In $\triangle {ABC}$, tangents of the circumcircle $\odot {O}$ at $B, C$ and at $A, B$ intersects at $X, Y$ respectively. $AX$ cuts $BC$ at ${D}$ and $CY$ cuts $AB$ at ${F}$. Ray $DF$ cuts arc $AB$ of the circumcircle at ${P}$. $Q, R$ are on segments $AB, AC$ such that $P, Q, R$ are collinear and $QR \parallel BO$. If $PQ^2=PR \cdot QR$, find $\angle ACB$.

Denote by $B_1$ and $C_1$, the midpoints of the sides $AB$ and $AC$ of the triangle $ABC$. Let the circles circumscribed around the triangles $ABC_1$ and $AB_1C$ intersect at points $A$ and $P$, and let the line $AP$ intersect the circle circumscribed around the triangle $ABC$ at points $A$ and $Q$. Find the ratio $\frac{AQ}{AP}$.

Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width? [asy]unitsize(8mm); defaultpen(linewidth(.8pt)); draw(scale(4)*unitsquare); draw((0,3)--(4,3)); draw((1,3)--(1,4)); draw((2,3)--(2,4)); draw((3,3)--(3,4));[/asy]$ \textbf{(A)}\ \frac {5}{4} \qquad \textbf{(B)}\ \frac {4}{3} \qquad \textbf{(C)}\ \frac {3}{2} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$

Find all integers of the form $\frac{m-6n}{m+2n}$ where $m,n$ are nonzero rational numbers satisfying $m^3=(27n^2+1)(m+2n)$.

Let $ABC$ be a triangle with $A = 90^o, AH$ the altitude, $P,Q$ the feet of the perpendiculars from $H$ to $AB,AC$ respectively. Let $M$ be a variable point on the line $PQ$. The line through $M$ perpendicular to $MH$ meets the lines $AB,AC$ at $R, S$ respectively. i) Prove that circumcircle of $ARS$ always passes the fixed point $H$. ii) Let $M_1$ be another position of $M$ with corresponding points $R_1, S_1$. Prove that the ratio $RR_1/SS_1$ is constant. iii) The point $K$ is symmetric to $H$ with respect to $M$. The line through $K$ perpendicular to the line $PQ$ meets the line $RS$ at $D$. Prove that$ \angle BHR = \angle DHR, \angle DHS = \angle CHS$.

A jacket and a shirt originally sold for $ \$80$ and $ \$40$, respectively. During a sale Chris bought the $ \$80$ jacket at a $40\%$ discount and the $ \$40$ shirt at a $55\%$ discount. The total amount saved was what percent of the total of the original prices? $\text{(A)}\ 45\% \qquad \text{(B)}\ 47\dfrac{1}{2}\% \qquad \text{(C)}\ 50\% \qquad \text{(D)}\ 79\dfrac{1}{6}\% \qquad \text{(E)}\ 95\%$.

Let $\vartriangle ABC$ be a triangle with $AB > AC$, its incircle is tangent to $BC$ at $D$. Let $DE$ be a diameter of the incircle, and let $F$ be the intersection between line $AE$ and side $BC$. Find the ratio between the areas of $\vartriangle DEF$ and $\vartriangle ABC$ in terms of the three side lengths of$\vartriangle ABC$.

Given $ \triangle ABC$ with point $ D$ inside. Let $ A_0\equal{}AD\cap BC$, $ B_0\equal{}BD\cap AC$, $ C_0 \equal{}CD\cap AB$ and $ A_1$, $ B_1$, $ C_1$, $ A_2$, $ B_2$, $ C_2$ are midpoints of $ BC$, $ AC$, $ AB$, $ AD$, $ BD$, $ CD$ respectively. Two lines parallel to $ A_1A_2$ and $ C_1C_2$ and passes through point $ B_0$ intersects $ B_1B_2$ in points $ A_3$ and $ C_3$respectively. Prove that $ \frac{A_3B_1}{A_3B_2}\equal{}\frac{C_3B_1}{C_3B_2}$.

