A triangle $KLM$ is given in the plane together with a point $A$ lying on the half-line opposite to $KL$. Construct a rectangle $ABCD$ whose vertices $B, C$ and $D$ lie on the lines $KM, KL$ and $LM$, respectively. (We allow the rectangle to be a square.)

[asy] draw(unitsquare);draw((0,0)--(.4,1)^^(0,.6)--(1,.2)); label("D",(0,1),NW);label("E",(.4,1),N);label("C",(1,1),NE); label("P",(0,.6),W);label("M",(.25,.55),E);label("Q",(1,.2),E); label("A",(0,0),SW);label("B",(1,0),SE); //Credit to Zimbalono for the diagram[/asy] Inside square $ABCD$ (See figure) with sides of length $12$ inches, segment $AE$ is drawn where $E$ is the point on $DC$ which is $5$ inches from $D$. The perpendicular bisector of $AE$ is drawn and intersects $AE$, $AD$, and $BC$ at points $M$, $P$, and $Q$ respectively. The ratio of segment $PM$ to $MQ$ is $\textbf{(A) }5:12\qquad\textbf{(B) }5:13\qquad\textbf{(C) }5:19\qquad\textbf{(D) }1:4\qquad \textbf{(E) }5:21$

The points $ (2,\minus{}3)$, $ (4,3)$, and $ (5, k/2)$ are on the same straight line. The value(s) of $ k$ is (are): $ \textbf{(A)}\ 12\qquad \textbf{(B)}\ \minus{}12\qquad \textbf{(C)}\ \pm 12\qquad \textbf{(D)}\ {12}\text{ or }{6}\qquad \textbf{(E)}\ {6}\text{ or }{6\frac{2}{3}}$

Prove that there exists an infinite set of points \[ \dots, \; P_{-3}, \; P_{-2},\; P_{-1},\; P_0,\; P_1,\; P_2,\; P_3,\; \dots \] in the plane with the following property: For any three distinct integers $a,b,$ and $c$, points $P_a$, $P_b$, and $P_c$ are collinear if and only if $a+b+c=2014$.

Let $P$ be a point in a square whose side are mirror. A ray of light comes from $P$ and with slope $\alpha$. We know that this ray of light never arrives to a vertex. We make an infinite sequence of $0,1$. After each contact of light ray with a horizontal side, we put $0$, and after each contact with a vertical side, we put $1$. For each $n\geq 1$, let $B_{n}$ be set of all blocks of length $n$, in this sequence. a) Prove that $B_{n}$ does not depend on location of $P$. b) Prove that if $\frac{\alpha}{\pi}$ is irrational, then $|B_{n}|=n+1$.

Let $n$ be a given positive integer. Solve the system \[x_1 + x_2^2 + x_3^3 + \cdots + x_n^n = n,\] \[x_1 + 2x_2 + 3x_3 + \cdots + nx_n = \frac{n(n+1)}{2}\] in the set of nonnegative real numbers.

Find polynomials $f(x)$, $g(x)$, and $h(x)$, if they exist, such that for all $x$, \[|f(x)|-|g(x)|+h(x)=\begin{cases}-1 & \text{if }x<-1\\3x+2 &\text{if }-1\leq x\leq 0\\-2x+2 & \text{if }x>0.\end{cases}\]

Triangle $ABC$ lies entirely in the first quadrant of the Cartesian plane, and its sides have slopes $63$, $73$, $97$. Suppose the curve $\mathcal V$ with equation $y=(x+3)(x^2+3)$ passes through the vertices of $ABC$. Find the sum of the slopes of the three tangents to $\mathcal V$ at each of $A$, $B$, $C$. [i]Proposed by Akshaj[/i]

Find the best approximation of $\sqrt{3}$ by a rational number with denominator less than or equal to $15$

If the normal to the curve $x^{\frac{2}{3}}+y^{\frac23}=a^{\frac23}$ at some point makes an angle $\theta$ with the $X$-axis, show that the equation of the normal is \[y\cos\theta-x\sin\theta=a\cos 2\theta\]

Consider the parallelogram with vertices $(10,45),$ $(10,114),$ $(28,153),$ and $(28,84).$ A line through the origin cuts this figure into two congruent polygons. The slope of the line is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Let $ ABCDE$ be a convex pentagon. If $ P$, $ Q$, $ R$ and $ S$ are the respective centroids of the triangles $ ABE$, $ BCE$, $ CDE$ and $ DAE$, show that $ PQRS$ is a parallelogram and its area is $ 2/9$ of that of $ ABCD$.

