Let $ABC$ be a right triangle where $\measuredangle A = 90^\circ$ and $M\in (AB)$ such that $\frac{AM}{MB}=3\sqrt{3}-4$. It is known that the symmetric point of $M$with respect to the line $GI$ lies on $AC$. Find the measure of $\measuredangle B$.

Rectangle $ ABCD$ has $ AB \equal{} 4$ and $ BC \equal{} 3$. Segment $ EF$ is constructed through $ B$ so that $ EF$ is perpendicular to $ DB$, and $ A$ and $ C$ lie on $ DE$ and $ DF$, respectively. What is $ EF$? $ \textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ \frac {125}{12}\qquad \textbf{(D)}\ \frac {103}{9}\qquad \textbf{(E)}\ 12$

Triangle $ABC$ has $AB = 4$, $AC = 5$, and $BC = 6$. An angle bisector is drawn from angle $A$, and meets $BC$ at $M$. What is the nearest integer to $100 \frac{AM}{CM}$?

Let $ABCD$ be a trapezoid such that $AB \parallel CD$, $|BC|+|AD| = 7$, $|AB| = 9$ and $|BC| = 14$. What is the ratio of the area of the triangle formed by $CD$, angle bisector of $\widehat{BCD}$ and angle bisector of $\widehat{CDA}$ over the area of the trapezoid? $ \textbf{a)}\ \dfrac{9}{14} \qquad\textbf{b)}\ \dfrac{5}{7} \qquad\textbf{c)}\ \sqrt 2 \qquad\textbf{d)}\ \dfrac{49}{69} \qquad\textbf{e)}\ \dfrac 13 $

The parallel sides of a trapezoid are $ 3$ and $ 9$. The non-parallel sides are $ 4$ and $ 6$. A line parallel to the bases divides the trapezoid into two trapezoids of equal perimeters. The ratio in which each of the non-parallel sides is divided is: [asy]defaultpen(linewidth(.8pt)); unitsize(2cm); pair A = origin; pair B = (2.25,0); pair C = (2,1); pair D = (1,1); pair E = waypoint(A--D,0.25); pair F = waypoint(B--C,0.25); draw(A--B--C--D--cycle); draw(E--F); label("6",midpoint(A--D),NW); label("3",midpoint(C--D),N); label("4",midpoint(C--B),NE); label("9",midpoint(A--B),S);[/asy]$ \textbf{(A)}\ 4: 3\qquad \textbf{(B)}\ 3: 2\qquad \textbf{(C)}\ 4: 1\qquad \textbf{(D)}\ 3: 1\qquad \textbf{(E)}\ 6: 1$

Let $\phi = \frac{1+\sqrt5}{2}$. Prove that a positive integer appears in the list $$\lfloor \phi \rfloor , \lfloor 2 \phi \rfloor, \lfloor 3\phi \rfloor ,... , \lfloor n\phi \rfloor , ... $$ if and only if it appears exactly twice in the list $$\lfloor 1/ \phi \rfloor , \lfloor 2/ \phi \rfloor, \lfloor 3/\phi \rfloor , ... ,\lfloor n/\phi \rfloor , ... $$

A pyramid with the base $ABCD$ and the top $V$ is inscribed in a sphere. Let $AD = 2BC$ and let the rays $AB$ and $DC$ intersect in point $E$. Compute the ratio of the volume of the pyramid $VAED$ to the volume of the pyramid $VABCD$.

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 $

Let $\triangle{ABC}$ be a triangle with circumcircle $\tau$. The tangentlines to $\tau$ through $A$ and $B$ intersect at $T$. The circle through $A$, $B$ and $T$ intersects $BC$ and $AC$ again at $D$ and $E$, respectively; $CT$ and $BE$ intersect at $F$. Suppose $D$ is the midpoint of $BC$. Calculate the ratio $BF:BE$. [i](Swiss Mathematical Olympiad 2011, Final round, problem 5)[/i]

Triangle $ABC$ has side-lengths $AB=12$, $BC=24$, and $AC=18$. The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N$. What is the perimeter of $\triangle AMN$? $ \textbf{(A)}\ 27 \qquad \textbf{(B)}\ 30 \qquad \textbf{(C)}\ 33 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 42 $

Find the ratio of two numbers if the ratio of their arithmetic mean to their geometric mean is $25 : 24$

A square sheet of paper $ABCD$ is folded straight in such a way that point $B$ hits to the midpoint of side $CD$. In what ratio does the fold line divide side $BC$?

Let $ ABC$ be a triangle, and $ I$ its incenter. Let the incircle of $ ABC$ touch side $ BC$ at $ D$, and let lines $ BI$ and $ CI$ meet the circle with diameter $ AI$ at points $ P$ and $ Q$, respectively. Given $ BI \equal{} 6, CI \equal{} 5, DI \equal{} 3$, determine the value of $ \left( DP / DQ \right)^2$.

