Let $ABC$ be a triangle with $AB>AC$. Let $D$ be the foot of the internal angle bisector of $A$. Points $F$ and $E$ are on $AC,AB$ respectively such that $B,C,F,E$ are concyclic. Prove that the circumcentre of $DEF$ is the incentre of $ABC$ if and only if $BE+CF=BC$.

An equilateral triangle has been drawn inside the circle. Split the triangle to two parts with equal area by a line segment parallel to the triangle side. Draw an inscribed circle inside this smaller triangle. What is the ratio of the area of this circle compared to the area of original circle.

Suppose line $\ell$ and four points $A,B,C,D$ lies on $\ell$. Suppose that circles $\omega_1 , \omega_2$ passes through $A,B$ and circles $\omega'_1 , \omega'_2$ passes through $C,D$. If $\omega_1 \perp \omega'_1$ and $\omega_2 \perp \omega'_2$ then prove that lines $O_1O'_2 , O_2O'_1 , \ell $ are concurrent where $O_1,O_2,O'_1,O'_2$ are center of $\omega_1 , \omega_2 , \omega'_1 , \omega'_2$.

In parallelogram $ ABCD$, point $ M$ is on $ \overline{AB}$ so that $ \frac{AM}{AB} \equal{} \frac{17}{1000}$ and point $ N$ is on $ \overline{AD}$ so that $ \frac{AN}{AD} \equal{} \frac{17}{2009}$. Let $ P$ be the point of intersection of $ \overline{AC}$ and $ \overline{MN}$. Find $ \frac{AC}{AP}$.

A triangle with side lengths in the ratio $ 3: 4: 5$ is inscribed in a circle of radius $ 3$. What is the area of the triangle? $ \textbf{(A)}\ 8.64 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 5\pi \qquad \textbf{(D)}\ 17.28 \qquad \textbf{(E)}\ 18$

There are four more girls than boys in Ms. Raub's class of $28$ students. What is the ratio of number of girls to the number of boys in her class? $\textbf{(A) }3 : 4\qquad\textbf{(B) }4 : 3\qquad\textbf{(C) }3 : 2\qquad\textbf{(D) }7 : 4\qquad \textbf{(E) }2 : 1$

In right triangle $ \triangle ACE$, we have $ AC \equal{} 12$, $ CE \equal{} 16$, and $ EA \equal{} 20$. Points $ B$, $ D$, and $ F$ are located on $ \overline{AC}$, $ \overline{CE}$, and $ \overline{EA}$, respectively, so that $ AB \equal{} 3$, $ CD \equal{} 4$, and $ EF \equal{} 5$. What is the ratio of the area of $ \triangle DBF$ to that of $ \triangle ACE$? [asy] size(200);defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair C = (0,0); pair E = (16,0); pair A = (0,12); pair F = waypoint(E--A,0.25); pair B = waypoint(A--C,0.25); pair D = waypoint(C--E,0.25); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("$A$",A,NW);label("$B$",B,W);label("$C$",C,SW);label("$D$",D,S);label("$E$",E,SE);label("$F$",F,NE); label("$3$",midpoint(A--B),W); label("$9$",midpoint(B--C),W); label("$4$",midpoint(C--D),S); label("$12$",midpoint(D--E),S); label("$5$",midpoint(E--F),NE); label("$15$",midpoint(F--A),NE); draw(A--C--E--cycle); draw(B--F--D--cycle);[/asy]$ \textbf{(A)}\ \frac {1}{4}\qquad \textbf{(B)}\ \frac {9}{25}\qquad \textbf{(C)}\ \frac {3}{8}\qquad \textbf{(D)}\ \frac {11}{25}\qquad \textbf{(E)}\ \frac {7}{16}$

$\Delta oa_1b_1$ is isosceles with $\angle a_1ob_1 = 36^\circ$. Construct $a_2,b_2,a_3,b_3,...$ as below, with $|oa_{i+1}| = |a_ib_i|$ and $\angle a_iob_i = 36^\circ$, Call the summed area of the first $k$ triangles $A_k$. Let $S$ be the area of the isocseles triangle, drawn in - - -, with top angle $108^\circ$ and $|oc|=|od|=|oa_1|$, going through the points $b_2$ and $a_2$ as shown on the picture. (yes, $cd$ is parallel to $a_1b_1$ there) Show $A_k < S$ for every positive integer $k$. [img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=284[/img]

Find the perimeter of a triangle whose altitudes are $3,4,$ and $6$. $ \textbf{(A)}\ 12\sqrt\frac35 \qquad \textbf{(B)}\ 16\sqrt\frac35 \qquad \textbf{(C)}\ 20\sqrt\frac35 \qquad \textbf{(D)}\ 24\sqrt\frac35 \qquad \textbf{(E)}\ \text{None}$

Point $ E$ is selected on side $ AB$ of triangle $ ABC$ in such a way that $ AE: EB \equal{} 1: 3$ and point $ D$ is selected on side $ BC$ such that $ CD: DB \equal{} 1: 2$. The point of intersection of $ AD$ and $ CE$ is $ F$. Then $ \frac {EF}{FC} \plus{} \frac {AF}{FD}$ is: $ \textbf{(A)}\ \frac {4}{5} \qquad \textbf{(B)}\ \frac {5}{4} \qquad \textbf{(C)}\ \frac {3}{2} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \frac {5}{2}$

Triangle $ABC$ is inscribed in circle $\omega$ with $AB = 5$, $BC = 7$, and $AC = 3$. The bisector of angle $A$ meets side $BC$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $DE$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where m and n are relatively prime positive integers. Find $m + n$.

