Let $ABC$ be a triangle inscribed in a circle and let \[ l_a = \frac{m_a}{M_a} \ , \ \ l_b = \frac{m_b}{M_b} \ , \ \ l_c = \frac{m_c}{M_c} \ , \] where $m_a$,$m_b$, $m_c$ are the lengths of the angle bisectors (internal to the triangle) and $M_a$, $M_b$, $M_c$ are the lengths of the angle bisectors extended until they meet the circle. Prove that \[ \frac{l_a}{\sin^2 A} + \frac{l_b}{\sin^2 B} + \frac{l_c}{\sin^2 C} \geq 3 \] and that equality holds iff $ABC$ is an equilateral triangle.

The figure below shows a large white circle with a number of smaller white and shaded circles in its interior. What fraction of the interior of the large white circle is shaded? [asy] size(5cm); defaultpen(linewidth(1pt)); draw(circle((3,3),3)); filldraw(circle((5.5,3),0.5),mediumgray*0.5 + lightgray*0.5); filldraw(circle((2,3),2),mediumgray*0.5 + lightgray*0.5); filldraw(circle((1,3),1),white); filldraw(circle((3,3),1),white); add(grid(6,6,mediumgray*0.5+gray*0.5+linetype("4 4"))); filldraw(circle((4.5,4.5),0.5),mediumgray*0.5 + lightgray*0.5); filldraw(circle((4.5,1.5),0.5),mediumgray*0.5 + lightgray*0.5); [/asy]$\textbf{(A) } \dfrac14\qquad\textbf{(B) } \dfrac{11}{36}\qquad\textbf{(C) } \dfrac13\qquad\textbf{(D) } \dfrac{19}{36}\qquad\textbf{(E) } \dfrac59$

Excircle of triangle $ABC$ to side $AB$ of triangle $ABC$ touches side $AB$ in point $D$. Determine ratio $AD : BD$ if $\angle CAB = 2 \angle ADC$

Four congruent rectangles are placed as shown. The area of the outer square is $ 4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side? [asy]unitsize(6mm); defaultpen(linewidth(.8pt)); path p=(1,1)--(-2,1)--(-2,2)--(1,2); draw(p); draw(rotate(90)*p); draw(rotate(180)*p); draw(rotate(270)*p);[/asy]$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ \sqrt {10} \qquad \textbf{(C)}\ 2 \plus{} \sqrt2 \qquad \textbf{(D)}\ 2\sqrt3 \qquad \textbf{(E)}\ 4$

[asy] draw(unitsquare);draw((0,0)--(.4,1)^^(0,.6)--(1,.2)); label("D",(0,1),NW);label("E",(.4,1),N);label("C",(1,1),NE); label("P",(0,.6),W);label("M",(.25,.55),E);label("Q",(1,.2),E); label("A",(0,0),SW);label("B",(1,0),SE); //Credit to Zimbalono for the diagram[/asy] Inside square $ABCD$ (See figure) with sides of length $12$ inches, segment $AE$ is drawn where $E$ is the point on $DC$ which is $5$ inches from $D$. The perpendicular bisector of $AE$ is drawn and intersects $AE$, $AD$, and $BC$ at points $M$, $P$, and $Q$ respectively. The ratio of segment $PM$ to $MQ$ is $\textbf{(A) }5:12\qquad\textbf{(B) }5:13\qquad\textbf{(C) }5:19\qquad\textbf{(D) }1:4\qquad \textbf{(E) }5:21$

Let $M$ be any point on the circumcircle of triangle $ABC$. Suppose the tangents from $M$ to the incircle meet $BC$ at two points $X_1$ and $X_2$. Prove that the circumcircle of triangle $MX_1X_2$ intersects the circumcircle of $ABC$ again at the tangency point of the $A$-mixtilinear incircle.

Prove that, for any integer $a_{1}>1$, there exist an increasing sequence of positive integers $a_{1}, a_{2}, a_{3}, \cdots$ such that \[a_{1}+a_{2}+\cdots+a_{n}\; \vert \; a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\] for all $n \in \mathbb{N}$.

