Let $\triangle ABC$ be an acute triangle, and let $D$ be the projection of $A$ on $BC$. Let $M,N$ be the midpoints of $AB$ and $AC$ respectively. Let $\Gamma_1$ and $\Gamma_2$ be the circumcircles of $\triangle BDM$ and $\triangle CDN$ respectively, and let $K$ be the other intersection point of $\Gamma_1$ and $\Gamma_2$. Let $P$ be an arbitrary point on $BC$ and $E,F$ are on $AC$ and $AB$ respectively such that $PEAF$ is a parallelogram. Prove that if $MN$ is a common tangent line of $\Gamma_1$ and $\Gamma_2$, then $K,E,A,F$ are concyclic.

In the adjoining figure $AB$ and $BC$ are adjacent sides of square $ABCD$; $M$ is the midpoint of $AB$; $N$ is the midpoint of $BC$; and $AN$ and $CM$ intersect at $O$. The ratio of the area of $AOCD$ to the area of $ABCD$ is [asy] draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--(2,1)--(2,2)--(1,0)); label("A", (0,0), S); label("B", (2,0), S); label("C", (2,2), N); label("D", (0,2), N); label("M", (1,0), S); label("N", (2,1), E); label("O", (1.2, .8)); [/asy] $ \textbf{(A)}\ \frac{5}{6} \qquad\textbf{(B)}\ \frac{3}{4} \qquad\textbf{(C)}\ \frac{2}{3} \qquad\textbf{(D)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(E)}\ \frac{(\sqrt{3}-1)}{2} $

Given a rhombus $ABCD$. A point $M$ is chosen on its side $BC$. The lines, which pass through $M$ and are perpendicular to $BD$ and $AC$, meet line $AD$ in points $P$ and $Q$ respectively. Suppose that the lines $PB,QC,AM$ have a common point. Find all possible values of a ratio $\frac{BM}{MC}$. [i]S. Berlov, F. Petrov, A. Akopyan[/i]

The ratio of the measures of two acute angles is $5:4$, and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles? $\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$

In triangle $ABC,$ point $D$ is on $\overline{BC}$ with $CD=2$ and $DB=5,$ point $E$ is on $\overline{AC}$ with $CE=1$ and $EA=3,$ $AB=8,$ and $\overline{AD}$ and $\overline{BE}$ intersect at $P.$ Points $Q$ and $R$ lie on $\overline{AB}$ so that $\overline{PQ}$ is parallel to $\overline{CA}$ and $\overline{PR}$ is parallel to $\overline{CB}.$ It is given that the ratio of the area of triangle $PQR$ to the area of triangle $ABC$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

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]

Six-hundred fifty students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti? [asy] size(200); defaultpen(linewidth(0.7)); defaultpen(fontsize(8)); draw(origin--(0,250)); int i; for(i=0; i<6; i=i+1) { draw((0,50*i)--(5,50*i)); } filldraw((25,0)--(75,0)--(75,150)--(25,150)--cycle, gray, black); filldraw((75,0)--(125,0)--(125,100)--(75,100)--cycle, gray, black); filldraw((125,0)--(175,0)--(175,150)--(125,150)--cycle, gray, black); filldraw((225,0)--(175,0)--(175,250)--(225,250)--cycle, gray, black); label("$50$", (0,50), W); label("$100$", (0,100), W); label("$150$", (0,150), W); label("$200$", (0,200), W); label("$250$", (0,250), W); label(rotate(90)*"Lasagna", (50,0), S); label(rotate(90)*"Manicotti", (100,0), S); label(rotate(90)*"Ravioli", (150,0), S); label(rotate(90)*"Spaghetti", (200,0), S); label(rotate(90)*"$\mbox{Number of People}$", (-40,140), W);[/asy] $\textbf{(A)} \: \frac25\qquad \textbf{(B)} \: \frac12\qquad \textbf{(C)} \: \frac54\qquad \textbf{(D)} \: \frac53\qquad \textbf{(E)} \: \frac52$

Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width? [asy]unitsize(8mm); defaultpen(linewidth(.8pt)); draw(scale(4)*unitsquare); draw((0,3)--(4,3)); draw((1,3)--(1,4)); draw((2,3)--(2,4)); draw((3,3)--(3,4));[/asy]$ \textbf{(A)}\ \frac {5}{4} \qquad \textbf{(B)}\ \frac {4}{3} \qquad \textbf{(C)}\ \frac {3}{2} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$

By adding the same constant to $20,50,100$ a geometric progression results. The common ratio is: $ \textbf{(A)}\ \frac53 \qquad\textbf{(B)}\ \frac43\qquad\textbf{(C)}\ \frac32\qquad\textbf{(D)}\ \frac12\qquad\textbf{(E)}\ \frac13 $

Given a parallelogram $ABCD$, join $A$ to the midpoints $E$ and $F$ of the opposite sides $BC$ and $CD$. $AE$ and $AF$ intersect the diagonal $BD$ in $M$ and $N$. Prove that $M$ and $N$ divide $BD$ into three equal parts.