All three vertices of an equilateral triangle are on the parabola $ y \equal{} x^2$, and one of its sides has a slope of 2. The x-coordinates of the three vertices have a sum of $ m/n$, where $ m$ and $ n$ are relatively prime positive integers. What is the value of $ m \plus{} n$? $ \textbf{(A)}\ 14\qquad \textbf{(B)}\ 15\qquad \textbf{(C)}\ 16\qquad \textbf{(D)}\ 17\qquad \textbf{(E)}\ 18$

A point $ (x,y)$ in the plane is called a lattice point if both $ x$ and $ y$ are integers. The area of the largest square that contains exactly three lattice points in its interior is closest to $ \textbf{(A)}\ 4.0\qquad \textbf{(B)}\ 4.2\qquad \textbf{(C)}\ 4.5\qquad \textbf{(D)}\ 5.0\qquad \textbf{(E)}\ 5.6$

The coordinate of $ P$ at time $ t$, moving on a plane, is expressed by $ x = f(t) = \cos 2t + t\sin 2t,\ y = g(t) = \sin 2t - t\cos 2t$. (1) Find the acceleration vector $ \overrightarrow{\alpha}$ of $ P$ at time $ t$ . (2) Let $ L$ denote the line passing through the point $ P$ for the time $ t%Error. "neqo" is a bad command. $, which is parallel to the acceleration vector $ \overrightarrow{\alpha}$ at the time. Prove that $ L$ always touches to the unit circle with center the origin, then find the point of tangency $ Q$. (3) Prove that $ f(t)$ decreases in the interval $ 0\leq t \leqq \frac {\pi}{2}$. (4) When $ t$ varies in the range $ \frac {\pi}{4}\leq t\leq \frac {\pi}{2}$, find the area $ S$ of the figure formed by moving the line segment $ PQ$.

A block of mass $m$ on a frictionless inclined plane of angle $\theta$ is connected by a cord over a small frictionless, massless pulley to a second block of mass $M$ hanging vertically, as shown. If $M=1.5m$, and the acceleration of the system is $\frac{g}{3}$, where $g$ is the acceleration of gravity, what is $\theta$, in degrees, rounded to the nearest integer? [asy]size(12cm); pen p=linewidth(1), dark_grey=gray(0.25), ll_grey=gray(0.90), light_grey=gray(0.75); pair B = (-1,-1); pair C = (-1,-7); pair A = (-13,-7); path inclined_plane = A--B--C--cycle; draw(inclined_plane, p); real r = 1; // for marking angles draw(arc(A, r, 0, degrees(B-A))); // mark angle label("$\theta$", A + r/1.337*(dir(C-A)+dir(B-A)), (0,0), fontsize(16pt)); // label angle as theta draw((C+(-r/2,0))--(C+(-r/2,r/2))--(C+(0,r/2))); // draw right angle real h = 1.2; // height of box real w = 1.9; // width of box path box = (0,0)--(0,h)--(w,h)--(w,0)--cycle; // the box // box on slope with label picture box_on_slope; filldraw(box_on_slope, box, light_grey, black); label(box_on_slope, "$m$", (w/2,h/2)); pair V = A + rotate(90) * (h/2 * dir(B-A)); // point with distance l/2 from AB pair T1 = dir(125); // point of tangency with pulley pair X1 = intersectionpoint(T1--(T1 - rotate(-90)*(2013*dir(T1))), V--(V+B-A)); // construct midpoint of right side of box draw(T1--X1); // string add(shift(X1-(w,h/2))*rotate(degrees(B-A), (w,h/2)) * box_on_slope); // picture for the hanging box picture hanging_box; filldraw(hanging_box, box, light_grey, black); label(hanging_box, "$M$", (w/2,h/2)); pair T2 = (1,0); pair X2 = (1,-3); draw(T2--X2); // string add(shift(X2-(w/2,h)) * hanging_box); // Draws the actual pulley filldraw(unitcircle, grey, p); // outer boundary of pulley wheel filldraw(scale(0.4)*unitcircle, light_grey, p); // inner boundary of pulley wheel path pulley_body=arc((0,0),0.3,-40,130)--arc((-1,-1),0.5,130,320)--cycle; // defines "arm" of pulley filldraw(pulley_body, ll_grey, dark_grey+p); // draws the arm filldraw(scale(0.18)*unitcircle, ll_grey, dark_grey+p); // inner circle of pulley[/asy][i](Proposed by Ahaan Rungta)[/i]

