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$

Four congruent rectangles are placed as shown. The area of the outer square is $ 4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side? [asy]unitsize(6mm); defaultpen(linewidth(.8pt)); path p=(1,1)--(-2,1)--(-2,2)--(1,2); draw(p); draw(rotate(90)*p); draw(rotate(180)*p); draw(rotate(270)*p);[/asy]$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ 2 \plus{} \sqrt2 \qquad \textbf{(D)}\ 2\sqrt3 \qquad \textbf{(E)}\ 4$

In a convex quadrilateral $ABCD$ the angles $\angle A$ and $\angle C$ are equal and the bisector of $\angle B$ passes through the midpoint of the side $CD$. If it is known that $CD = 3AD$, calculate $\frac{AB}{BC}$.

In the triangle $ABC$ , the angle bisector $AD$ divides the side $BC$ into the ratio $BD: DC = 2: 1$. In what ratio, does the median $CE$ divide this bisector?

After a cyclist has gone $ \frac{2}{3}$ of his route, he gets a flat tire. Finishing on foot, he spends twice as long walking as he did riding. How many times as fast does he ride as walk?

Given distinct straight lines $ OA$ and $ OB$. From a point in $ OA$ a perpendicular is drawn to $ OB$; from the foot of this perpendicular a line is drawn perpendicular to $ OA$. From the foot of this second perpendicular a line is drawn perpendicular to $ OB$; and so on indefinitely. The lengths of the first and second perpendiculars are $ a$ and $ b$, respectively. Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds. This limit is: $ \textbf{(A)}\ \frac {b}{a \minus{} b} \qquad \textbf{(B)}\ \frac {a}{a \minus{} b} \qquad \textbf{(C)}\ \frac {ab}{a \minus{} b} \qquad \textbf{(D)}\ \frac {b^2}{a \minus{} b} \qquad \textbf{(E)}\ \frac {a^2}{a \minus{} b}$

An equilateral triangle is originally painted black. Each time the triangle is changed, the middle fourth of each black triangle turns white. After five changes, what fractional part of the original area of the black triangle remains black? [asy] unitsize(36); fill((0,0)--(2,0)--(1,sqrt(3))--cycle,gray); draw((0,0)--(2,0)--(1,sqrt(3))--cycle,linewidth(1)); fill((4,0)--(6,0)--(5,sqrt(3))--cycle,gray); fill((5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--cycle,white); draw((5,sqrt(3))--(4,0)--(5,0)--(9/2,sqrt(3)/2)--(11/2,sqrt(3)/2)--(5,0)--(6,0)--cycle,linewidth(1)); fill((8,0)--(10,0)--(9,sqrt(3))--cycle,gray); fill((9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--cycle,white); fill((17/2,0)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--cycle,white); fill((9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--cycle,white); fill((19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--cycle,white); draw((9,sqrt(3))--(35/4,3*sqrt(3)/4)--(37/4,3*sqrt(3)/4)--(9,sqrt(3)/2)--(35/4,3*sqrt(3)/4)--(33/4,sqrt(3)/4)--(35/4,sqrt(3)/4)--(17/2,0)--(33/4,sqrt(3)/4)--(8,0)--(9,0)--(17/2,sqrt(3)/2)--(19/2,sqrt(3)/2)--(9,0)--(19/2,0)--(37/4,sqrt(3)/4)--(39/4,sqrt(3)/4)--(19/2,0)--(10,0)--cycle,linewidth(1)); label("Change 1",(3,3*sqrt(3)/4),N); label("$\Longrightarrow $",(3,5*sqrt(3)/8),S); label("Change 2",(7,3*sqrt(3)/4),N); label("$\Longrightarrow $",(7,5*sqrt(3)/8),S); [/asy] $\text{(A)}\ \frac{1}{1024} \qquad \text{(B)}\ \frac{15}{64} \qquad \text{(C)}\ \frac{243}{1024} \qquad \text{(D)}\ \frac{1}{4} \qquad \text{(E)}\ \frac{81}{256}$

The lengths of the sides of an acute angled triangle are successive integers. Prove that the altitude to the second longest side divides this side into two segments whose difference in length equals $4$.

The median $BM$ is drawn in the triangle $ABC$. It is known that $\angle ABM = 40 {} ^ \circ$ and $\angle CBM = 70 {} ^ \circ $ Find the ratio $AB: BM$.

After elections, every parliament member (PM), has his own absolute rating. When the parliament set up, he enters in a group and gets a relative rating. The relative rating is the ratio of its own absolute rating to the sum of all absolute ratings of the PMs in the group. A PM can move from one group to another only if in his new group his relative rating is greater. In a given day, only one PM can change the group. Show that only a finite number of group moves is possible. [i](A rating is positive real number.)[/i]

Let $ABCD$ be a quadrilateral, and let $E$ and $F$ be points on sides $AD$ and $BC$, respectively, such that $\frac{AE}{ED} = \frac{BF}{FC}$. Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T$, respectively. Prove that the circumcircles of triangles $SAE$, $SBF$, $TCF$, and $TDE$ pass through a common point.

