Let $F$ be the focus of parabola $y^2=2px(p>0)$, with directrix $l$ and two points $A,B$ on it. Knowing that $\angle AFB=\frac{\pi}{3}$, find the maximal value of $\frac{|MN|}{|AB|}$, where $M$ is the midpoint of $AB$ and $N$ is the projection of $M$ to $l$.

Let $ABCD$ be a square of side length $13$. Let $E$ and $F$ be points on rays $AB$ and $AD$ respectively, so that the area of square $ABCD$ equals the area of triangle $AEF$. If $EF$ intersects $BC$ at $X$ and $BX = 6$, determine $DF$.

In triangle $ABC$ shown in the adjoining figure, $M$ is the midpoint of side $BC$, $AB=12$ and $AC=16$. Points $E$ and $F$ are taken on $AC$ and $AB$, respectively, and lines $EF$ and $AM$ intersect at $G$. If $AE=2AF$ then $\frac{EG}{GF}$ equals [asy] draw((0,0)--(12,0)--(14,7.75)--(0,0)); draw((0,0)--(13,3.875)); draw((5,0)--(8.75,4.84)); label("A", (0,0), S); label("B", (12,0), S); label("C", (14,7.75), E); label("E", (8.75,4.84), N); label("F", (5,0), S); label("M", (13,3.875), E); label("G", (7,1)); [/asy] $ \textbf{(A)}\ \frac{3}{2} \qquad\textbf{(B)}\ \frac{4}{3} \qquad\textbf{(C)}\ \frac{5}{4} \qquad\textbf{(D)}\ \frac{6}{5} \\ \qquad\textbf{(E)}\ \text{not enough information to solve the problem} $

The ratio of the number of games won to the number of games lost (no ties) by the Middle School Middies is $11:4$ . To the nearest whole percent, what percent of its games did the team lose? $ \text{(A)}\ 24\qquad\text{(B)}\ 27\qquad\text{(C)}\ 36\qquad\text{(D)}\ 45\qquad\text{(E)}\ 73 $

In his triangle $ABC$ Serge made some measurements and informed Ilias about the lengths of median $AD$ and side $AC$. Based on these data Ilias proved the assertion: angle $CAB$ is obtuse, while angle $DAB$ is acute. Determine a ratio $AD/AC$ and prove Ilias' assertion (for any triangle with such a ratio).

An amusement park has a collection of scale models, with ratio $1 : 20$, of buildings and other sights from around the country. The height of the United States Capitol is 289 feet. What is the height in feet of its replica to the nearest whole number? $\textbf{(A) }14\qquad\textbf{(B) }15\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad\textbf{(E) }20$

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 $\angle BOC=\angle AOD$.

Given an acute angled triangle $ ABC$ , $ O$ is the circumcenter and $ H$ is the orthocenter.Let $ A_1$,$ B_1$,$ C_1$ be the midpoints of the sides $ BC$,$ AC$ and $ AB$ respectively. Rays $ [HA_1$,$ [HB_1$,$ [HC_1$ cut the circumcircle of $ ABC$ at $ A_0$,$ B_0$ and $ C_0$ respectively.Prove that $ O$,$ H$ and $ H_0$ are collinear if $ H_0$ is the orthocenter of $ A_0B_0C_0$

The ratio of the length to the width of a rectangle is $4:3$. If the rectangle has diagonal of length $d$, then the area may be expressed as $kd^2$ for some constant $k$. What is $k$? $\textbf{(A) }\dfrac27\qquad\textbf{(B) }\dfrac37\qquad\textbf{(C) }\dfrac{12}{25}\qquad\textbf{(D) }\dfrac{16}{25}\qquad\textbf{(E) }\dfrac34$

In the figure, $ \overline{CD}, \overline{AE}$ and $ \overline{BF}$ are one-third of their respective sides. It follows that $ \overline{AN_2}: \overline{N_2N_1}: \overline{N_1D} \equal{} 3: 3: 1$, and similarly for lines $ BE$ and $ CF.$ Then the area of triangle $ N_1N_2N_3$ is: [asy]unitsize(27); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A,B,C,D,E,F,X,Y,Z; A=(3,3); B=(0,0); C=(6,0); D=(4,0); E=(4,2); F=(1,1); draw(A--B--C--cycle); draw(A--D); draw(B--E); draw(C--F); X=intersectionpoint(A--D,C--F); Y=intersectionpoint(B--E,A--D); Z=intersectionpoint(B--E,C--F); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,NE); label("$F$",F,NW); label("$N_1$",X,NE); label("$N_2$",Y,WNW); label("$N_3$",Z,S);[/asy]$ \textbf{(A)}\ \frac {1}{10} \triangle ABC \qquad\textbf{(B)}\ \frac {1}{9} \triangle ABC \qquad\textbf{(C)}\ \frac {1}{7} \triangle ABC \qquad\textbf{(D)}\ \frac {1}{6} \triangle ABC \qquad\textbf{(E)}\ \text{none of these}$

In a triangle $ABC, X$ and $Y$ are points on the segments $AB$ and $AC$, respectively, such that $AX : XB = 1 : 2$ and $AY :YC = 2:1$. If the area of triangle $AXY$ is $10$, then what is the area of triangle $ABC$?

Let $z$ be a complex number satisfying $(z+\tfrac{1}{z})(z+\tfrac{1}{z}+1)=1$. Evaluate $(3z^{100}+\tfrac{2}{z^{100}}+1)(z^{100}+\tfrac{2}{z^{100}}+3)$.

