Let $ABC$ be a triangle, and let the tangent to the circumcircle of the triangle $ABC$ at $A$ meet the line $BC$ at $D$. The perpendicular to $BC$ at $B$ meets the perpendicular bisector of $AB$ at $E$. The perpendicular to $BC$ at $C$ meets the perpendicular bisector of $AC$ at $F$. Prove that the points $D$, $E$ and $F$ are collinear. [i]Valentin Vornicu[/i]

Let $ A, B, D, E, F, C $ be six points lie on a circle (in order) satisfy $ AB=AC $ . Let $ P=AD \cap BE, R=AF \cap CE, Q=BF \cap CD, S=AD \cap BF, T=AF \cap CD $ . Let $ K $ be a point lie on $ ST $ satisfy $ \angle QKS=\angle ECA $ . Prove that $ \frac{SK}{KT}=\frac{PQ}{QR} $

There are $270$ students at Colfax Middle School, where the ratio of boys to girls is $5 : 4$. There are $180$ students at Winthrop Middle School, where the ratio of boys to girls is $4 : 5$. The two schools hold a dance and all students from both schools attend. What fraction of the students at the dance are girls? $ \textbf{(A)} \dfrac7{18} \qquad\textbf{(B)} \dfrac7{15} \qquad\textbf{(C)} \dfrac{22}{45} \qquad\textbf{(D)} \dfrac12 \qquad\textbf{(E)} \dfrac{23}{45} $

$O$ is the circumcenter of acute $\Delta ABC$, $H$ is the Orthocenter. $AD \bot BC$, $EF$ is the perpendicular bisector of $AO$,$D,E$ on the $BC$. Prove that the circumcircle of $\Delta ADE$ through the midpoint of $OH$.

Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra? $ \textbf{(A)}\ \frac{1}{12}\qquad\textbf{(B)}\ \frac{\sqrt{2}}{12}\qquad\textbf{(C)}\ \frac{\sqrt{3}}{12}\qquad\textbf{(D)}\ \frac{1}{6}\qquad\textbf{(E)}\ \frac{\sqrt{2}}{6} $

Let $O$ be the intersection point of medians $AP$ and $CQ$ of triangle $ABC$. If $OQ$ is $3$ inches, then $OP$, in inches, is: $\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \dfrac{9}{2} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ \text{undetermined}$

Given the circumradius $R$ of a triangle, a side length $c$, and the ratio $a/b$ of the other two side lengths, determine all three sides and angles of this triangle.

Let $AEF$ be a triangle with $EF = 20$ and $AE = AF = 21$. Let $B$ and $D$ be points chosen on segments $AE$ and $AF,$ respectively, such that $BD$ is parallel to $EF.$ Point $C$ is chosen in the interior of triangle $AEF$ such that $ABCD$ is cyclic. If $BC = 3$ and $CD = 4,$ then the ratio of areas $\tfrac{[ABCD]}{[AEF]}$ can be written as $\tfrac{a}{b}$ for relatively prime positive integers $a, b$. Compute $100a + b$.

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} $

In the right triangle $ABC,$ $CF$ is the altitude based on the hypotenuse $AB.$ The circle centered at $B$ and passing through $F$ and the circle with centre $A$ and the same radius intersect at a point of $CB.$ Determine the ratio $FB : BC.$

We call a subset $B$ of natural numbers [i]loyal[/i] if there exists natural numbers $i\le j$ such that $B=\{i,i+1,\ldots,j\}$. Let $Q$ be the set of all [i]loyal[/i] sets. For every subset $A=\{a_1<a_2<\ldots<a_k\}$ of $\{1,2,\ldots,n\}$ we set \[f(A)=\max_{1\le i \le k-1}{a_{i+1}-a_i}\qquad\text{and}\qquad g(A)=\max_{B\subseteq A, B\in Q} |B|.\] Furthermore, we define \[F(n)=\sum_{A\subseteq \{1,2,\ldots,n\}} f(A)\qquad\text{and}\qquad G(n)=\sum_{A\subseteq \{1,2,\ldots,n\}} g(A).\] Prove that there exists $m\in \mathbb N$ such that for each natural number $n>m$ we have $F(n)>G(n)$. (By $|A|$ we mean the number of elements of $A$, and if $|A|\le 1$, we define $f(A)$ to be zero). [i]Proposed by Javad Abedi[/i]

A small point-like object is thrown horizontally off of a $50.0$-$\text{m}$ high building with an initial speed of $10.0 \text{ m/s}$. At any point along the trajectory there is an acceleration component tangential to the trajectory and an acceleration component perpendicular to the trajectory. How many seconds after the object is thrown is the tangential component of the acceleration of the object equal to twice the perpendicular component of the acceleration of the object? Ignore air resistance. $ \textbf{(A)}\ 2.00\text{ s}$ $\textbf{(B)}\ 1.50\text{ s}$ $\textbf{(C)}\ 1.00\text{ s}$ $\textbf{(D)}\ 0.50\text{ s}$ $\textbf{(E)}\ \text{The building is not high enough for this to occur.} $

