A circle is inscribed in an equilateral triangle, and a square is inscribed in the circle. The ratio of the area of the triangle to the area of the square is: $\textbf{(A)}\ \sqrt{3}:1 \qquad \textbf{(B)}\ \sqrt{3}:\sqrt{2} \qquad \textbf{(C)}\ 3\sqrt{3}:2 \qquad \textbf{(D)}\ 3:\sqrt{2} \qquad \textbf{(E)}\ 3:2\sqrt{2}$

In the plane, there are two circles $\Gamma_1, \Gamma_2$ intersecting each other at two points $A$ and $B$. Tangents of $\Gamma_1$ at $A$ and $B$ meet each other at $K$. Let us consider an arbitrary point $M$ (which is different of $A$ and $B$) on $\Gamma_1$. The line $MA$ meets $\Gamma_2$ again at $P$. The line $MK$ meets $\Gamma_1$ again at $C$. The line $CA$ meets $\Gamma_2 $ again at $Q$. Show that the midpoint of $PQ$ lies on the line $MC$ and the line $PQ$ passes through a fixed point when $M$ moves on $\Gamma_1$. [color=red][Moderator edit: This problem was also discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=21414 .][/color]

The area of a pizza with radius $4$ inches is $N$ percent larger than the area of a pizza with radius $3$ inches. What is the integer closest to $N$? $\textbf{(A) } 25 \qquad\textbf{(B) } 33 \qquad\textbf{(C) } 44\qquad\textbf{(D) } 66 \qquad\textbf{(E) } 78$

Let $A$, $B$ and $C$ be three points on a line (in this order). For each circle $k$ through the points $B$ and $C$, let $D$ be one point of intersection of the perpendicular bisector of $BC$ with the circle $k$. Further, let $E$ be the second point of intersection of the line $AD$ with $k$. Show that for each circle $k$, the ratio of lengths $\overline{BE}:\overline{CE}$ is the same.

Consider the triangle $ABC$ and the points $D \in (BC),E \in (CA), F \in (AB)$, such that $\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}$. Prove that if the circumcenters of triangles $DEF$ and $ABC$ coincide, then the triangle $ABC$ is equilateral.

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$? [asy] size(200); defaultpen(linewidth(.6pt)+fontsize(12pt)); dotfactor=4; draw((0,0)--(0,2)); draw((0,0)--(1,0)); draw((1,0)--(1,2)); draw((0,1)--(2,1)); draw((0,0)--(1,2)); draw((0,2)--(2,1)); draw((0,2)--(2,2)); draw((2,1)--(2,2)); label("$A$",(0,2),NW); label("$B$",(1,2),N); label("$C$",(4/5,1.55),W); dot((0,2)); dot((1,2)); dot((4/5,1.6)); dot((2,1)); dot((0,0)); [/asy] $ \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{5}\qquad\textbf{(C)}\ \frac{2}{9}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{\sqrt2}{4} $

The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle? $\textbf{(A)} \text{ } \frac{\sqrt{3}-1}{2} \qquad \textbf{(B)} \text{ } \frac{1}{2} \qquad \textbf{(C)} \text{ } \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)} \text{ } \frac{\sqrt{2}}{2} \qquad \textbf{(E)} \text{ } \frac{\sqrt{6}-1}{2}$

Alex has a piece of cheese. He chooses a positive number $a\neq 1$ and cuts the piece into several pieces one by one. Every time he chooses a piece and cuts it in the same ratio $1:a.$ His goal is to divide the cheese into two piles of equal masses. Can he do it?

Let $ABC$ be a triangle inscribed in a circle and let \[ l_a = \frac{m_a}{M_a} \ , \ \ l_b = \frac{m_b}{M_b} \ , \ \ l_c = \frac{m_c}{M_c} \ , \] where $m_a$,$m_b$, $m_c$ are the lengths of the angle bisectors (internal to the triangle) and $M_a$, $M_b$, $M_c$ are the lengths of the angle bisectors extended until they meet the circle. Prove that \[ \frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B} + \frac{l_c}{\sin^2 C} \geq 3 \] and that equality holds iff $ABC$ is an equilateral triangle.

The sides of triangle $BAC$ are in the ratio $2: 3: 4$. $BD$ is the angle-bisector drawn to the shortest side $AC$, dividing it into segments $AD$ and $CD$. If the length of $AC$ is $10$, then the length of the longer segment of $AC$ is: $\text{(A)} \ 3\frac12 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 5\frac57 \qquad \text{(D)} \ 6 \qquad \text{(E)} \ 7\frac12$

[asy] fill(circle((4,0),4),grey); fill((0,0)--(8,0)--(8,-4)--(0,-4)--cycle,white); fill(circle((7,0),1),white); fill(circle((3,0),3),white); draw((0,0)--(8,0),black+linewidth(1)); draw((6,0)--(6,sqrt(12)),black+linewidth(1)); MP("A", (0,0), W); MP("B", (8,0), E); MP("C", (6,0), S); MP("D",(6,sqrt(12)), N); [/asy] In this diagram semi-circles are constructed on diameters $\overline{AB}$, $\overline{AC}$, and $\overline{CB}$, so that they are mutually tangent. If $\overline{CD} \bot \overline{AB}$, then the ratio of the shaded area to the area of a circle with $\overline{CD}$ as radius is: $\textbf{(A)}\ 1:2\qquad \textbf{(B)}\ 1:3\qquad \textbf{(C)}\ \sqrt{3}:7\qquad \textbf{(D)}\ 1:4\qquad \textbf{(E)}\ \sqrt{2}:6$

