In trapezoid $ ABCD$ with bases $ AB$ and $ CD$, we have $ AB\equal{}52$, $ BC\equal{}12$, $ CD\equal{}39$, and $ DA\equal{}5$. The area of $ ABCD$ is [asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W);[/asy] $ \text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$

A balance has a left pan, a right pan, and a pointer that moves along a graduated ruler. Like many other grocer balances, this one works as follows: An object of weight $ L$ is placed in the left pan and another of weight $ R$ in the right pan, the pointer stops at the number $ R \minus{} L$ on the graduated ruler. There are $ n, (n \geq 2)$ bags of coins, each containing $ \frac{n(n\minus{}1)}{2} \plus{} 1$ coins. All coins look the same (shape, color, and so on). $ n\minus{}1$ bags contain real coins, all with the same weight. The other bag (we don’t know which one it is) contains false coins. All false coins have the same weight, and this weight is different from the weight of the real coins. A legal weighing consists of placing a certain number of coins in one of the pans, putting a certain number of coins in the other pan, and reading the number given by the pointer in the graduated ruler. With just two legal weighings it is possible to identify the bag containing false coins. Find a way to do this and explain it.

Let $ABC$ be a triangle with sides of length $a$, $b$ and $c$. Let the bisector of the angle $C$ cut $AB$ in $D$. Prove that the length of $CD$ is \[ \frac{2ab\cos \frac{C}{2}}{a+b}. \]

The sizes of the freshmen class and the sophomore class are in the ratio $5:4$. The sizes of the sophomore class and the junior class are in the ratio $7:8$. The sizes of the junior class and the senior class are in the ratio $9:7$. If these four classes together have a total of $2158$ students, how many of the students are freshmen?

On sides $BC$, $CA$, $AB$ of triangle $ABC$, points $A_1$, $B_1$, $C_1$ are chosen, respectively, so that the medians $A_1A_2$, $B_1B_2$, $C_1C_2$ of the triangle $A_1B_1C_1$ are respectively parallel to straight lines $AB$, $BC$, $CA$. Determine in what ratio points $A_1$, $B_1$, $C_1$ divide the sides of the triangle $ABC$.

Suppose that a triangle $ABC$ with incenter $I$ satisfies $CA+AI=BC$. Find the ratio between the measures of the angles $\angle BAC$ and $\angle CBA$.

A square and equilateral triangle have the same perimeter. If the triangle has area $16\sqrt3$, what is the area of the square? [i]Proposed by Evan Chen[/i]

In the plane are given a circle with center $ O$ and radius $ r$ and a point $ A$ outside the circle. For any point $ M$ on the circle, let $ N$ be the diametrically opposite point. Find the locus of the circumcenter of triangle $ AMN$ when $ M$ describes the circle.

Let $\gamma,\Gamma$ be two concentric circles with radii $r,R$ with $r<R$. Let $ABCD$ be a cyclic quadrilateral inscribed in $\gamma$. If $\overrightarrow{AB}$ denotes the Ray starting from $A$ and extending indefinitely in $B's$ direction then Let $\overrightarrow{AB}, \overrightarrow{BC}, \overrightarrow{CD} , \overrightarrow{DA}$ meet $\Gamma$ at the points $C_1,D_1,A_1,B_1$ respectively. Prove that \[\frac{[A_1B_1C_1D_1]}{[ABCD]} \ge \frac{R^2}{r^2}\] where $[.]$ denotes area.

In the figure below, $N$ congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let $A$ be the combined area of the small semicircles and $B$ be the area of the region inside the large semicircle but outside the semicircles. The ratio $A:B$ is $1:18$. What is $N$? [asy] draw((0,0)--(18,0)); draw(arc((9,0),9,0,180)); filldraw(arc((1,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((3,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((5,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((7,0),1,0,180)--cycle,gray(0.8)); label("...",(9,0.5)); filldraw(arc((11,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((13,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((15,0),1,0,180)--cycle,gray(0.8)); filldraw(arc((17,0),1,0,180)--cycle,gray(0.8)); [/asy] $\textbf{(A) } 16 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 36$

You are given a pair of triangles for which two sides of one triangle are equal in length to two sides of the second triangle, and the triangles are similar, but not necessarily congruent. Prove that the ratio of the sides that correspond under the similarity is a number between $ \frac {1}{2}(\sqrt {5} \minus{} 1)$ and $ \frac {1}{2}(\sqrt {5} \plus{} 1)$.

Circle $\Gamma$ has radius $10$, center $O$, and diameter $AB$. Point $C$ lies on $\Gamma$ such that $AC = 12$. Let $P$ be the circumcenter of $\vartriangle AOC$. Line $AP$ intersects $\Gamma$ at $Q$, where $Q$ is different from $A$. Then the value of $\frac{AP}{AQ}$ can be expressed in the form $\frac{m}{n}$, where m and n are relatively prime positive integers. Compute $m + n$.

