Given triangle $ABC$ and point $P$ on the circumcircle of triangle $ABC$. Suppose the line $CP$ intersects line $AB$ at point $E$ and line $BP$ intersect line $AC$ at point $F$. Suppose also the perpendicular bisector of $AB$ intersects $AC$ at point $K$ and the perpendicular bisector of $AC$ intersects $AB$ at point $J$. Prove that $$\left( \frac{CE}{BF}\right)^2= \frac{AJ \cdot JE }{ AK \cdot KF}$$

The geometric series $ a \plus{} ar \plus{} ar^{2} \plus{} ...$ has a sum of $ 7$, and the terms involving odd powers of $ r$ have a sum of $ 3$. What is $ a \plus{} r$? $ \textbf{(A)}\ \frac {4}{3}\qquad \textbf{(B)}\ \frac {12}{7}\qquad \textbf{(C)}\ \frac {3}{2}\qquad \textbf{(D)}\ \frac {7}{3}\qquad \textbf{(E)}\ \frac {5}{2}$

Rectangle $ ABCD$ and a semicircle with diameter $ AB$ are coplanar and have nonoverlapping interiors. Let $ \mathcal{R}$ denote the region enclosed by the semicircle and the rectangle. Line $ \ell$ meets the semicircle, segment $ AB$, and segment $ CD$ at distinct points $ N$, $ U$, and $ T$, respectively. Line $ \ell$ divides region $ \mathcal{R}$ into two regions with areas in the ratio $ 1: 2$. Suppose that $ AU \equal{} 84$, $ AN \equal{} 126$, and $ UB \equal{} 168$. Then $ DA$ can be represented as $ m\sqrt {n}$, where $ m$ and $ n$ are positive integers and $ n$ is not divisible by the square of any prime. Find $ m \plus{} n$.

The point $M$ is inside the tetrahedron $ABCD$ and the intersection points of the lines $AM,BM,CM$ and $DM$ with the opposite walls are denoted with $A_1,B_1,C_1,D_1$ respectively. It is given also that the ratios $\frac{MA}{MA_1}$, $\frac{MB}{MB_1}$, $\frac{MC}{MC_1}$, and $\frac{MD}{MD_1}$ are equal to the same number $k$. Find all possible values of $k$. [i]K. Petrov[/i]

Find all pairwise relatively prime positive integers $l, m, n$ such that \[(l+m+n)\left( \frac{1}{l}+\frac{1}{m}+\frac{1}{n}\right)\] is an integer.

The edges of a regular tetrahedron with vertices $A ,~ B,~ C$, and $D$ each have length one. Find the least possible distance between a pair of points $P$ and $Q$, where $P$ is on edge $AB$ and $Q$ is on edge $CD$. $\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{3}{4}\qquad\textbf{(C) }\frac{\sqrt{2}}{2}\qquad\textbf{(D) }\frac{\sqrt{3}}{2}\qquad\textbf{(E) }\frac{\sqrt{3}}{3}$ [asy] size(150); import patterns; pair D=(0,0),C=(1,-1),B=(2.5,-0.2),A=(1,2),AA,BB,CC,DD,P,Q,aux; add("hatch",hatch()); //AA=new A and etc. draw(rotate(100,D)*(A--B--C--D--cycle)); AA=rotate(100,D)*A; BB=rotate(100,D)*D; CC=rotate(100,D)*C; DD=rotate(100,D)*B; aux=midpoint(AA--BB); draw(BB--DD); P=midpoint(AA--aux); aux=midpoint(CC--DD); Q=midpoint(CC--aux); draw(AA--CC,dashed); dot(P); dot(Q); fill(DD--BB--CC--cycle,pattern("hatch")); label("$A$",AA,W); label("$B$",BB,S); label("$C$",CC,E); label("$D$",DD,N); label("$P$",P,S); label("$Q$",Q,E); //Credit to TheMaskedMagician for the diagram [/asy]

The summation $\sum_{k=1}^{360} \frac{1}{k \sqrt{k+1} + (k+1)\sqrt{k}}$ is the ratio of two relatively prime positive integers $m$ and $n$. Find $m + n$.