In a triangle $ABC$ with $AC = b \ne BC = a$, points $E,F$ are taken on the sides $AC,BC$ respectively such that $AE = BF =\frac{ab}{a+b}$. Let $M$ and $N$ be the midpoints of $AB$ and $EF$ respectively, and $P$ be the intersection point of the segment $EF$ with the bisector of $\angle ACB$. Find the ratio of the area of $CPMN$ to that of $ABC$.

Given three distinct unit circles, each of which is tangent to the other two, find the radii of the circles which are tangent to all three circles.

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$. Suppose that the radius of the circle is 5, that $BC = 6$, and that $AD$ is bisected by $BC$. Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$. It follows that the sine of the minor arc $AB$ is a rational number. If this fraction is expressed as a fraction $m/n$ in lowest terms, what is the product $mn$? [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=dir(200), D=dir(95), M=midpoint(A--D), C=dir(30), BB=C+2*dir(C--M), B=intersectionpoint(M--BB, Circle(origin, 1)); draw(Circle(origin, 1)^^A--D^^B--C); real r=0.05; pair M1=midpoint(M--D), M2=midpoint(M--A); draw((M1+0.1*dir(90)*dir(A--D))--(M1+0.1*dir(-90)*dir(A--D))); draw((M2+0.1*dir(90)*dir(A--D))--(M2+0.1*dir(-90)*dir(A--D))); pair point=origin; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D));[/asy]

Distinct points $A$ and $B$ are given on the plane. Consider all triangles $ABC$ in this plane on whose sides $BC,CA$ points $D,E$ respectively can be taken so that (i) $\frac{BD}{BC}=\frac{CE}{CA}=\frac{1}{3}$; (ii) points $A,B,D,E$ lie on a circle in this order. Find the locus of the intersection points of lines $AD$ and $BE$.

The points with integer coordinates are painted by red if the product of $x$ and $y$ coordinates is divisible by $6$. Otherwise the points with integer coordinates are painted by white. Consider a very big square whose sides are parallel to the axis of the $xy-$plane. The ratio of white points over red points inside this square will be closer to $\textbf{(A)}\ \frac75 \qquad\textbf{(B)}\ \frac32 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac43 \qquad\textbf{(E)}\ \frac54$

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]

The diagram shows a parabola, a line perpendicular to the parabola's axis of symmetry, and three similar isosceles triangles each with a base on the line and vertex on the parabola. The two smaller triangles are congruent and each have one base vertex on the parabola and one base vertex shared with the larger triangle. The ratio of the height of the larger triangle to the height of the smaller triangles is $\tfrac{a+\sqrt{b}}{c}$ where $a$, $b$, and $c$ are positive integers, and $a$ and $c$ are relatively prime. Find $a + b + c$. [asy] size(200); real f(real x) {return 1.2*exp(2/3*log(16-x^2));} path Q=graph(f,-3.99999,3.99999); path [] P={(-4,0)--(-2,0)--(-3,f(-3))--cycle,(-2,0)--(2,0)--(0,f(0))--cycle,(4,0)--(2,0)--(3,f(3))--cycle}; for(int k=0;k<3;++k) { fill(P[k],grey); draw(P[k]); } draw((-6,0)--(6,0),linewidth(1)); draw(Q,linewidth(1));[/asy]

Let $ABC$ be a triangle with $\angle B - \angle C = 30^{\circ}$. Let $D$ be the point where the $A$-excircle touches line $BC$, $O$ the circumcenter of triangle $ABC$, and $X,Y$ the intersections of the altitude from $A$ with the incircle with $X$ in between $A$ and $Y$. Suppose points $A$, $O$ and $D$ are collinear. If the ratio $\frac{AO}{AX}$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$ for positive integers $a,b,c,d$ with $\gcd(a,b,d)=1$ and $c$ not divisible by the square of any prime, find $a+b+c+d$. [i]James Tao[/i]

[asy] size(2.5inch); pair A, B, C, E, F, G; A = (0,3); B = (-1,0); C = (3,0); E = (0,0); F = (1,2); G = intersectionpoint(B--F,A--E); draw(A--B--C--cycle); draw(A--E); draw(B--F); label("$A$",A,N); label("$B$",B,W); label("$C$",C,dir(0)); label("$E$",E,S); label("$F$",F,NE); label("$G$",G,SE); //Credit to chezbgone2 for the diagram[/asy] In triangle $ABC$, point $F$ divides side $AC$ in the ratio $1:2$. Let $E$ be the point of intersection of side $BC$ and $AG$ where $G$ is the midpoints of $BF$. The point $E$ divides side $BC$ in the ratio $\textbf{(A) }1:4\qquad\textbf{(B) }1:3\qquad\textbf{(C) }2:5\qquad\textbf{(D) }4:11\qquad \textbf{(E) }3:8$