In terms of the fixed non-negative integers $\alpha$ and $\beta$ determine the least upper bound of the ratio (or show that it is unbounded) \[ \frac{S(n)}{S(2^{\alpha}5^{\beta}n)} \] as $n$ varies through the positive integers, where $S(\cdot)$ denotes sum of digits in decimal representation.

Let $S$ be a finite subset of a straight line. Say that $S$ has the [i]repeated distance property [/i] if every value of the distance between two points of $S$ (except the longest) occurs at least twice. Show that if $S$ has the [i]repeated distance property [/i] then the ratio of any two distances between two points of $S$ is rational.

Let $ ABCD$ be a trapezoid with $ AB\parallel{}CD$, $ AB\equal{}11$, $ BC\equal{}5$, $ CD\equal{}19$, and $ DA\equal{}7$. Bisectors of $ \angle A$ and $ \angle D$ meet at $ P$, and bisectors of $ \angle B$ and $ \angle C$ meet at $ Q$. What is the area of hexagon $ ABQCDP$? $ \textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textbf{(D)}\ 35\sqrt{3}\qquad \textbf{(E)}\ 36\sqrt{3}$

Three points $A,B,C$ are such that $B\in AC$. On one side of $AC$, draw the three semicircles with diameters $AB,BC,CA$. The common interior tangent at $B$ to the first two semicircles meets the third circle $E$. Let $U,V$ be the points of contact of the common exterior tangent to the first two semicircles. Evaluate the ratio $R=\frac{[EUV]}{[EAC]}$ as a function of $r_{1} = \frac{AB}{2}$ and $r_2 = \frac{BC}{2}$, where $[X]$ denotes the area of polygon $X$.

Larry and Julius are playing a game, taking turns throwing a ball at a bottle sitting on a ledge. Larry throws first. The winner is the first person to knock the bottle off the ledge. At each turn the probability that a player knocks the bottle off the ledge is $\frac{1}{2}$, independently of what has happened before. What is the probability that Larry wins the game? $\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{3}{5}\qquad\textbf{(C) }\frac{2}{3}\qquad\textbf{(D) }\frac{3}{4}\qquad\textbf{(E) }\frac{4}{5}$

A square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is $ 2: 1$. The ratio of the rectangle's length to its width is $ 2: 1$. What percent of the rectangle's area is inside the square? $ \textbf{(A)}\ 12.5 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 87.5$

Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE \equal{} CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF \equal{} FG$.

Jesse cuts a circular paper disk of radius $12$ along two radii to form two sectors, the smaller having a central angle of $120$ degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger? $ \textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{1}{4} \qquad\textbf{(C)}\ \frac{\sqrt{10}}{10} \qquad\textbf{(D)}\ \frac{\sqrt{5}}{6} \qquad\textbf{(E)}\ \frac{\sqrt{10}}{5} $

Older television screens have an aspect ratio of $ 4: 3$. That is, the ratio of the width to the height is $ 4: 3$. The aspect ratio of many movies is not $ 4: 3$, so they are sometimes shown on a television screen by 'letterboxing' - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of $ 2: 1$ and is shown on an older television screen with a $ 27$-inch diagonal. What is the height, in inches, of each darkened strip? [asy]unitsize(1mm); defaultpen(linewidth(.8pt)); filldraw((0,0)--(21.6,0)--(21.6,2.7)--(0,2.7)--cycle,grey,black); filldraw((0,13.5)--(21.6,13.5)--(21.6,16.2)--(0,16.2)--cycle,grey,black); draw((0,2.7)--(0,13.5)); draw((21.6,2.7)--(21.6,13.5));[/asy]$ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 2.25 \qquad \textbf{(C)}\ 2.5 \qquad \textbf{(D)}\ 2.7 \qquad \textbf{(E)}\ 3$

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$

The number $2.5081081081081\ldots$ can be written as $\frac{m}{n}$ where $m$ and $n$ are natural numbers with no common factors. Find $m + n$.

A square sheet of paper $ABCD$ is so folded that $B$ falls on the mid point of $M$ of $CD$. Prove that the crease will divide $BC$ in the ration $5 : 3$.

Quadrilateral $XABY$ is inscribed in the semicircle $\omega$ with diameter $XY$. Segments $AY$ and $BX$ meet at $P$. Point $Z$ is the foot of the perpendicular from $P$ to line $XY$. Point $C$ lies on $\omega$ such that line $XC$ is perpendicular to line $AZ$. Let $Q$ be the intersection of segments $AY$ and $XC$. Prove that \[\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.\]

Let $M$, $N$ and $P$ be midpoints of sides $BC, AC$ and $AB$, respectively, and let $O$ be circumcenter of acute-angled triangle $ABC$. Circumcircles of triangles $BOC$ and $MNP$ intersect at two different points $X$ and $Y$ inside of triangle $ABC$. Prove that \[\angle BAX=\angle CAY.\]