Consider an equilateral triangle $\triangle ABC$. The points $K$ and $L$ divide the leg $BC$ into three equal parts, the point $M$ divides the leg $AC$ in the ratio $1:2$, counting from the vertex $A$. Prove that $\angle AKM+\angle ALM=30^{\circ}$. Proposed by V. Proizvolov

From the vertex $C$ in triangle $ABC$, draw a straight line that bisects the median from $A$. In what ratio does this line divide the segment $AB$? [img]https://1.bp.blogspot.com/-SxWIQ12DIvs/XzcJv5xoV0I/AAAAAAAAMY4/Ezfe8bd7W-Mfp2Qi4qE_gppbh9Fzvb4XwCLcBGAsYHQ/s0/1995%2BMohr%2Bp3.png[/img]

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] $\textbf{(A)}\ \displaystyle \frac{1}{2} \qquad \textbf{(B)}\ \displaystyle \frac{\sqrt 3}{3} \qquad \textbf{(C)}\ \displaystyle \frac{2}{3} \qquad \textbf{(D)}\ \displaystyle \frac{3}{4} \qquad \textbf{(E)}\ \displaystyle \frac{\sqrt 3}{2}$

Define a sequence of real numbers $ a_1$, $ a_2$, $ a_3$, $ \dots$ by $ a_1 = 1$ and $ a_{n + 1}^3 = 99a_n^3$ for all $ n \ge 1$. Then $ a_{100}$ equals $ \textbf{(A)}\ 33^{33} \qquad \textbf{(B)}\ 33^{99} \qquad \textbf{(C)}\ 99^{33} \qquad \textbf{(D)}\ 99^{99} \qquad \textbf{(E)}\ \text{none of these}$

Prove that there is a constant $c>0$ such that for any $n>3$ there exists a planar graph $G$ with $n$ vertices such that every straight-edged plane embedding of $G$ has a pair of edges with ratio of lengths at least $cn$.

A polyhedron has faces that all either triangles or squares. No two square faces share an edge, and no two triangular faces share an edge. What is the ratio of the number of triangular faces to the number of square faces?

In the diagram below $ \angle CAB, \angle CBD$, and $\angle CDE$ are all right angles with side lengths $AC = 3$, $BC = 5$, $BD = 12$, and $DE = 84$. The distance from point $E$ to the line $AB$ can be expressed as the ratio of two relatively prime positive integers, $m$ and $n$. Find $m + n$. [asy] size(300); defaultpen(linewidth(0.8)); draw(origin--(3,0)--(0,4)--cycle^^(0,4)--(6,8)--(3,0)--(30,-4)--(6,8)); label("$A$",origin,SW); label("$B$",(0,4),dir(160)); label("$C$",(3,0),S); label("$D$",(6,8),dir(80)); label("$E$",(30,-4),E);[/asy]

In the figure, what is the ratio of the area of the gray squares to the area of the white squares? [asy] size((70)); draw((10,0)--(0,10)--(-10,0)--(0,-10)--(10,0)); draw((-2.5,-7.5)--(7.5,2.5)); draw((-5,-5)--(5,5)); draw((-7.5,-2.5)--(2.5,7.5)); draw((-7.5,2.5)--(2.5,-7.5)); draw((-5,5)--(5,-5)); draw((-2.5,7.5)--(7.5,-2.5)); fill((-10,0)--(-7.5,2.5)--(-5,0)--(-7.5,-2.5)--cycle, gray); fill((-5,0)--(0,5)--(5,0)--(0,-5)--cycle, gray); fill((5,0)--(7.5,2.5)--(10,0)--(7.5,-2.5)--cycle, gray); [/asy] $ \textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1 $

