If the point $(x,-4)$ lies on the straight line joining the points $(0,8)$ and $(-4,0)$ in the xy-plane, then $x$ is equal to $\textbf{(A) }-2\qquad\textbf{(B) }2\qquad\textbf{(C) }-8\qquad\textbf{(D) }6\qquad \textbf{(E) }-6$

Draw a tangent with positive slope to a parabola $y=x^2+1$. Find the $x$-coordinate such that the area of the figure bounded by the parabola, the tangent and the coordinate axisis is $\frac{11}{3}.$

Two rays emanate from the origin $O$ and form a $45^\circ$ angle in the first quadrant of the Cartesian coordinate plane. For some positive numbers $X$, $Y$, and $S$, the ray with the larger slope passes through point $A = (X, S)$, and the ray with the smaller slope passes through point $B = (S, Y)$. If $6X + 6Y + 5S = 600$, then determine the maximum possible area of $\triangle OAB$.

The graph of $ 2x^2 \plus{} xy \plus{} 3y^2 \minus{} 11x \minus{} 20y \plus{} 40 \equal{} 0$ is an ellipse in the first quadrant of the $ xy$-plane. Let $ a$ and $ b$ be the maximum and minimum values of $ \frac {y}{x}$ over all points $ (x, y)$ on the ellipse. What is the value of $ a \plus{} b$? $ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ \frac72 \qquad \textbf{(D)}\ \frac92 \qquad \textbf{(E)}\ 2\sqrt {14}$

Let $ ABC$ be a triangle with $ AB \equal{} 5$, $ BC \equal{} 4$ and $ AC \equal{} 3$. Let $ \mathcal P$ and $ \mathcal Q$ be squares inside $ ABC$ with disjoint interiors such that they both have one side lying on $ AB$. Also, the two squares each have an edge lying on a common line perpendicular to $ AB$, and $ \mathcal P$ has one vertex on $ AC$ and $ \mathcal Q$ has one vertex on $ BC$. Determine the minimum value of the sum of the areas of the two squares. [asy]import olympiad; import math; import graph; unitsize(1.5cm); pair A, B, C; A = origin; B = A + 5 * right; C = (9/5, 12/5); pair X = .7 * A + .3 * B; pair Xa = X + dir(135); pair Xb = X + dir(45); pair Ya = extension(X, Xa, A, C); pair Yb = extension(X, Xb, B, C); pair Oa = (X + Ya)/2; pair Ob = (X + Yb)/2; pair Ya1 = (X.x, Ya.y); pair Ya2 = (Ya.x, X.y); pair Yb1 = (Yb.x, X.y); pair Yb2 = (X.x, Yb.y); draw(A--B--C--cycle); draw(Ya--Ya1--X--Ya2--cycle); draw(Yb--Yb1--X--Yb2--cycle); label("$A$", A, W); label("$B$", B, E); label("$C$", C, N); label("$\mathcal P$", Oa, origin); label("$\mathcal Q$", Ob, origin);[/asy]

For every pair of reals $0 < a < b < 1$, we define sequences $\{x_n\}_{n \ge 0}$ and $\{y_n\}_{n \ge 0}$ by $x_0 = 0$, $y_0 = 1$, and for each integer $n \ge 1$: \begin{align*} x_n & = (1 - a) x_{n - 1} + a y_{n - 1}, \\ y_n & = (1 - b) x_{n - 1} + b y_{n - 1}. \end{align*} The [i]supermean[/i] of $a$ and $b$ is the limit of $\{x_n\}$ as $n$ approaches infinity. Over all pairs of real numbers $(p, q)$ satisfying $\left (p - \textstyle\frac{1}{2} \right)^2 + \left (q - \textstyle\frac{1}{2} \right)^2 \le \left(\textstyle\frac{1}{10}\right)^2$, the minimum possible value of the supermean of $p$ and $q$ can be expressed as $\textstyle\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Compute $100m + n$. [i]Proposed by Lewis Chen[/i]

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 $ABC$ be an acute-angled triangle with $AB<AC$ and let $O$ be the centre of its circumcircle $\omega$. Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M$, $N$ and $P$ are the midpoints of the line segments $BE$, $OD$ and $AC$, respectively, show that the points $M$, $N$ and $P$ are collinear.

Consider the $ 12$-sided polygon $ ABCDEFGHIJKL$, as shown. Each of its sides has length $ 4$, and each two consecutive sides form a right angle. Suppose that $ \overline{AG}$ and $ \overline{CH}$ meet at $ M$. What is the area of quadrilateral $ ABCM$? [asy]unitsize(13mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); dot(M); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,NE); label("$D$",D,NE); label("$E$",Ep,SE); label("$F$",F,SE); label("$G$",G,SE); label("$H$",H,SW); label("$I$",I,SW); label("$J$",J,SW); label("$K$",K,NW); label("$L$",L,NW); label("$M$",M,W);[/asy]$ \textbf{(A)}\ \frac {44}{3}\qquad \textbf{(B)}\ 16 \qquad \textbf{(C)}\ \frac {88}{5}\qquad \textbf{(D)}\ 20 \qquad \textbf{(E)}\ \frac {62}{3}$

A sled loaded with children starts from rest and slides down a snowy $25^\circ$ (with respect to the horizontal) incline traveling $85$ meters in $17$ seconds. Ignore air resistance. What is the coefficient of kinetic friction between the sled and the slope? $ \textbf {(A) } 0.36 \qquad \textbf {(B) } 0.40 \qquad \textbf {(C) } 0.43 \qquad \textbf {(D) } 1.00 \qquad \textbf {(E) } 2.01 $

Square $\mathcal S$ has vertices $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$. Points $P$ and $Q$ are independently selected, uniformly at random, from the perimeter of $\mathcal S$. Determine, with proof, the probability that the slope of line $PQ$ is positive. [i]Proposed by Isabella Grabski[/i]

Find the number of finite sequences $ \{a_1,a_2,\ldots,a_{2n\plus{}1}\}$, formed with nonnegative integers, for which $ a_1\equal{}a_{2n\plus{}1}\equal{}0$ and $ |a_k \minus{}a_{k\plus{}1}|\equal{}1$, for all $ k\in\{1,2,\ldots,2n\}$.

Draw a tangent with positive slope to a parabola $y=x^2+1$. Find the $x$-coordinate such that the area of the figure bounded by the parabola, the tangent and the coordinate axisis is $\frac{11}{3}.$