In triangle $ABC$, point $I$ is the center, point $I_a$ is the center of the excircle tangent to the side $BC$. From the vertex $A$ inside the angle $BAC$ draw rays $AX$ and $AY$. The ray $AX$ intersects the lines $BI$, $CI$, $BI_a$, $CI_a$ at points $X_1$, $...$, $X_4$, respectively, and the ray $AY$ intersects the same lines at points $Y_1$, $...$, $Y_4$ respectively. It turned out that the points $X_1,X_2,Y_1,Y_2$ lie on the same circle. Prove the equality $$\frac{X_1X_2}{X_3X_4}= \frac{Y_1Y_2}{Y_3Y_4}.$$

Let $ ABC$ be an acute triangle. Points $ D$, $ E$, and $ F$ lie on segments $ BC$, $ CA$, and $ AB$, respectively, and each of the three segments $ AD$, $ BE$, and $ CF$ contains the circumcenter of $ ABC$. Prove that if any two of the ratios $ \frac{BD}{DC}$, $ \frac{CE}{EA}$, $ \frac{AF}{FB}$, $ \frac{BF}{FA}$, $ \frac{AE}{EC}$, $ \frac{CD}{DB}$ are integers, then triangle $ ABC$ is isosceles.

Points $ B$ and $ C$ lie on $ \overline{AD}$. The length of $ \overline{AB}$ is $ 4$ times the length of $ \overline{BD}$, and the length of $ \overline{AC}$ is $ 9$ times the length of $ \overline{CD}$. The length of $ \overline{BC}$ is what fraction of the length of $ \overline{AD}$? $ \textbf{(A)}\ \frac{1}{36} \qquad \textbf{(B)}\ \frac{1}{13} \qquad \textbf{(C)}\ \frac{1}{10} \qquad \textbf{(D)}\ \frac{5}{36} \qquad \textbf{(E)}\ \frac{1}{5}$

Equiangular hexagon $ ABCDEF$ has side lengths $ AB \equal{} CD \equal{} EF \equal{} 1$ and $ BC \equal{} DE \equal{} FA \equal{} r$. The area of $ \triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $ r$? $ \textbf{(A)}\ \frac {4\sqrt {3}}{3} \qquad \textbf{(B)}\ \frac {10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac {17}{4} \qquad \textbf{(E)}\ 6$

Let $ ABC$ be an acute triangle, let $ M,N$ be the midpoints of minor arcs $ \widehat{CA},\widehat{AB}$ of the circumcircle of triangle $ ABC,$ point $ D$ is the midpoint of segment $ MN,$ point $ G$ lies on minor arc $ \widehat{BC}.$ Denote by $ I,I_{1},I_{2}$ the incenters of triangle $ ABC,ABG,ACG$ respectively.Let $ P$ be the second intersection of the circumcircle of triangle $ GI_{1}I_{2}$ with the circumcircle of triangle $ ABC.$ Prove that three points $ D,I,P$ are collinear.

$E$ and $F$ are the midpoints of the sides $BC$ and $AD$ of the convex quadrilateral $ABCD$. Prove that the segment $EF$ divides the diagonals $AC$ and $BD$ in the same ratio.

Let $AB=AC$ in $\triangle ABC$, and let $D$ be a point on segment $AB$. The tangent at $D$ to the circumcircle $\omega$ of $BCD$ hits $AC$ at $E$. The other tangent from $E$ to $\omega$ touches it at $F$, and $G=BF \cap CD$, $H=AG \cap BC$. Prove that $BH=2HC$. [i]Proposed by David Stoner[/i]

Given triangle $ABC$. Suppose $P$ and $P_1$ are points on $BC, Q$ lies on $CA, R$ lies on $AB$, such that $$\frac{AR}{RB}=\frac{BP}{PC}=\frac{CQ}{QA}=\frac{CP_1}{P_1B}$$ Let $G$ be the centroid of triangle $ABC$ and $K = AP_1 \cap RQ$. Prove that points $P,G$, and $K$ are collinear.

In triangle $ABC$, $\measuredangle CBA=72^\circ$, $E$ is the midpoint of side $AC$, and $D$ is a point on side $BC$ such that $2BD=DC$; $AD$ and $BE$ intersect at $F$. The ratio of the area of triangle $BDF$ to the area of quadrilateral $FDCE$ is [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E); draw(E--B--A--C--B^^A--D); label("$A$", A, dir(D--A)); label("$B$", B, dir(E--B)); label("$C$", C, dir(0)); label("$D$", D, SE); label("$E$", E, N); label("$F$", F, dir(80));[/asy] $\text{(A)} \ \frac 15 \qquad \text{(B)} \ \frac 14 \qquad \text{(C)} \ \frac 13 \qquad \text{(D)} \ \frac 25 \qquad \text{(E)} \ \text{none of these}$

Suppose that $a$ and $b$ are two distinct positive real numbers such that $\lfloor na\rfloor$ divides $\lfloor nb\rfloor$ for any positive integer $n$. Prove that $a$ and $b$ are positive integers.

Let $ABCD$ be a square with side length $43$ and points $X$ and $Y$ lies on sides $AD$ and $BC$ respectively such that the ratio of the area of $ABYX$ to the area of $CDXY$ is $20 : 23$ . Find the maximum possible length of $XY$.

A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and yellow when in the sun. Initially the ratio of green to yellow frogs was 3:1. Then 3 green frogs moved to the sunny side and 5 yellow frogs moved to the shady side. Now the ratio is 4:1. What is the difference between the number of green frogs and yellow frogs now? $\textbf{(A) } 10\qquad\textbf{(B) } 12\qquad\textbf{(C) } 16\qquad\textbf{(D) } 20\qquad\textbf{(E) } 24$