A cuboctahedron is a solid with 6 square faces and 8 equilateral triangle faces, with each edge adjacent to both a square and a triangle (see picture). Suppose the ratio of the volume of an octahedron to a cuboctahedron with the same side length is $r$. Find $100r^2$. [asy] // dragon96, replacing // [img]http://i.imgur.com/08FbQs.png[/img] size(140); defaultpen(linewidth(.7)); real alpha=10, x=-0.12, y=0.025, r=1/sqrt(3); path hex=rotate(alpha)*polygon(6); pair A = shift(x,y)*(r*dir(330+alpha)), B = shift(x,y)*(r*dir(90+alpha)), C = shift(x,y)*(r*dir(210+alpha)); pair X = (-A.x, -A.y), Y = (-B.x, -B.y), Z = (-C.x, -C.y); int i; pair[] H; for(i=0; i<6; i=i+1) { H[i] = dir(alpha+60*i);} fill(X--Y--Z--cycle, rgb(204,255,255)); fill(H[5]--Y--Z--H[0]--cycle^^H[2]--H[3]--X--cycle, rgb(203,153,255)); fill(H[1]--Z--X--H[2]--cycle^^H[4]--H[5]--Y--cycle, rgb(255,203,153)); fill(H[3]--X--Y--H[4]--cycle^^H[0]--H[1]--Z--cycle, rgb(153,203,255)); draw(hex^^X--Y--Z--cycle); draw(H[1]--B--H[2]^^H[3]--C--H[4]^^H[5]--A--H[0]^^A--B--C--cycle, linewidth(0.6)+linetype("5 5")); draw(H[0]--Z--H[1]^^H[2]--X--H[3]^^H[4]--Y--H[5]);[/asy]

A flea moves in the positive direction on the real Ox axis, starting from the origin. He can only jump over distances equal with $\sqrt 2$ or $\sqrt{2005}$. Prove that there exists $n_0$ such that the flea can reach any interval $[n,n+1]$ with $n\geq n_0$.

Prove that we can fill in the three dimensional space with regular tetrahedrons and regular octahedrons, all of which have the same edge-lengths. Also find the ratio of the number of the regular tetrahedrons used and the number of the regular octahedrons used.

Points $M$ and $N$ are considered in the interior of triangle $ABC$ such that $\angle MAB = \angle NAC$ and $\angle MBA = \angle NBC$. Prove that $$\frac{AM \cdot AN}{AB \cdot AC}+ \frac{BM\cdot BN}{BA \cdot BC}+ \frac{CM \cdot CN }{CA \cdot CB}=1$$

Let $ABC $ be a triangle and $D $ be the feet of $A $-altitude.\\ $E,F $ are defined on segments $AD,BC $,respectively such that $\frac {AE}{DE}=\frac{BF}{CF} $.\\ Assume that $G $ lies on $AF $ such that $BG\perp AF $.Prove that $EF $ is tangent to the circumcircle of $CFG $. [i]Proposed by Mehdi Etesami Fard[/i]

Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $ 4 : 3$. The horizontal length of a “$ 27$-inch” television screen is closest, in inches, to which of the following? [asy]import math; unitsize(7mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(4,0)--(4,3)--(0,3)--(0,0)--(4,3)); fill((0,0)--(4,0)--(4,3)--cycle,mediumgray); label(rotate(aTan(3.0/4.0))*"Diagonal",(2,1.5),NW); label(rotate(90)*"Height",(4,1.5),E); label("Length",(2,0),S);[/asy]$ \textbf{(A)}\ 20 \qquad \textbf{(B)}\ 20.5 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 21.5 \qquad \textbf{(E)}\ 22$

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.)

After elections, every parliament member (PM), has his own absolute rating. When the parliament set up, he enters in a group and gets a relative rating. The relative rating is the ratio of its own absolute rating to the sum of all absolute ratings of the PMs in the group. A PM can move from one group to another only if in his new group his relative rating is greater. In a given day, only one PM can change the group. Show that only a finite number of group moves is possible. [i](A rating is positive real number.)[/i]

