Given a square cardboard of area $\frac{1}{4}$, and a paper triangle of area $\frac{1}{2}$ such that the square of its sidelength is a positive integer. Prove that the triangle can be folded in some ways such that the squace can be placed inside the folded figure so that both of its faces are completely covered with paper. [i]Proposed by N.Beluhov, Bulgaria[/i]

In circle $\theta_1$ with radius $1$, circles $\phi_1, \phi_2, \dots, \phi_8$, with equal radii, are drawn such that for $1 \le i \le 8$, $\phi_i$ is tangent to $\omega_1$, $\phi_{i-1}$, and $\phi_{i+1}$, where $\phi_0 = \phi_8$ and $\phi_1 = \phi_9$. There exists a circle $\omega_2$ such that $\omega_1 \neq \omega_2$ and $\omega_2$ is tangent to $\phi_i$ for $1 \le i \le 8$. The radius of $\omega_2$ can be expressed in the form $a - b\sqrt{c} -d\sqrt{e - \sqrt{f}} + g \sqrt{h - j \sqrt{k}}$ such that $a, b, \dots, k$ are positive integers and the numbers $e, f, k, \gcd(h, j)$ are squarefree. What is $a+b+c+d+e+f+g+h+j+k$. [i]Proposed by Eugene Chen [/i]

Determine the triangle with sides $a,b,c$ and circumradius $R$ for which $R(b+c) = a\sqrt{bc}$. [i]Romania[/i]

Given circle $O$ with radius $R$, the inscribed triangle $ABC$ is an acute scalene triangle, where $AB$ is the largest side. $AH_A, BH_B,CH_C$ are heights on $BC,CA,AB$. Let $D$ be the symmetric point of $H_A$ with respect to $H_BH_C$, $E$ be the symmetric point of $H_B$ with respect to $H_AH_C$. $P$ is the intersection of $AD,BE$, $H$ is the orthocentre of $\triangle ABC$. Prove: $OP\cdot OH$ is fixed, and find this value in terms of $R$. (Edited)

Square $ ABCD$ has side length $ s$, a circle centered at $ E$ has radius $ r$, and $ r$ and $ s$ are both rational. The circle passes through $ D$, and $ D$ lies on $ \overline{BE}$. Point $ F$ lies on the circle, on the same side of $ \overline{BE}$ as $ A$. Segment $ AF$ is tangent to the circle, and $ AF \equal{} \sqrt {9 \plus{} 5\sqrt {2}}$. What is $ r/s$? [asy]unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B=(0,0), C=(3,0), D=(3,3), A=(0,3); pair Ep=(3+5*sqrt(2)/6,3+5*sqrt(2)/6); pair F=intersectionpoints(Circle(A,sqrt(9+5*sqrt(2))),Circle(Ep,5/3))[0]; pair[] dots={A,B,C,D,Ep,F}; draw(A--F); draw(Circle(Ep,5/3)); draw(A--B--C--D--cycle); dot(dots); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,SW); label("$E$",Ep,E); label("$F$",F,NW);[/asy]$ \textbf{(A) } \frac {1}{2}\qquad \textbf{(B) } \frac {5}{9}\qquad \textbf{(C) } \frac {3}{5}\qquad \textbf{(D) } \frac {5}{3}\qquad \textbf{(E) } \frac {9}{5}$

Given a regular pentagon, find the ratio of its diagonal, $d$, to its side, $a$

$ABCD$ is a parallelogram in which angle $DAB$ is acute. Points $A, P, B, D$ lie on one circle in exactly this order. Lines $AP$ and $CD$ intersect in $Q$. Point $O$ is the circumcenter of the triangle $CPQ$. Prove that if $D \neq O$ then the lines $AD$ and $DO$ are perpendicular.

The triangle $ABC$ has $CA = CB$. $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $AB$ to $C$). $D$ is the foot of the perpendicular from $C$ to $PB$. Show that $PA + PB = 2 \cdot PD$.

Consider a triangle $ABC$ and $I$ its incenter. The line $(AI)$ meets the circumcircle of $ABC$ in $D$. Let $E$ and $F$ be the orthogonal projections of $I$ on $(BD)$ and $(CD)$ respectively. Assume that $IE+IF=\frac{1}{2}AD$. Calculate $\angle{BAC}$. [color=red][Moderator edited: Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=5088 .][/color]

A circle passes through the three vertices of an isosceles triangle that has two sides of length $ 3$ and a base of length $ 2$. What is the area of this circle? $ \textbf{(A)}\ 2\pi\qquad \textbf{(B)}\ \frac {5}{2}\pi\qquad \textbf{(C)}\ \frac {81}{32}\pi\qquad \textbf{(D)}\ 3\pi\qquad \textbf{(E)}\ \frac {7}{2}\pi$

In triangle $ABC$ let $A'$, $B'$, $C'$ respectively be the midpoints of the sides $BC$, $CA$, $AB$. Furthermore let $L$, $M$, $N$ be the projections of the orthocenter on the three sides $BC$, $CA$, $AB$, and let $k$ denote the nine-point circle. The lines $AA'$, $BB'$, $CC'$ intersect $k$ in the points $D$, $E$, $F$. The tangent lines on $k$ in $D$, $E$, $F$ intersect the lines $MN$, $LN$ and $LM$ in the points $P$, $Q$, $R$. Prove that $P$, $Q$ and $R$ are collinear.

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]

In quadrilateral $ABCD$, we have $AB = 5$, $BC = 6$, $CD = 5$, $DA = 4$, and $\angle ABC = 90^\circ$. Let $AC$ and $BD$ meet at $E$. Compute $\dfrac{BE}{ED}$.

Let $ABC$ be a triangle with $AB=13$, $BC=14$, and $AC=15$. Let $M$ be the midpoint of $BC$ and let $\Gamma$ be the circle passing through $A$ and tangent to line $BC$ at $M$. Let $\Gamma$ intersect lines $AB$ and $AC$ at points $D$ and $E$, respectively, and let $N$ be the midpoint of $DE$. Suppose line $MN$ intersects lines $AB$ and $AC$ at points $P$ and $O$, respectively. If the ratio $MN:NO:OP$ can be written in the form $a:b:c$ with $a,b,c$ positive integers satisfying $\gcd(a,b,c)=1$, find $a+b+c$. [i]James Tao[/i]