In a triangle $ABC$, the median $AD$ divides $\angle{BAC}$ in the ratio $1:2$. Extend $AD$ to $E$ such that $EB$ is perpendicular $AB$. Given that $BE=3,BA=4$, find the integer nearest to $BC^2$.

Let $M$ be any point on the circumcircle of triangle $ABC$. Suppose the tangents from $M$ to the incircle meet $BC$ at two points $X_1$ and $X_2$. Prove that the circumcircle of triangle $MX_1X_2$ intersects the circumcircle of $ABC$ again at the tangency point of the $A$-mixtilinear incircle.

The points $C_1, A_1, B_1$ belong to $[AB], [BC], [CA]$ sides, respectively, of the triangle $ABC$ . $$\frac{|AC_1|}{|C_1B| }=\frac{ |BA_1|}{|A_1C| }= \frac{|CB_1|}{|B_1A| }= \frac{1}{3}$$ Prove that the perimeter $P$ of the triangle $ABC$ and the perimeter $p$ of the triangle $A_1B_1C_1$ , satisfy inequality $$\frac{P}{2} < p < \frac{3P}{4}$$

If an arc of $ 45^\circ$ on circle $ A$ has the same length as an arc of $ 30^\circ$ on circle $ B$, then the ratio of the area of circle $ A$ to the area of circle $ B$ is $ \textbf{(A)}\ \frac {4}{9} \qquad \textbf{(B)}\ \frac {2}{3} \qquad \textbf{(C)}\ \frac {5}{6} \qquad \textbf{(D)}\ \frac {3}{2} \qquad \textbf{(E)}\ \frac {9}{4}$

A mink is standing in the center of a field shaped like a regular polygon. The field is surrounded by a fence, and the mink can only exit through the vertices of the polygon. A dog is standing on one of the vertices, and can move along the fence. The mink wants to escape the field, while the dog tries to prevent it. Each of them moves with constant velocity. For what ratio of velocities could the mink escape if: a. The field is a regular triangle? b. The field is a square?

A triangle $ABC$ is given in the plane with $AB = \sqrt{7},$ $BC = \sqrt{13}$ and $CA = \sqrt{19},$ circles are drawn with centers at $A,B$ and $C$ and radii $\frac{1}{3},$ $\frac{2}{3}$ and $1,$ respectively. Prove that there are points $A',B',C'$ on these three circles respectively such that triangle $ABC$ is congruent to triangle $A'B'C'.$

On triangle $ABC$ let $D$ be the point on $AB$ so that $CD$ is an altitude of the triangle, and $E$ be the point on $BC$ so that $AE$ bisects angle $BAC.$ Let $G$ be the intersection of $AE$ and $CD,$ and let point $F$ be the intersection of side $AC$ and the ray $BG.$ If $AB$ has length $28,$ $AC$ has length $14,$ and $CD$ has length $10,$ then the length of $CF$ can be written as $\tfrac{k-m\sqrt{p}}{n}$ where $k, m, n,$ and $p$ are positive integers, $k$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $k - m + n + p.$

In the diagram below, three squares are inscribed in right triangles. Their areas are $A$, $M$, and $N$, as indicated in the diagram. If $M = 5$ and $N = 12$, then $A$ can be expressed as $a + b\sqrt{c}$, where $a$, $b$, and $c$ are positive integers and $c$ is not divisible by the square of any prime. Compute $a + b + c$. [asy] size(250); defaultpen (linewidth (0.7) + fontsize (10)); pair O = origin, A = (1, 1), B = (4/3, 1/3), C = (2/3, 5/3), P = (3/2, 0), Q = (0,3); draw (P--O--Q--cycle^^(0, 5/3)--C--(2/3,1)^^(0,1)--A--(1,0)^^(1,1/3)--B--(4/3,0)); label("$A$", (.5,.5)); label("$M$", (7/6, 1/6)); label("$N$", (1/3, 4/3));[/asy] [i]Proposed by Aaron Lin[/i]

A blackboard contains 68 pairs of nonzero integers. Suppose that for each positive integer $k$ at most one of the pairs $(k, k)$ and $(-k, -k)$ is written on the blackboard. A student erases some of the 136 integers, subject to the condition that no two erased integers may add to 0. The student then scores one point for each of the 68 pairs in which at least one integer is erased. Determine, with proof, the largest number $N$ of points that the student can guarantee to score regardless of which 68 pairs have been written on the board.

In the adjoining diagram, $BO$ bisects $\angle CBA$, $CO$ bisects $\angle ACB$, and $MN$ is parallel to $BC$. If $AB=12$, $BC=24$, and $AC=18$, then the perimeter of $\triangle AMN$ is [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(24,0), A=intersectionpoints(Circle(B,12), Circle(C,18))[0], O=incenter(A,B,C), M=intersectionpoint(A--B, O--O+40*dir(180)), N=intersectionpoint(A--C, O--O+40*dir(0)); draw(B--M--O--B--C--O--N--C^^N--A--M); label("$A$", A, dir(90)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$M$", M, dir(90)*dir(B--A)); label("$N$", N, dir(90)*dir(A--C)); label("$O$", O, dir(90));[/asy] $\textbf {(A) } 30 \qquad \textbf {(B) } 33 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 39 \qquad \textbf {(E) } 42$

$A$ and $B$ move uniformly along two straight paths intersecting at right angles in point $O$. When $A$ is at $O$, $B$ is $500$ yards short of $O$. In $2$ minutes, they are equidistant from $O$, and in $8$ minutes more they are again equidistant from $O$. Then the ratio of $A'$s speed to $B'$s speed is: $\textbf{(A)}\ 4:5 \qquad\textbf{(B)}\ 5:6 \qquad\textbf{(C)}\ 2:3 \qquad\textbf{(D)}\ 5:8 \qquad\textbf{(E)}\ 1:2$

Susie pays for $4$ muffins and $3$ bananas. Calvin spends twice as much paying for $2$ muffins and $16$ bananas. A muffin is how many times as expensive as a banana? $ \textbf {(A) } \frac{3}{2} \qquad \textbf {(B) } \frac{5}{3} \qquad \textbf {(C) } \frac{7}{4} \qquad \textbf {(D) } 2 \qquad \textbf {(E) } \frac{13}{4}$

Given the line $y = \dfrac{3}{4}x + 6$ and a line $L$ parallel to the given line and $4$ units from it. A possible equation for $L$ is: $\textbf{(A)}\ y = \dfrac{3}{4}x + 1 \qquad \textbf{(B)}\ y = \dfrac{3}{4}x\qquad \textbf{(C)}\ y = \dfrac{3}{4}x -\dfrac{2}{3} \qquad$ $ \textbf{(D)}\ y = \dfrac{3}{4}x -1 \qquad \textbf{(E)}\ y = \dfrac{3}{4}x + 2$

The sum to infinity of the terms of an infinite geometric progression is $ 6$. The sum of the first two terms is $ 4\frac {1}{2}$. The first term of the progression is: $ \textbf{(A)}\ 3 \text{ or } 1\frac {1}{2} \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2\frac {1}{2} \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 9 \text{ or } 3$

From the vertex $C$ in triangle $ABC$, draw a straight line that bisects the median from $A$. In what ratio does this line divide the segment $AB$? [img]https://1.bp.blogspot.com/-SxWIQ12DIvs/XzcJv5xoV0I/AAAAAAAAMY4/Ezfe8bd7W-Mfp2Qi4qE_gppbh9Fzvb4XwCLcBGAsYHQ/s0/1995%2BMohr%2Bp3.png[/img]