In the isosceles triangle $ ABC$ the angle $ BAC$ is a right angle. Point $ D$ lies on the side $ BC$ and satisfies $ BD \equal{} 2 \cdot CD$. Point $ E$ is the foot of the perpendicular of the point $ B$ on the line $ AD$. Find the angle $ CED$.

Let $ABC$ be a nondegenerate acute triangle with circumcircle $\omega$ and let its incircle $\gamma$ touch $AB, AC, BC$ at $X, Y, Z$ respectively. Let $XY$ hit arcs $AB, AC$ of $\omega$ at $M, N$ respectively, and let $P \neq X, Q \neq Y$ be the points on $\gamma$ such that $MP=MX, NQ=NY$. If $I$ is the center of $\gamma$, prove that $P, I, Q$ are collinear if and only if $\angle BAC=90^\circ$. [i]Proposed by David Stoner[/i]

Let $ABC$ be an acute-angled triangle with $AB\not= AC$. Let $\Gamma$ be the circumcircle, $H$ the orthocentre and $O$ the centre of $\Gamma$. $M$ is the midpoint of $BC$. The line $AM$ meets $\Gamma$ again at $N$ and the circle with diameter $AM$ crosses $\Gamma$ again at $P$. Prove that the lines $AP,BC,OH$ are concurrent if and only if $AH=HN$.

The circumcenter of a triangle lies on its incircle. Prove that the ratio of its greatest and smallest sides is less than two. [i]Proposed by B.Frenkin[/i]

Let $AB$ and $FD$ be chords in circle, which does not intersect and $P$ point on arc $AB$ which does not contain chord $FD$. Lines $PF$ and $PD$ intersect chord $AB$ in $Q$ and $R$. Prove that $\frac{AQ* RB}{QR}$ is constant, while point $P$ moves along the ray $AB$.

The diagonals of convex quadrilateral $BSCT$ meet at the midpoint $M$ of $\overline{ST}$. Lines $BT$ and $SC$ meet at $A$, and $AB = 91$, $BC = 98$, $CA = 105$. Given that $\overline{AM} \perp \overline{BC}$, find the positive difference between the areas of $\triangle SMC$ and $\triangle BMT$. [i]Proposed by Evan Chen[/i]

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 an acute-angled triangle $ABC$, we consider the feet $H_a$ and $H_b$ of the altitudes from $A$ and $B$, and the intersections $W_a$ and $W_b$ of the angle bisectors from $A$ and $B$ with the opposite sides $BC$ and $CA$ respectively. Show that the centre of the incircle $I$ of triangle $ABC$ lies on the segment $H_aH_b$ if and only if the centre of the circumcircle $O$ of triangle $ABC$ lies on the segment $W_aW_b$.

In acute-angled triangle $ABC$, a semicircle with radius $r_a$ is constructed with its base on $BC$ and tangent to the other two sides. $r_b$ and $r_c$ are defined similarly. $r$ is the radius of the incircle of $ABC$. Show that \[ \frac{2}{r} = \frac{1}{r_a} + \frac{1}{r_b} + \frac{1}{r_c}. \]

Prove that in all triangles $\vartriangle ABC$ with $\angle A = 2 \angle B$ it holds that, if $D$ is the foot of the perpendicular from $C$ to the perpendicular bisector of $AB$, $\frac{AC}{DC}$ is constant for any value of $\angle B$.

Incircle of triangle $ ABC$ touches $ AB,AC$ at $ P,Q$. $ BI, CI$ intersect with $ PQ$ at $ K,L$. Prove that circumcircle of $ ILK$ is tangent to incircle of $ ABC$ if and only if $ AB\plus{}AC\equal{}3BC$.

Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta who did not attempt 300 points on the first day, had a positive integer score on each of the two days, and Beta's daily success rate (points scored divided by points attempted) on each day was less than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is $m/n$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

Circles $W_1,W_2$ intersect at $P,K$. $XY$ is common tangent of two circles which is nearer to $P$ and $X$ is on $W_1$ and $Y$ is on $W_2$. $XP$ intersects $W_2$ for the second time in $C$ and $YP$ intersects $W_1$ in $B$. Let $A$ be intersection point of $BX$ and $CY$. Prove that if $Q$ is the second intersection point of circumcircles of $ABC$ and $AXY$ \[\angle QXA=\angle QKP\]

Given is an acute triangle $ ABC$ and the points $ A_1,B_1,C_1$, that are the feet of its altitudes from $ A,B,C$ respectively. A circle passes through $ A_1$ and $ B_1$ and touches the smaller arc $ AB$ of the circumcircle of $ ABC$ in point $ C_2$. Points $ A_2$ and $ B_2$ are defined analogously. Prove that the lines $ A_1A_2$, $ B_1B_2$, $ C_1C_2$ have a common point, which lies on the Euler line of $ ABC$.

In $\triangle ABC$, $AB= 425$, $BC=450$, and $AC=510$. An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle. If these three segments are of an equal length $d$, find $d$.

Prove that for each natural number $n$ there is a polynomial $f$ with real coefficients and degree $n$ such that $ p(x)=f(x^2-1)$ is divisible by $f(x)$ over the ring $\mathbb{R}[x]$.