If $AB$ and $CD$ are perpendicular diameters of circle $Q$, $P$ in $\overline{AQ}$, and $\measuredangle QPC = 60^\circ$, then the length of $PQ$ divided by the length of $AQ$ is [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=(-1,0), B=(1,0), C=(0,1), D=(0,-1), Q=origin, P=(-0.5,0); draw(P--C--D^^A--B^^Circle(Q,1)); label("$A$", A, W); label("$B$", B, E); label("$C$", C, N); label("$D$", D, S); label("$P$", P, S); label("$Q$", Q, SE); label("$60^\circ$", P+0.0.5*dir(30), dir(30));[/asy] $ \textbf{(A)} \ \frac{\sqrt{3}}{2} \qquad \textbf{(B)} \ \frac{\sqrt{3}}{3} \qquad \textbf{(C)} \ \frac{\sqrt{2}}{2} \qquad \textbf{(D)} \ \frac12 \qquad \textbf{(E)} \ \frac23 $

Let $M$ be the midpoint of a given segment $BC$. Point $A$ is chosen to maximize $\angle ABC$ while subject to the condition that $\angle MAC = 20^o$ . What is the ratio $BC/BA$ ?

The first three terms of a geometric progression are $\sqrt{2}, \sqrt[3]{2}, \sqrt[6]{2}$. Find the fourth term. ${{ \textbf{(A)}\ 1 \qquad\textbf{(B)}\ \sqrt[7]{2} \qquad\textbf{(C)}\ \sqrt[8]{2} \qquad\textbf{(D)}\ \sqrt[9]{2} }\qquad\textbf{(E)}\ \sqrt[10]{2} } $

A square and equilateral triangle have the same perimeter. If the triangle has area $16\sqrt3$, what is the area of the square? [i]Proposed by Evan Chen[/i]

Michael walks at the rate of $ 5$ feet per second on a long straight path. Trash pails are located every $ 200$ feet along the path. A garbage truck travels at $ 10$ feet per second in the same direction as Michael and stops for $ 30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leaving the next pail. How many times will Michael and the truck meet? $ \textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$

A circle that passes through the vertex $A$ of a rectangle $ABCD$ intersects the side $AB$ at a second point $E$ different from $B.$ A line passing through $B$ is tangent to this circle at a point $T,$ and the circle with center $B$ and passing through $T$ intersects the side $BC$ at the point $F.$ Show that if $\angle CDF= \angle BFE,$ then $\angle EDF=\angle CDF.$

Given a triangle $ ABC$. The angle bisectors of the angles $ ABC$ and $ BCA$ intersect the sides $ CA$ and $ AB$ at the points $ B_1$ and $ C_1$, and intersect each other at the point $ I$. The line $ B_1C_1$ intersects the circumcircle of triangle $ ABC$ at the points $ M$ and $ N$. Prove that the circumradius of triangle $ MIN$ is twice as long as the circumradius of triangle $ ABC$.

Let $P,Q,R$ be points on the sides of $\triangle ABC$ such that $P \in [AB],Q\in[BC],R\in[CA]$ and $\frac{|AP|}{|AB|} = \frac {|BQ|}{|BC|} =\frac{|CR|}{|CA|} =k < \frac 12$ If $G$ is the centroid of $\triangle ABC$, find the ratio $\frac{Area(\triangle PQG)}{Area(\triangle PQR)}$ .

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$.

In isosceles right triangle $ ABC$ a circle is inscribed. Let $ CD$ be the hypotenuse height ($ D\in AB$), and let $ P$ be the intersection of inscribed circle and height $ CD$. In which ratio does the circle divide segment $ AP$?

Given are two points $B,C$. Consider point $A$ not lying on the line $BC$ and draw the circles $C_1(K_1,R_1)$ (with center $K_1$ and radius $R_1$) and $C_2(K_2,R_2)$ with chord $AB, AC$ respectively such that their centers lie on the interior of the triangle $ABC$ and also $R_1 \cdot AC= R_2 \cdot AB$. Let $T$ be the intersection point of the two circles, different from $A$, and M be a random pointof line $AT$, prove that $TC \cdot S_{(MBT)}=TB \cdot S_{(MCT)}$

Let \begin{eqnarray*} f(x) & = & a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 \ \ \mbox{and} \\ g(x) & = & c_{n+1} x^{n+1} + c_n x^n + \cdots + c_0 \end{eqnarray*} be non-zero polynomials with real coefficients such that $g(x) = (x+r)f(x)$ for some real number $r$. If $a = \max(|a_n|, \ldots, |a_0|)$ and $c = \max(|c_{n+1}|, \ldots, |c_0|)$, prove that $\frac{a}{c} \leq n+1$.

In a convex quadrilateral $ABCD$ we have $\angle ADB=\angle ACD$ and $AC=CD=DB$. If the diagonals $AC$ and $BD$ intersect at $X$, prove that $\frac{CX}{BX}-\frac{AX}{DX}=1$.

In the diagram below, three squares are inscribed in right triangles. Their areas are $A$, $M$, and $N$, as indicated in the diagram. If $M = 5$ and $N = 12$, then $A$ can be expressed as $a + b\sqrt{c}$, where $a$, $b$, and $c$ are positive integers and $c$ is not divisible by the square of any prime. Compute $a + b + c$. [asy] size(250); defaultpen (linewidth (0.7) + fontsize (10)); pair O = origin, A = (1, 1), B = (4/3, 1/3), C = (2/3, 5/3), P = (3/2, 0), Q = (0,3); draw (P--O--Q--cycle^^(0, 5/3)--C--(2/3,1)^^(0,1)--A--(1,0)^^(1,1/3)--B--(4/3,0)); label("$A$", (.5,.5)); label("$M$", (7/6, 1/6)); label("$N$", (1/3, 4/3));[/asy] [i]Proposed by Aaron Lin[/i]