Triangle $ ABC$ has a right angle at $ B$. Point $ D$ is the foot of the altitude from $ B$, $ AD\equal{}3$, and $ DC\equal{}4$. What is the area of $ \triangle{ABC}$? [asy]unitsize(5mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)); pair D=foot(B,A,C); pair[] ps={B,C,A,D}; draw(A--B--C--cycle); draw(B--D); draw(rightanglemark(B,D,C)); dot(ps); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$3$",midpoint(A--D),NE); label("$4$",midpoint(D--C),NE);[/asy]$ \textbf{(A)}\ 4\sqrt3 \qquad \textbf{(B)}\ 7\sqrt3 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 14\sqrt3 \qquad \textbf{(E)}\ 42$

Two medians of a triangle with unequal sides are $ 3$ inches and $ 6$ inches. Its area is $ 3 \sqrt{15}$ square inches. The length of the third median in inches, is: $ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \sqrt{3} \qquad \textbf{(C)}\ 3 \sqrt{6} \qquad \textbf{(D)}\ 6 \sqrt{3} \qquad \textbf{(E)}\ 6 \sqrt{6}$

$\dfrac{2+4+6+\cdots + 34}{3+6+9+\cdots+51}=$ $\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{2}{3} \qquad \text{(C)}\ \dfrac{3}{2} \qquad \text{(D)}\ \dfrac{17}{3} \qquad \text{(E)}\ \dfrac{34}{3}$

Let $E$ and $F$ be the midpoints of the lines $AB$ and $DA$ of a square $ABCD$, respectively and let $G$ be the intersection of $DE$ with $CF$. Find the aspect ratio of sidelengths of the triangle $EGC$, $| EG | : | GC | : | CE |$.

In the adjoining figure $AB$ and $BC$ are adjacent sides of square $ABCD$; $M$ is the midpoint of $AB$; $N$ is the midpoint of $BC$; and $AN$ and $CM$ intersect at $O$. The ratio of the area of $AOCD$ to the area of $ABCD$ is [asy] draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(2,1)--(2,2)--(1,0)); label("A", (0,0), S); label("B", (2,0), S); label("C", (2,2), N); label("D", (0,2), N); label("M", (1,0), S); label("N", (2,1), E); label("O", (1.2, .8)); [/asy] $ \textbf{(A)}\ \frac{5}{6} \qquad\textbf{(B)}\ \frac{3}{4} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(E)}\ \frac{(\sqrt{3}-1)}{2} $

(1) For integer $n=0,\ 1,\ 2,\ \cdots$ and positive number $a_{n},$ let $f_{n}(x)=a_{n}(x-n)(n+1-x).$ Find $a_{n}$ such that the curve $y=f_{n}(x)$ touches to the curve $y=e^{-x}.$ (2) For $f_{n}(x)$ defined in (1), denote the area of the figure bounded by $y=f_{0}(x), y=e^{-x}$ and the $y$-axis by $S_{0},$ for $n\geq 1,$ the area of the figure bounded by $y=f_{n-1}(x),\ y=f_{n}(x)$ and $y=e^{-x}$ by $S_{n}.$ Find $\lim_{n\to\infty}(S_{0}+S_{1}+\cdots+S_{n}).$

A parallelogram with an angle of $60^o$ has $a$ as the longest side and a shortest side $b$. Let's take the perpendiculars down from the vertices of the obtuse angles to the longest diagonal, then it is divided into three equal parts. Determine the ratio $\frac{a}{b}$.

Prove that the average of the numbers $n \sin n^{\circ} \; (n = 2,4,6,\ldots,180)$ is $\cot 1^{\circ}$.

Let $ABCD$ be a quadrilateral $AB = BC = CD = DA$. Let $MN$ and $PQ$ be two segments perpendicular to the diagonal $BD$ and such that the distance between them is $d > \frac{BD}{2}$, with $M \in AD$, $N \in DC$, $P \in AB$, and $Q \in BC$. Show that the perimeter of hexagon $AMNCQP$ does not depend on the position of $MN$ and $PQ$ so long as the distance between them remains constant.