As illustrated below, we can dissect every triangle $ABC$ into four pieces so that piece 1 is a triangle similar to the original triangle, while the other three pieces can be assembled into a triangle also similar to the original triangle. Determine the ratios of the sizes of the three triangles and verify that the construction works. [asy] import olympiad;size(350);defaultpen(linewidth(0.7)+fontsize(10)); path p=origin--(13,0)--(9,8)--cycle; path p2=rotate(180)*p, p3=shift(-26,0)*scale(2)*p, p4=shift(-27,-24)*scale(3)*p, p1=shift(-53,-24)*scale(4)*p; pair A=(-53,-24), B=(-8,16), C=(12,-24), D=(-17,8), E=(-1,-24), F=origin, G=(-13,0), H=(-9,-8); label("1", centroid(A,D,E)); label("2", centroid(F,G,H)); label("3", (-10,6)); label("4", (0,-15)); draw(p2^^p3^^p4); filldraw(p1, white, black); pair point = centroid(F,G,H); label("$\mathbf{A}$", A, dir(point--A)); label("$\mathbf{B}$", B, dir(point--B)); label("$\mathbf{C}$", C, dir(point--C)); label("$\mathbf{D}$", D, dir(point--D)); label("$\mathbf{E}$", E, dir(point--E)); label("$\mathbf{F}$", F, dir(point--F)); label("$\mathbf{G}$", G, dir(point--G)); label("$\mathbf{H}$", H, dir(point--H)); real x=90; draw(shift(x)*p1); label("1", centroid(shift(x)*A,shift(x)*D,shift(x)*E)); draw(shift(130,0)*p4); draw(shift(130,0)*shift(-27,-24)*p); draw(shift(130,0)*shift(-1,-24)*p3); label("2", shift(130,0)*shift(-27,-24)*centroid(F,(9,8),(13,0))); label("3", shift(130,0)*shift(-1,-24)*(-10,6)); label("4", shift(130,0)*(0,-15)); label("Piece 2 rotated $180^\circ$", (130,10));[/asy]

Let $ ABC$ a triangle and $ X$, $ Y$ and $ Z$ points at the segments $ BC$, $ AC$ and $ AB$, respectively.Let $ A'$, $ B'$ and $ C'$ the circuncenters of triangles $ AZY$,$ BXZ$,$ CYX$, respectively.Prove that $ 4(A'B'C')\geq(ABC)$ with equality if and only if $ AA'$, $ BB'$ and $ CC'$ are concurrents. Note: $ (XYZ)$ denotes the area of $ XYZ$

A recipe that makes $ 5$ servings of hot chocolate requires $ 2$ squares of chocolate, $ \frac{1}{4}$ cup sugar, $ 1$ cup water and $ 4$ cups milk. Jordan has $ 5$ squares of chocolate, $ 2$ cups of sugar, lots of water and $ 7$ cups of milk. If she maintains the same ratio of ingredients, what is the greatest number of servings of hot chocolate she can make? $ \textbf{(A)}\ 5 \frac18 \qquad \textbf{(B)}\ 6\frac14 \qquad \textbf{(C)}\ 7\frac12 \qquad \textbf{(D)}\ 8 \frac34 \qquad \textbf{(E)}\ 9\frac78$

The triangle $ABC$ is $| BC | = a$ and $| AC | = b$. On the ray starting from vertex $C$ and passing the midpoint of side $AB$ , choose any point $D$ other than vertex $C$. Let $K$ and $L$ be the projections of $D$ on the lines $AC$ and $BC$, respectively, $K$ and $L$. Find the ratio $| DK | : | DL |$.

Let $E$ and $F$ be the intersections of opposite sides of a convex quadrilateral $ABCD$. The two diagonals meet at $P$. Let $O$ be the foot of the perpendicular from $P$ to $EF$. Show that $\angle BOC=\angle AOD$.

Let the circle $ {\omega}_{1}$ be internally tangent to another circle $ {\omega}_{2}$ at $ N$.Take a point $ K$ on $ {\omega}_{1}$ and draw a tangent $ AB$ which intersects $ {\omega}_{2}$ at $ A$ and $ B$. Let $M$ be the midpoint of the arc $ AB$ which is on the opposite side of $ N$. Prove that, the circumradius of the $ \triangle KBM$ doesnt depend on the choice of $ K$.

Given a triangle $ABC$ with angles $\angle A = 60^{\circ}, \angle B = 75^{\circ}, \angle C = 45^{\circ}$, let $H$ be its orthocentre, and $O$ be its circumcenter. Let $F$ be the midpoint of side $AB$, and $Q$ be the foot of the perpendicular from $B$ onto $AC$. Denote by $X$ the intersection point of the lines $FH$ and $QO$. Suppose the ratio of the length of $FX$ and the circumradius of the triangle is given by $\dfrac{a + b \sqrt{c}}{d}$, then find the value of $1000a + 100b + 10c + d$.

A massless elastic cord (that obeys Hooke's Law) will break if the tension in the cord exceeds $T_{max}$. One end of the cord is attached to a fixed point, the other is attached to an object of mass $3m$. If a second, smaller object of mass m moving at an initial speed $v_0$ strikes the larger mass and the two stick together, the cord will stretch and break, but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one-dimensional collision, the cord will still break, and the larger mass will move off with a final speed of $v_f$. All motion occurs on a horizontal, frictionless surface. Find the ratio of the total kinetic energy of the system of two masses after the perfectly elastic collision and the cord has broken to the initial kinetic energy of the smaller mass prior to the collision. $ \textbf{(A)}\ 1/4 \qquad\textbf{(B)}\ 1/3 \qquad\textbf{(C)}\ 1/2 \qquad\textbf{(D)}\ 3/4 \qquad\textbf{(E)}\ \text{none of the above} $

In a certain population the ratio of the number of women to the number of men is 11 to 10. If the average (arithmetic mean) age of the women is 34 and the average age of the men is 32, then the average age of the population is $ \textbf{(A)}\ 32\frac{9}{10}\qquad\textbf{(B)}\ 32\frac{20}{21}\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 33\frac{1}{21}\qquad\textbf{(E)}\ 33\frac{1}{10} $