In a triangle $ABC$, the angle bisector at $A,B,C$ meet the opposite sides at $A_1,B_1,C_1$, respectively. Prove that if the quadrilateral $BA_1B_1C_1$ is cyclic, then $$\frac{AC}{AB+BC}=\frac{AB}{AC+CB}+\frac{BC}{BA+AC}.$$

In quadrilateral $ ABCD$, it is given that $ \angle A \equal{} 120^\circ$, angles $ B$ and $ D$ are right angles, $ AB \equal{} 13$, and $ AD \equal{} 46$. Then $ AC \equal{}$ $ \textbf{(A)}\ 60 \qquad \textbf{(B)}\ 62 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 65 \qquad \textbf{(E)}\ 72$

Isosceles trapezoid $ ABCD$, with $ AB \parallel CD$, is such that there exists a circle $ \Gamma$ tangent to its four sides. Let $ T \equal{} \Gamma \cap BC$, and $ P \equal{} \Gamma \cap AT$ ($ P \neq T$). If $ \frac{AP}{AT} \equal{} \frac{2}{5}$, compute $ \frac{AB}{CD}$.

If 64 is divided into three parts proportional to 2, 4, and 6, the smallest part is: $\textbf{(A)}\ 5\dfrac{1}{3} \qquad \textbf{(B)}\ 11 \qquad \textbf{(C)}\ 10\dfrac{2}{3} \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{None of these answers}$

The centers of two circles are $ 41$ inches apart. The smaller circle has a radius of $ 4$ inches and the larger one has a radius of $ 5$ inches. The length of the common internal tangent is: $ \textbf{(A)}\ 41\text{ inches} \qquad\textbf{(B)}\ 39\text{ inches} \qquad\textbf{(C)}\ 39.8\text{ inches} \qquad\textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches}$

Let be a sequence $ \left( s_n\right)_{n\geqslant 0} $ of positive real numbers, with $ s_0 $ being the golden ratio, and defined as $$ s_{n+2}=\frac{1+s_{n+1}}{s_n} . $$ Establish the necessary and sufficient condition under which $ \left( s_n\right)_{n\geqslant 0} $ is convergent.

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $\mathcal{C}$ to the area of shaded region $\mathcal{B}$ is $11/5$. Find the ratio of shaded region $\mathcal{D}$ to the area of shaded region $\mathcal{A}$. [asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy]

A triangle $ABC$ is inscribed in a circle of radius $R.$ Let $BD$ and $CE$ be the bisectors of the angles $B$ and $C$ respectively and let the line $DE$ meet the arc $AB$ not containing $C$ at point $K.$ Let $A_1, B_1, C_1$ be the feet of perpendiculars from $K$ to $BC, AC, AB,$ and $x, y$ be the distances from $D$ and $E$ to $BC,$ respectively. (a) Express the lengths of $KA_1, KB_1, KC_1$ in terms of $x, y$ and the ratio $l = KD/ED.$ (b) Prove that $\frac{1}{KB}=\frac{1}{KA}+\frac{1}{KC}.$

On an exam with $80$ problems, Roger solved $68$ of them. What percentage of the problems did he solve?

Triangle $ABC$ is given with its centroid $G$ and cicumcentre $O$ is such that $GO$ is perpendicular to $AG$. Let $A'$ be the second intersection of $AG$ with circumcircle of triangle $ABC$. Let $D$ be the intersection of lines $CA'$ and $AB$ and $E$ the intersection of lines $BA'$ and $AC$. Prove that the circumcentre of triangle $ADE$ is on the circumcircle of triangle $ABC$.

A convex quadrilateral $ABC$ is given. $A',B',C',D'$ are the orthocenters of triangles $BCD, CDA, DAB, ABC$ respectively. Prove that in the quadrilaterals $ABCP$ and $A'B'C'D'$, the corresponding diagonals share the intersection points in the same ratio.

[asy] size(150); pair A=(0,0),B=(1,0),C=(0,1),D=(-1,0),E=(0,.5),F=(sqrt(2)/2,.25); draw(circle(A,1)^^D--B); draw(circle(E,.5)^^circle( F ,.25)); label("$A$", D, W); label("$K$", A, S); label("$B$", B, dir(0)); label("$L$", E, N); label("$M$",shift(-.05,.05)*F); //Credit to Klaus-Anton for the diagram[/asy] In the adjoining figure, circle $\mathit{K}$ has diameter $\mathit{AB}$; cirlce $\mathit{L}$ is tangent to circle $\mathit{K}$ and to $\mathit{AB}$ at the center of circle $\mathit{K}$; and circle $\mathit{M}$ tangent to circle $\mathit{K}$, to circle $\mathit{L}$ and $\mathit{AB}$. The ratio of the area of circle $\mathit{K}$ to the area of circle $\mathit{M}$ is $\textbf{(A) }12\qquad\textbf{(B) }14\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad \textbf{(E) }\text{not an integer}$

Given a circle $G$ with center $O$ and radius $r$. Let $AB$ be a fixed diameter of $G$. Let $K$ be a fixed point of segment $AO$. Denote by $t$ the line tangent to at $A$. For any chord $CD$ (other than $AB$) passing through $K$. Let $P$ and $Q$ be the points of intersection of lines $BC$ and $BD$ with $t$. Prove that the product $AP\cdot AQ$ remains costant as the chord $CD$ varies.