Let $ABC$ be a triangle inscribed in circle $\omega$, and let the medians from $B$ and $C$ intersect $\omega$ at $D$ and $E$ respectively. Let $O_1$ be the center of the circle through $D$ tangent to $AC$ at $C$, and let $O_2$ be the center of the circle through $E$ tangent to $AB$ at $B$. Prove that $O_1$, $O_2$, and the nine-point center of $ABC$ are collinear. [i]Proposed by Michael Kural[/i]

Let $w_1, w_2, \dots, w_n$ be complex numbers. A line $L$ in the complex plane is called a mean line for the points $w_1, w_2, \dots, w_n$ if $L$ contains points (complex numbers) $z_1, z_2, \dots, z_n$ such that \[ \sum_{k = 1}^n (z_k - w_k) = 0. \] For the numbers $w_1 = 32 + 170i$, $w_2 = -7 + 64i$, $w_3 = -9 +200i$, $w_4 = 1 + 27i$, and $w_5 = -14 + 43i$, there is a unique mean line with $y$-intercept 3. Find the slope of this mean line.

There is a square sheet of paper. How to get a rectangular sheet of paper with an aspect ratio equal to $\sqrt2$? (There are no tools, the sheet can only be bent.)

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$

In $\triangle ABC$, consider points $D$ and $E$ on $AC$ and $AB$, respectively, and assume that they do not coincide with any of the vertices $A$, $B$, $C$. If the segments $BD$ and $CE$ intersect at $F$, consider areas $w$, $x$, $y$, $z$ of the quadrilateral $AEFD$ and the triangles $BEF$, $BFC$, $CDF$, respectively. [list=a] [*] Prove that $y^2 > xz$. [*] Determine $w$ in terms of $x$, $y$, $z$. [/list] [asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(12); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.8465032978885407, xmax = 9.445649196374966, ymin = -1.7618066305534972, ymax = 4.389732795464592; /* image dimensions */ draw((3.8295013012181283,2.816337276198864)--(-0.7368327629589799,-0.5920813291311117)--(5.672613975760373,-0.636902634996282)--cycle, linewidth(0.5)); /* draw figures */ draw((3.8295013012181283,2.816337276198864)--(-0.7368327629589799,-0.5920813291311117), linewidth(0.5)); draw((-0.7368327629589799,-0.5920813291311117)--(5.672613975760373,-0.636902634996282), linewidth(0.5)); draw((5.672613975760373,-0.636902634996282)--(3.8295013012181283,2.816337276198864), linewidth(0.5)); draw((-0.7368327629589799,-0.5920813291311117)--(4.569287648059735,1.430279997142299), linewidth(0.5)); draw((5.672613975760373,-0.636902634996282)--(1.8844000180622977,1.3644681598392678), linewidth(0.5)); label("$y$",(2.74779188172294,0.23771684184669772),SE*labelscalefactor); label("$w$",(3.2941097703568736,1.8657441499758196),SE*labelscalefactor); label("$x$",(1.6660824622277512,1.0025618859342047),SE*labelscalefactor); label("$z$",(4.288408327670633,0.8168138037986672),SE*labelscalefactor); /* dots and labels */ dot((3.8295013012181283,2.816337276198864),dotstyle); label("$A$", (3.8732067323088435,2.925600853925651), NE * labelscalefactor); dot((-0.7368327629589799,-0.5920813291311117),dotstyle); label("$B$", (-1.1,-0.7565817154670613), NE * labelscalefactor); dot((5.672613975760373,-0.636902634996282),dotstyle); label("$C$", (5.763466626982254,-0.7784344310124186), NE * labelscalefactor); dot((4.569287648059735,1.430279997142299),dotstyle); label("$D$", (4.692683565259744,1.5051743434774234), NE * labelscalefactor); dot((1.8844000180622977,1.3644681598392678),dotstyle); label("$E$", (1.775346039954538,1.4942479857047448), NE * labelscalefactor); dot((2.937230516274804,0.8082418657164665),linewidth(4.pt) + dotstyle); label("$F$", (2.889834532767763,0.954), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

The ratio $ \dfrac{10^{2000}\plus{}10^{2002}}{10^{2001}\plus{}10^{2001}}$ is closest to which of the following numbers? $ \text{(A)}\ 0.1\qquad \text{(B)}\ 0.2\qquad \text{(C)}\ 1\qquad \text{(D)}\ 5\qquad \text{(E)}\ 10$

Points $A,B,C,D$ are on a circle with radius $R$. $|AC|=|AB|=500$, while the ratio between $|DC|, |DA|, |DB|$ is $1,5,7$. Find $R$.

Each point of the plane is coloured in one of two colours. Given an odd integer number $n \geq 3$, prove that there exist (at least) two similar triangles whose similitude ratio is $n$, each of which has a monochromatic vertex-set. [i]Vasile Pop[/i]