Show that there exists a triangle $ABC$ such that $a \ne b$ and $a+t_a = b+t_b$, where $t_a,t_b$ are the medians corresponding to $a,b$, respectively. Also prove that there exists a number $k$ such that every such triangle satisfies $a+t_a = b+t_b = k(a+b)$. Finally, find all possible ratios $a : b$ in such triangles.

A square piece of paper, 4 inches on a side, is folded in half vertically. Both layers are then cut in half parallel to the fold. Three new rectangles are formed, a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle? [asy] draw((0,8)--(0,0)--(4,0)--(4,8)--(0,8)--(3.5,8.5)--(3.5,8)); draw((2,-1)--(2,9),dashed);[/asy] $ \text{(A)}\ \frac{1}{3}\qquad\text{(B)}\ \frac{1}{2}\qquad\text{(C)}\ \frac{3}{4}\qquad\text{(D)}\ \frac{4}{5}\qquad\text{(E)}\ \frac{5}{6} $

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 $AB \cdot CD = AC \cdot BD$.

Let $K$ be a point in the interior of an acute triangle $ABC$ and $ARBPCQ$ be a convex hexagon whose vertices lie on the circumcircle $\Gamma$ of the triangle $ABC.$ Let $A_1$ be the second point where the circle passing through $K$ and tangent to $\Gamma$ at $A$ intersects the line $AP.$ The points $B_1$ and $C_1$ are defined similarly. Prove that \[ \min\left\{\frac{PA_1}{AA_1}, \: \frac{QB_1}{BB_1}, \: \frac{RC_1}{CC_1}\right\} \leq 1.\]

A paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width? [asy]unitsize(15mm); defaultpen(linewidth(.8pt)); path P=(-sqrt(2)/2,1)--(0,1-sqrt(2)/2)--(sqrt(2)/2,1); path Pc=(-sqrt(2)/2,1)--(0,1-sqrt(2)/2)--(sqrt(2)/2,1)--cycle; path S=(-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle; fill(S,gray); for(int i=0;i<4;++i) { fill(rotate(90*i)*Pc,white); draw(rotate(90*i)*P); } draw(S);[/asy]$ \textbf{(A)}\ 2\sqrt {2} \plus{} 1 \qquad \textbf{(B)}\ 3\sqrt {2}\qquad \textbf{(C)}\ 2\sqrt {2} \plus{} 2 \qquad \textbf{(D)}\ 3\sqrt {2} \plus{} 1 \qquad \textbf{(E)}\ 3\sqrt {2} \plus{} 2$

Let $P,Q,R$ be points on the sides of $\triangle ABC$ such that $P \in [AB],Q\in[BC],R\in[CA]$ and $\frac{|AP|}{|AB|} = \frac {|BQ|}{|BC|} =\frac{|CR|}{|CA|} =k < \frac 12$ If $G$ is the centroid of $\triangle ABC$, find the ratio $\frac{Area(\triangle PQG)}{Area(\triangle PQR)}$ .

On circle there are points $A$, $B$ and $C$ such that they divide circle in ratio $3:5:7$. Find angles of triangle $ABC$

Prove that for any tetrahedron in space, it is possible to find two perpendicular planes such that ratio between the projections of the tetrahedron on the two planes lies in the interval $[\frac{1}{\sqrt{2}}, \sqrt{2}].$

The base of a triangle is twice as long as a side of a square and their areas are the same. Then the ratio of the altitude of the triangle to the side of the square is: $ \textbf{(A)}\ \frac{1}{4} \qquad \textbf{(B)}\ \frac{1}{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 4$

From a point within a triangle, line segments are drawn to the vertices. A necessary and sufficient condition that the three triangles thus formed have equal areas is that the point be: $ \textbf{(A)}\ \text{the center of the inscribed circle} \qquad \\ \textbf{(B)}\ \text{the center of the circumscribed circle}\qquad \\ \textbf{(C)}\ \text{such that the three angles fromed at the point each be }{120^\circ}\qquad \\ \textbf{(D)}\ \text{the intersection of the altitudes of the triangle}\qquad \\ \textbf{(E)}\ \text{the intersection of the medians of the triangle}$

Given Triangle $ABC$, $\angle B= 2 \angle C$, and $\angle A>90^\circ$. Let $M$ be midpoint of $BC$. Perpendicular of $AC$ at $C$ intersects $AB$ at $D$. Show $\angle AMB = \angle DMC$ [hide]If possible, don't use projective geometry[/hide]

Let $ABC$ be a triangle, let $B_{1}$ be the midpoint of the side $AB$ and $C_{1}$ the midpoint of the side $AC$. Let $P$ be the point of intersection, other than $A$, of the circumscribed circles around the triangles $ABC_{1}$ and $AB_{1}C$. Let $P_{1}$ be the point of intersection, other than $A$, of the line $AP$ with the circumscribed circle around the triangle $AB_{1}C_{1}$. Prove that $2AP=3AP_{1}$.

In the plane, two intersecting lines $a$ and $b$ are given, along with a circle $\omega$ that has no common points with these lines. For any line $\ell||b$, define $A=\ell\cap a$, and $\{B,C\}=\ell\cap \omega$ such that $B$ is on segment $AC$. Construct the line $\ell$ such that the ratio $\frac{|BC|}{|AB|}$ is maximal.