Two beads, each of mass $m$, are free to slide on a rigid, vertical hoop of mass $m_h$. The beads are threaded on the hoop so that they cannot fall off of the hoop. They are released with negligible velocity at the top of the hoop and slide down to the bottom in opposite directions. The hoop remains vertical at all times. What is the maximum value of the ratio $m/m_h$ such that the hoop always remains in contact with the ground? Neglect friction. [asy] pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((0,0)--(3,0)); draw(circle((1.5,1),1)); filldraw(circle((1.4,1.99499),0.1), gray(.3)); filldraw(circle((1.6,1.99499),0.1), gray(.3)); [/asy]

The width of rectangle $ABCD$ is twice its height, and the height of rectangle $EFCG$ is twice its width. The point $E$ lies on the diagonal $BD$. Which fraction of the area of the big rectangle is that of the small one? [img]https://1.bp.blogspot.com/-aeqefhbBh5E/XzcBjhgg7sI/AAAAAAAAMXM/B0qSgWDBuqc3ysd-mOitP1LarOtBdJJ3gCLcBGAsYHQ/s0/2004%2BMohr%2Bp1.png[/img]

The diagram below shows a regular hexagon with an inscribed square where two sides of the square are parallel to two sides of the hexagon. There are positive integers $m$, $n$, and $p$ such that the ratio of the area of the hexagon to the area of the square can be written as $\tfrac{m+\sqrt{n}}{p}$ where $m$ and $p$ are relatively prime. Find $m + n + p$. [asy] import graph; size(4cm); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); draw((0,1)--(1,1)--(1.5,1.87)--(1,2.73)--(0,2.73)--(-0.5,1.87)--cycle); filldraw((1.13,2.5)--(-0.13,2.5)--(-0.13,1.23)--(1.13,1.23)--cycle,grey); draw((0,1)--(1,1)); draw((1,1)--(1.5,1.87)); draw((1.5,1.87)--(1,2.73)); draw((1,2.73)--(0,2.73)); draw((0,2.73)--(-0.5,1.87)); draw((-0.5,1.87)--(0,1)); draw((1.13,2.5)--(-0.13,2.5)); draw((-0.13,2.5)--(-0.13,1.23)); draw((-0.13,1.23)--(1.13,1.23)); draw((1.13,1.23)--(1.13,2.5)); [/asy]

A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone? [asy] real r=(3+sqrt(5))/2; real s=sqrt(r); real Brad=r; real brad=1; real Fht = 2*s; import graph3; import solids; currentprojection=orthographic(1,0,.2); currentlight=(10,10,5); revolution sph=sphere((0,0,Fht/2),Fht/2); //draw(surface(sph),green+white+opacity(0.5)); //triple f(pair t) {return (t.x*cos(t.y),t.x*sin(t.y),t.x^(1/n)*sin(t.y/n));} triple f(pair t) { triple v0 = Brad*(cos(t.x),sin(t.x),0); triple v1 = brad*(cos(t.x),sin(t.x),0)+(0,0,Fht); return (v0 + t.y*(v1-v0)); } triple g(pair t) { return (t.y*cos(t.x),t.y*sin(t.x),0); } surface sback=surface(f,(3pi/4,0),(7pi/4,1),80,2); surface sfront=surface(f,(7pi/4,0),(11pi/4,1),80,2); surface base = surface(g,(0,0),(2pi,Brad),80,2); draw(sback,rgb(0,1,0)); draw(sfront,rgb(.3,1,.3)); draw(base,rgb(.4,1,.4)); draw(surface(sph),rgb(.3,1,.3)); [/asy] $ \textbf {(A) } \dfrac {3}{2} \qquad \textbf {(B) } \dfrac {1+\sqrt{5}}{2} \qquad \textbf {(C) } \sqrt{3} \qquad \textbf {(D) } 2 \qquad \textbf {(E) } \dfrac {3+\sqrt{5}}{2} $

Let $ ABC$ be a triangle, and let $ P$, $ Q$, $ R$ be three points in the interiors of the sides $ BC$, $ CA$, $ AB$ of this triangle. Prove that the area of at least one of the three triangles $ AQR$, $ BRP$, $ CPQ$ is less than or equal to one quarter of the area of triangle $ ABC$. [i]Alternative formulation:[/i] Let $ ABC$ be a triangle, and let $ P$, $ Q$, $ R$ be three points on the segments $ BC$, $ CA$, $ AB$, respectively. Prove that $ \min\left\{\left|AQR\right|,\left|BRP\right|,\left|CPQ\right|\right\}\leq\frac14\cdot\left|ABC\right|$, where the abbreviation $ \left|P_1P_2P_3\right|$ denotes the (non-directed) area of an arbitrary triangle $ P_1P_2P_3$.