Call a $ 3$-digit number [i]geometric[/i] if it has $ 3$ distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

A student performed an experiment to determine the ratio of $\text{H}_2\text{O}$ to $\text{CuSO}_4$ in a sample of hydrated copper(II) sulfate by heating it to drive off the water and weighing the solid before and after heating. The formula obtained experimentally was $\text{CuSO}_4 \bullet 5.5\text{H}_2\text{O}$ but the accepted formula is $\text{CuSO}_4 \bullet 5 \text{H}_2\text{O}$. Which error best accounts for the difference in results? $ \textbf{(A)}\ \text{During heating some of the hydrated copper(II) sulfate was lost} \qquad$ $\textbf{(B)}\ \text{The hydrated sample was not heated long enough to drive off all the water}\qquad$ $\textbf{(C)}\ \text{The student weighed out too much sample initially.} \qquad$ $\textbf{(D)}\ \text{The student used a balance that gave weights that were consistently too high by 0.10 g }\qquad$

Assume that $C$ is a convex subset of $\mathbb R^{d}$. Suppose that $C_{1},C_{2},\dots,C_{n}$ are translations of $C$ that $C_{i}\cap C\neq\emptyset$ but $C_{i}\cap C_{j}=\emptyset$. Prove that \[n\leq 3^{d}-1\] Prove that $3^{d}-1$ is the best bound. P.S. In the exam problem was given for $n=3$.

A circle passes through the vertices of a triangle with side-lengths of $7\tfrac{1}{2},10,12\tfrac{1}{2}$. The radius of the circle is: $\textbf{(A)}\ \dfrac{15}{4} \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ \dfrac{25}{4} \qquad \textbf{(D)}\ \dfrac{35}{4} \qquad \textbf{(E)}\ \dfrac{15\sqrt2}{2} $

Let $P$ be an interior point of a triangle $ABC$ and let $BP$ and $CP$ meet $AC$ and $AB$ in $E$ and $F$ respectively. IF $S_{BPF} = 4$,$S_{BPC} = 8$ and $S_{CPE} = 13$, find $S_{AFPE}.$

The perimeter of triangle $ABC$ is $k$ times larger than side $BC$, $AB \ne AC$. In what ratio does the median to side $BC$ divide the diameter of the circle inscribed in this triangle, perpendicular to this side?

In the figure, AQPB and ASRC are squares, and AQS is an equilateral triangle. If QS  = 4 and BC  = x, what is the value of x? [asy] unitsize(16); pair A,B,C,P,Q,R,T; A=(3.4641016151377544, 2); B=(0, 0); C=(6.928203230275509, 0); P=(-1.9999999999999991, 3.464101615137755); Q=(1.4641016151377544, 5.464101615137754); R=(8.928203230275509, 3.4641016151377544); T=(5.464101615137754, 5.464101615137754); dot(A);dot(B);dot(C);dot(P); dot(Q);dot(R);dot(T); label("$A$", (3.4641016151377544, 2),E); label("$B$", (0, 0),S); label("$C$", (6.928203230275509, 0),S); label("$P$", (-1.9999999999999991, 3.464101615137755), W); label("$Q$", (1.4641016151377544, 5.464101615137754),N); label("$R$", (8.928203230275509, 3.4641016151377544),E); label("$S$", (5.464101615137754, 5.464101615137754),N); draw(B--C--A--B); draw(B--P--Q--A--B); draw(A--C--R--T--A); draw(Q--T--A--Q); label("$x$", (3.4641016151377544, 0), S); label("$4$", (Q+T)/2, N);[/asy]

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.)

Let $ABC$ be an equilateral triangle, and $P$ be an arbitrary point within the triangle. Perpendiculars $PD,PE,PF$ are drawn to the three sides of the triangle. Show that, no matter where $P$ is chosen, \[ \frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}. \]

Two circles intersect at two points $A$ and $B$. A line $\ell$ which passes through the point $A$ meets the two circles again at the points $C$ and $D$, respectively. Let $M$ and $N$ be the midpoints of the arcs $BC$ and $BD$ (which do not contain the point $A$) on the respective circles. Let $K$ be the midpoint of the segment $CD$. Prove that $\measuredangle MKN = 90^{\circ}$.

Let $ABC$ be a triangle and $D$ a point inside the triangle, located on the median of $A$. Prove that if $\angle BDC = 180^o - \angle BAC$, then $AB \cdot CD = AC \cdot BD$.

Give a geometric series $(a_n)$ with common ratio of $q$, let $b_1=a_1+a_2+a_3,b_2=a_4+a_5+a_6,\cdots,b_n=a_{3n}+a_{3n+1}+a_{3n+2}$, then sequence $(b_n)$ $\text{(A)}$ is an arithmetic sequence $\text{(B)}$ is a geometric series with common ratio of $q$ $\text{(C)}$ is a geometric series with common ratio of $q^3$ $\text{(D)}$ is neither an arithmetic sequence nor a geometric series

In quadrilateral $ABCD$, we have $AB = 5$, $BC = 6$, $CD = 5$, $DA = 4$, and $\angle ABC = 90^\circ$. Let $AC$ and $BD$ meet at $E$. Compute $\dfrac{BE}{ED}$.

Determine all composite positive integers $n$ with the following property: If $1 = d_1 < d_2 < \cdots < d_k = n$ are all the positive divisors of $n$, then $$(d_2 - d_1) : (d_3 - d_2) : \cdots : (d_k - d_{k-1}) = 1:2: \cdots :(k-1)$$ (Walther Janous)

Let $ ABC$ be an acute triangle, and $ M_a$, $ M_b$, $ M_c$ be the midpoints of the sides $ a$, $ b$, $ c$. The perpendicular bisectors of $ a$, $ b$, $ c$ (passing through $ M_a$, $ M_b$, $ M_c$) intersect the boundary of the triangle again in points $ T_a$, $ T_b$, $ T_c$. Show that if the set of points $ \left\{A,B,C\right\}$ can be mapped to the set $ \left\{T_a, T_b, T_c\right\}$ via a similitude transformation, then two feet of the altitudes of triangle $ ABC$ divide the respective triangle sides in the same ratio. (Here, "ratio" means the length of the shorter (or equal) part divided by the length of the longer (or equal) part.) Does the converse statement hold?