Let $ S$ be the set of points $ (a,b)$ in the coordinate plane, where each of $ a$ and $ b$ may be $ \minus{} 1$, $ 0$, or $ 1$. How many distinct lines pass through at least two members of $ S$? $ \textbf{(A)}\ 8 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$

A frog begins at $P_0 = (0,0)$ and makes a sequence of jumps according to the following rule: from $P_n=(x_n,y_n)$, the frog jumps to $P_{n+1}$, which may be any of the points $(x_n+7, y_n+2)$, $(x_n+2,y_n+7)$, $(x_n-5, y_n-10)$, or $(x_n-10,y_n-5)$. There are $M$ points $(x,y)$ with $|x|+|y| \le 100$ that can be reached by a sequence of such jumps. Find the remainder when $M$ is divided by $1000$.

A point $(x,y)$ in the first quadrant lies on a line with intercepts $(a,0)$ and $(0,b)$, with $a,b > 0$. Rectangle $M$ has vertices $(0,0)$, $(x,0)$, $(x,y)$, and $(0,y)$, while rectangle $N$ has vertices $(x,y)$, $(x,b)$, $(a,b)$, and $(a,y)$. What is the ratio of the area of $M$ to that of $N$? [i]Proposed by Eugene Chen[/i]

Let $ n$ be a positive integer. Consider a rectangle $ (90n\plus{}1)\times(90n\plus{}5)$ consisting of unit squares. Let $ S$ be the set of the vertices of these squares. Prove that the number of distinct lines passing through at least two points of $ S$ is divisible by $ 4$.

Prove the following inequality. \[ \frac{2}{2\plus{}e^{\frac 12}}<\int_0^1 \frac{dx}{1\plus{}xe^{x}}<\frac{2\plus{}e}{2(1\plus{}e)}\]

In triangle $ABC$, $AB=\frac{20}{11} AC$. The angle bisector of $\angle A$ intersects $BC$ at point $D$, and point $M$ is the midpoint of $AD$. Let $P$ be the point of the intersection of $AC$ and $BM$. The ratio of $CP$ to $PA$ can be expressed in the form $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

The solution of the equations \begin{align*} 2x-3y&=7 \\ 4x-6y &=20 \\ \end{align*} is: $ \textbf{(A)}\ x=18, y=12 \qquad \textbf{(B)}\ x=0, y=0 \qquad \textbf{(C)}\ \text{There is no solution} \\ \textbf{(D)}\ \text{There are an unlimited number of solutions} \qquad \textbf{(E)}\ x=8, y=5$

A triangle has vertices $P=(-8,5)$, $Q=(-15,-19)$, and $R=(1,-7)$. The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0$. Find $a+c$.

Cities $A$, $B$, $C$, $D$, and $E$ are connected by roads $\widetilde{AB}$, $\widetilde{AD}$, $\widetilde{AE}$, $\widetilde{BC}$, $\widetilde{BD}$, $\widetilde{CD}$, $\widetilde{DE}$. How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit cities more than once.) [asy]unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08); dot (A); dot (B); dot (C); dot (D); dot (E); label("$A$",A,S); label("$B$",B,SE); label("$C$",C,E); label("$D$",D,N); label("$E$",E,W); guide squiggly(path g, real stepsize, real slope=45) { real len = arclength(g); real step = len / round(len / stepsize); guide squig; for (real u = 0; u < len; u += step){ real a = arctime(g, u); real b = arctime(g, u + step / 2); pair p = point(g, a); pair q = point(g, b); pair np = unit( rotate(slope) * dir(g,a)); pair nq = unit( rotate(0 - slope) * dir(g,b)); squig = squig .. p{np} .. q{nq}; } squig = squig .. point(g, length(g)){unit(rotate(slope)*dir(g,length(g)))}; return squig; } pen pp = defaultpen + 2.718; draw(squiggly(A--B, 4.04, 30), pp); draw(squiggly(A--D, 7.777, 20), pp); draw(squiggly(A--E, 5.050, 15), pp); draw(squiggly(B--C, 5.050, 15), pp); draw(squiggly(B--D, 4.04, 20), pp); draw(squiggly(C--D, 2.718, 20), pp); draw(squiggly(D--E, 2.718, -60), pp); [/asy] $ \textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 $