Square ABCD has side length 7. Let $A_1$, $B_1$, $C_1$, and $D_1$ be points on rays $\overrightarrow{AB}$, $\overrightarrow{BC}$, $\overrightarrow{CD}$, and $\overrightarrow{DA}$, respectively, where each point is $3$ units from the end of the ray so that $A_1B_1C_1D_1$ forms a second square as shown. SImilarly, let $A_2$, $B_2$, $C_2$, and $D_2$ be points on segments $A_1B_1$, $B_1C_1$, $C_1D_1$, and $D_1A_1$, respectively, forming another square where $A_2$ divides segment $A_1B_1$ into two pieces whose lengths are in the same ratio as $AA_1$ is to $A_1B$. Continue this process to construct square $A_nB_nC_nD_n$ for each positive integer $n$. Find the total of all the perimeters of all the squares. [asy] size(180); pair[] A={(-1,-1),(-1,1),(1,1),(1,-1),(-1,-1)}; string[] X={"A","B","C","D"}; for(int k=0;k<10;++k) { for(int m=0;m<4;++m) { if(k==0) label("$"+X[m]+"$",A[m],A[m]); if(k==1) label("$"+X[m]+"_1$",A[m],A[m]); draw(A[m]--A[m+1]); A[m]+=3/7*(A[m+1]-A[m]); } A[4]=A[0]; }[/asy]

In triangle $ABC$, $D$ is the midpoint of $AB$; $E$ is the midpoint of $DB$; and $F$ is the midpoint of $BC$. If the area of $\triangle ABC$ is $96$, then the area of $\triangle AEF$ is $\textbf{(A) }16\qquad\textbf{(B) }24\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad \textbf{(E) }48$

The ratio of $w$ to $x$ is $4 : 3$, the ratio of $y$ to $z$ is $3 : 2$, and the ratio of $z$ to $x$ is $1 : 6$. What is the ratio of $w$ to $y$? $\textbf{(A) }4:3 \qquad \textbf{(B) }3:2 \qquad \textbf{(C) } 8:3 \qquad \textbf{(D) } 4:1 \qquad \textbf{(E) } 16:3 $

A point is chosen at random from within a circular region. What is the probability that the point is closer to the center of the region than it is to the boundary of the region? $\text{(A)}\ 1/4 \qquad \text{(B)}\ 1/3 \qquad \text{(C)}\ 1/2 \qquad \text{(D)}\ 2/3 \qquad \text{(E)}\ 3/4$

Each diagonal of a convex pentagon is parallel to one of its sides. Prove that the ratio of the length of each diagonal to the length of the corresponding parallel side is the same, and find this ratio.

Find an upper bound for the ratio $$\frac{x_1x_2+2x_2x_3+x_3x_4}{x_1^2+x_2^2+x_3^2+x_4^2}$$ over all quadruples of real numbers $(x_1,x_2,x_3,x_4)\neq (0,0,0,0)$. [i]Note.[/i] The smaller the bound, the better the solution.

A right circular cone and a right cylinder with same height $20$ does not have same circular base but the circles are coplanar and their centers are same. If the cone and the cylinder are at the same side of the plane and their base radii are $20$ and $10$, respectively, what is the ratio of the volume of the part of the cone inside the cylinder over the volume of the part of the cone outside the cylinder? $ \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ \frac{5}{3} \qquad\textbf{(D)}\ \frac{4}{3} \qquad\textbf{(E)}\ 1 $

Let $ABC$ be an acute triangle with orthocenter $H$, and $P$ a point interior to the triangle. Points $D,E,F$ are the reflections of $P$ about $BC,CA,AB$. If $Q$ is the intersection of $HD$ and $EF$, prove that the ratio $HQ/HD$ is independent of the choice of $P$. Luis González

The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the sides $BC$ at $D$. let $X$ is a point on arc $BC$ from circumcircle of triangle $ABC$ such that if $E,F$ are feet of perpendicular from $X$ on $BI,CI$ and $M$ is midpoint of $EF$ we have $MB=MC$. prove that $\widehat{BAD}=\widehat{CAX}$

Suppose that $ ABCD$ is a convex quadrilateral. Let $ F \equal{} AB\cap CD$, $ E \equal{} AD\cap BC$ and $ T \equal{} AC\cap BD$. Suppose that $ A,B,T,E$ lie on a circle which intersects with $ EF$ at $ P$. Prove that if $ M$ is midpoint of $ AB$, then $ \angle APM \equal{} \angle BPT$.