E)}\ 30 : \pi$

Let $ABC$ be a triangle and let $P$ be a point in the plane of $ABC$ that is inside the region of the angle $BAC$ but outside triangle $ABC$. [b](a)[/b] Prove that any two of the following statements imply the third. [list] [b](i)[/b] the circumcentre of triangle $PBC$ lies on the ray $\stackrel{\to}{PA}$. [b](ii)[/b] the circumcentre of triangle $CPA$ lies on the ray $\stackrel{\to}{PB}$. [b](iii)[/b] the circumcentre of triangle $APB$ lies on the ray $\stackrel{\to}{PC}$.[/list] [b](b)[/b] Prove that if the conditions in (a) hold, then the circumcentres of triangles $BPC,CPA$ and $APB$ lie on the circumcircle of triangle $ABC$.

In rectangle $ABCD$, $DC = 2CB$ and points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$ as shown. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD$? [asy] draw((0, 0)--(0, 1)--(2, 1)--(2, 0)--cycle); draw((0, 0)--(sqrt(3)/3, 1)); draw((0, 0)--(sqrt(3), 1)); label("A", (0, 1), N); label("B", (2, 1), N); label("C", (2, 0), S); label("D", (0, 0), S); label("E", (sqrt(3)/3, 1), N); label("F", (sqrt(3), 1), N); [/asy] ${ \textbf{(A)}\ \ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{\sqrt{6}}{8}\qquad\textbf{(C)}\ \frac{3\sqrt{3}}{16}\qquad\textbf{(D)}}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$

Points $A$, $B$, and $O$ lie in the plane such that $\measuredangle AOB = 120^\circ$. Circle $\omega_0$ with radius $6$ is constructed tangent to both $\overrightarrow{OA}$ and $\overrightarrow{OB}$. For all $i \ge 1$, circle $\omega_i$ with radius $r_i$ is constructed such that $r_i < r_{i - 1}$ and $\omega_i$ is tangent to $\overrightarrow{OA}$, $\overrightarrow{OB}$, and $\omega_{i - 1}$. If \[ S = \sum_{i = 1}^\infty r_i, \] then $S$ can be expressed as $a\sqrt{b} + c$, where $a, b, c$ are integers and $b$ is not divisible by the square of any prime. Compute $100a + 10b + c$. [i]Proposed by Aaron Lin[/i]

A circle is inscribed in a square of side $m$, then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If $S_n$ is the sum of the areas of the first $n$ circles so inscribed, then, as $n$ grows beyond all bounds, $S_n$ approaches: $\textbf{(A)}\ \frac{\pi m^2}{2}\qquad \textbf{(B)}\ \frac{3\pi m^2}{8}\qquad \textbf{(C)}\ \frac{\pi m^2}{3}\qquad \textbf{(D)}\ \frac{\pi m^2}{4}\qquad \textbf{(E)}\ \frac{\pi m^2}{8}$

Let $ABCDEF$ be a convex hexagon. Segments $AE$ and $BF$ intersect at $X$ and segments $BD$ and $CE$ intersect in $Y.$ It's known that $$ \angle XBC = \angle XDE = \angle YAB = \angle YEF = 80^\circ \text{ and } \angle XCB = \angle XED = \angle YBA = \angle YFE = \angle 70^\circ.$$ Let $P$ and $Q$ be such points on line $XY$ that segments $PX$ and $AF$ intersect, segments $QY$ and $CD$ intersect and $\angle APF = \angle CQD = 30 ^\circ.$ Estimate the sum: \[ \frac{BX}{BF} + \frac{BY}{BD} + \frac{EX}{EA} + \frac{EY}{EC} + \frac{PX}{PY} + \frac{QY}{QX} \] [i]Proposed by Gogi Khimshiashvili, Georgia [/i]

In $ \triangle ABC$ we have $ AB \equal{} 7$, $ AC \equal{} 8$, and $ BC \equal{} 9$. Point $ D$ is on the circumscribed circle of the triangle so that $ \overline{AD}$ bisects $ \angle BAC$. What is the value of $ AD/CD$? $ \textbf{(A)}\ \frac{9}{8}\qquad \textbf{(B)}\ \frac{5}{3}\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ \frac{17}{7}\qquad \textbf{(E)}\ \frac{5}{2}$

Let $M, N$ be points inside the angle $\angle BAC$ usch that $\angle MAB\equiv \angle NAC$. If $M_{1}, M_{2}$ and $N_{1}, N_{2}$ are the projections of $M$ and $N$ on $AB, AC$ respectively then prove that $M, N$ and $P$ the intersection of $M_{1}N_{2}$ with $N_{1}M_{2}$ are collinear.

Two circles $ C_1,C_2$ with different radii are given in the plane, they touch each other externally at $ T$. Consider any points $ A\in C_1$ and $ B\in C_2$, both different from $ T$, such that $ \angle ATB \equal{} 90^{\circ}$. (a) Show that all such lines $ AB$ are concurrent. (b) Find the locus of midpoints of all such segments $ AB$.

Is there a polyhedron whose area ratio of any two faces is at least $2$ ?

Let $M$ and $N$ be the midpoints of the respective sides $AB$ and $AC$ of an acute-angled triangle $ABC$. Let $P$ be the foot of the perpendicular from $N$ onto $BC$ and let $A_1$ be the midpoint of $MP$. Points $B_1$ and $C_1$ are obtained similarly. If $AA_1$, $BB_1$ and $CC_1$ are concurrent, show that the triangle $ABC$ is isosceles. [i]Mircea Becheanu[/i]

You are given a pair of triangles for which two sides of one triangle are equal in length to two sides of the second triangle, and the triangles are similar, but not necessarily congruent. Prove that the ratio of the sides that correspond under the similarity is a number between $ \frac {1}{2}(\sqrt {5} \minus{} 1)$ and $ \frac {1}{2}(\sqrt {5} \plus{} 1)$.

Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB' = 3AB$. Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC' = 3BC$, and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA' = 3CA$. What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$? $\textbf{(A) }9:1\qquad\textbf{(B) }16:1\qquad\textbf{(C) }25:1\qquad\textbf{(D) }36:1\qquad\textbf{(E) }37:1$