Points $L$ and $H$ are marked on the sides $AB$ of an acute-angled triangle ABC so that $CL$ is a bisector and $CH$ is an altitude. Let $P,Q$ be the feet of the perpendiculars from $L$ to $AC$ and $BC$ respectively. Prove that $AP \cdot BH = BQ \cdot AH$. I. Gorodnin

An isosceles triangle $ABC$ is inscribed in a circle with $\angle ACB = 90^o$ and $EF$ is a chord of the circle such that neither E nor $F$ coincide with $C$. Lines $CE$ and $CF$ meet $AB$ at $D$ and $G$ respectively. Prove that $|CE|\cdot |DG| = |EF| \cdot |CG|$.

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.

Let $ABC$ be a triangle. Points $D$, $E$, and $F$ are respectively on the sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ of $\triangle ABC$. Suppose that \[ \frac{AE}{AC} = \frac{CD}{CB} = \frac{BF}{BA} = x \] for some $x$ with $\frac{1}{2} < x < 1$. Segments $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ cut the triangle into 7 nonoverlapping regions: 4 triangles and 3 quadrilaterals. The total area of the 4 triangles equals the total area of the 3 quadrilaterals. Compute the value of $x$.

Suppose that $ ABCD$ is a convex quadrilateral. Let $ F \equal{} AB\cap CD$, $ E \equal{} AD\cap BC$ and $ T \equal{} AC\cap BD$. Suppose that $ A,B,T,E$ lie on a circle which intersects with $ EF$ at $ P$. Prove that if $ M$ is midpoint of $ AB$, then $ \angle APM \equal{} \angle BPT$.

The ratio of $3$ to the positive number $n$ is the same as the ratio of $n$ to $192.$ Find $n.$

Let be given a triangle $ ABC$ and its internal angle bisector $ BD$ $ (D\in BC)$. The line $ BD$ intersects the circumcircle $ \Omega$ of triangle $ ABC$ at $ B$ and $ E$. Circle $ \omega$ with diameter $ DE$ cuts $ \Omega$ again at $ F$. Prove that $ BF$ is the symmedian line of triangle $ ABC$.

In the rectangle there is a broken line, the neighboring links of which are perpendicular and equal to the smaller side of the rectangle (see the figure). Find the ratio of the sides of the rectangle. [img]https://2.bp.blogspot.com/-QYj53KiPTJ8/XT_mVIw876I/AAAAAAAAKbE/gJ1roU4Bx-kfGVfJxYMAuLE0Ax0glRbegCK4BGAYYCw/s1600/oral%2Bmoscow%2B2016%2B8.9%2Bp2.png[/img]

$M$ is any point on the side $AB$ of the triangle $ABC$. $r,r_1,r_2$ are the radii of the circles inscribed in $ABC,AMC,BMC$. $q$ is the radius of the circle on the opposite side of $AB$ to $C$, touching the three sides of $AB$ and the extensions of $CA$ and $CB$. Similarly, $q_1$ and $q_2$. Prove that $r_1r_2q=rq_1q_2$.

In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below. \[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} \text{Row 0: } & & & & & & & 1 & & & & & & \\\vspace{4pt} \text{Row 1: } & & & & & & 1 & & 1 & & & & & \\\vspace{4pt} \text{Row 2: } & & & & & 1 & & 2 & & 1 & & & & \\\vspace{4pt} \text{Row 3: } & & & & 1 & & 3 & & 3 & & 1 & & & \\\vspace{4pt} \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt} \text{Row 5: } & & 1 & & 5 & &10& &10 & & 5 & & 1 & \\\vspace{4pt} \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 \end{array}\] In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3: 4: 5$?

Acute triangle $ABP$, where $AB > BP$, has altitudes $BH$, $PQ$, and $AS$. Let $C$ denote the intersection of lines $QS$ and $AP$, and let $L$ denote the intersection of lines $HS$ and $BC$. If $HS = SL$ and $HL$ is perpendicular to $BC$, find the value of $\frac{SL}{SC}$.

$\overline{AB}$ is a diameter of circle $C_1$. Point $P$ is on $C_1$ such that $AP>BP$. Circle $C_2$ is centered at $P$ with radius $PB$. The extension of $\overline{AP}$ past $P$ meets $C_2$ at $Q$. Circle $C_3$ is centered at $A$ and is externally tangent to $C_2$. Circle $C_4$ passes through $A$, $Q$, and $R$. Find, with proof, the ratio between the area of $C_4$ and the area of $C_1$, and show that this ratio is the same for all points $P$ on $C_1$ such that $AP>BP$.

In triangle $ ABC$ the medians $ \overline{AD}$ and $ \overline{CE}$ have lengths 18 and 27, respectively, and $ AB \equal{} 24$. Extend $ \overline{CE}$ to intersect the circumcircle of $ ABC$ at $ F$. The area of triangle $ AFB$ is $ 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$.