Sides $AB,~ BC,$ and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5,$ and $20$, respectively. If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5} $, then side $AD$ has length $\textbf{(A) }24\qquad\textbf{(B) }24.5\qquad\textbf{(C) }24.6\qquad\textbf{(D) }24.8\qquad\textbf{(E) }25$ [size=70]*A polygon is called “simple” if it is not self intersecting.[/size]

Given a triangle $ABC$, let $D$ and $E$ be points on the side $BC$ such that $\angle BAD = \angle CAE$. If $M$ and $N$ are, respectively, the points of tangency of the incircles of the triangles $ABD$ and $ACE$ with the line $BC$, then show that \[\frac{1}{MB}+\frac{1}{MD}= \frac{1}{NC}+\frac{1}{NE}. \]

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.

Let $A_1, A_2, \ldots, A_n$ be an $n$ -sided regular polygon. If $\frac{1}{A_1 A_2} = \frac{1}{A_1 A_3} + \frac{1}{A_1A_4}$, find $n$.

Quadrilateral $ABCD$ is both cyclic and circumscribed. Its incircle touches its sides $AB$ and $CD$ at points $X$ and $Y$, respectively. The perpendiculars to $AB$ and $CD$ drawn at $A$ and $D$, respectively, meet at point $U$; those drawn at $X$ and $Y$ meet at point $V$, and finally, those drawn at $B$ and $C$ meet at point $W$. Prove that points $U$, $V$ and $W$ are collinear. [i]Proposed by A. Golovanov[/i]

Find all ordered triples $ (x, y, z)$ of real numbers such that \[ 5 \left(x \plus{} \frac{1}{x} \right) \equal{} 12 \left(y \plus{} \frac{1}{y} \right) \equal{} 13 \left(z \plus{} \frac{1}{z} \right),\] and \[ xy \plus{} yz \plus{} zy \equal{} 1.\]

Let $ABC$ be a triangle with $\angle{BAC} = 40^{\circ}$ and $\angle{ABC}=60^{\circ}$. Let $D$ and $E$ be the points lying on the sides $AC$ and $AB$, respectively, such that $\angle{CBD} = 40^{\circ}$ and $\angle{BCE} = 70^{\circ}$. Let $F$ be the point of intersection of the lines $BD$ and $CE$. Show that the line $AF$ is perpendicular to the line $BC$.

Let $ABCD$ be a tetrahedron with $\angle BAD = 60^{\cdot}$, $\angle BAC = 40^{\cdot}$, $\angle ABD = 80^{\cdot}$, $\angle ABC = 70^{\cdot}$. Prove that the lines $AB$ and $CD$ are perpendicular.

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]

Let $ABCD$ be a circumscribed quadrilateral and let $P$ be the orthogonal projection of its in center on $AC$. Prove that $\angle {APB}=\angle {APD}$

Two circles $S_1$ and $S_2$ with centers $O_1$ and $O_2$ respectively intersect at $A$ and $B$. The tangents at $A$ to $S_1$ and $S_2$ meet segments $BO_2$ and $BO_1$ at $K$ and $L$ respectively. Show that $KL \parallel O_1O_2.$

In convex quadrilateral $ ABCD$, $ CB,DA$ are external angle bisectors of $ \angle DCA,\angle CDB$, respectively. Points $ E,F$ lie on the rays $ AC,BD$ respectively such that $ CEFD$ is cyclic quadrilateral. Point $ P$ lie in the plane of quadrilateral $ ABCD$ such that $ DA,CB$ are external angle bisectors of $ \angle PDE,\angle PCF$ respectively. $ AD$ intersects $ BC$ at $ Q.$ Prove that $ P$ lies on $ AB$ if and only if $ Q$ lies on segment $ EF$.

Given a triangle $ ABC$, let $ r$ be the external bisector of $ \angle ABC$. $ P$ and $ Q$ are the feet of the perpendiculars from $ A$ and $ C$ to $ r$. If $ CP \cap BA \equal{} M$ and $ AQ \cap BC\equal{}N$, show that $ MN$, $ r$ and $ AC$ concur.

On a semicircle with unit radius four consecutive chords $AB,BC, CD,DE$ with lengths $a, b, c, d$, respectively, are given. Prove that \[a^2 + b^2 + c^2 + d^2 + abc + bcd < 4.\]

Two congruent $ 30^{\circ}$-$ 60^{\circ}$-$ 90^{\circ}$ are placed so that they overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the area common to both triangles is $ \textbf{(A)}\ 6\sqrt3 \qquad \textbf{(B)}\ 8\sqrt3 \qquad \textbf{(C)}\ 9\sqrt3 \qquad \textbf{(D)}\ 12\sqrt3 \qquad \textbf{(E)}\ 24$

Convex quadrilateral $ ABCD$ is inscribed in a circle, $ \angle{A}\equal{}60^o$, $ BC\equal{}CD\equal{}1$, rays $ AB$ and $ DC$ intersect at point $ E$, rays $ BC$ and $ AD$ intersect each other at point $ F$. It is given that the perimeters of triangle $ BCE$ and triangle $ CDF$ are both integers. Find the perimeter of quadrilateral $ ABCD$.

Convex quadrilateral $ ABCD$ is inscribed in a circle, $ \angle{A}\equal{}60^o$, $ BC\equal{}CD\equal{}1$, rays $ AB$ and $ DC$ intersect at point $ E$, rays $ BC$ and $ AD$ intersect each other at point $ F$. It is given that the perimeters of triangle $ BCE$ and triangle $ CDF$ are both integers. Find the perimeter of quadrilateral $ ABCD$.

Consider a regular heptagon ( polygon of $7$ equal sides and angles) $ABCDEFG$ as in the figure below:- $(a).$ Prove $\frac{1}{\sin\frac{\pi}{7}}=\frac{1}{\sin\frac{2\pi}{7}}+\frac{1}{\sin\frac{3\pi}{7}}$ $(b).$ Using $(a)$ or otherwise, show that $\frac{1}{AG}=\frac{1}{AF}+\frac{1}{AE}$ [asy] draw(dir(360/7)..dir(2*360/7),blue); draw(dir(2*360/7)..dir(3*360/7),blue); draw(dir(3*360/7)..dir(4*360/7),blue); draw(dir(4*360/7)..dir(5*360/7),blue); draw(dir(5*360/7)..dir(6*360/7),blue); draw(dir(6*360/7)..dir(7*360/7),blue); draw(dir(7*360/7)..dir(360/7),blue); draw(dir(2*360/7)..dir(4*360/7),blue); draw(dir(4*360/7)..dir(1*360/7),blue); label("$A$",dir(4*360/7),W); label("$B$",dir(5*360/7),S); label("$C$",dir(6*360/7),S); label("$D$",dir(7*360/7),E); label("$E$",dir(1*360/7),E); label("$F$",dir(2*360/7),N); label("$G$",dir(3*360/7),W); [/asy]

Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length $1$. The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles? $\textbf{(A) }\dfrac{\sqrt3}4\qquad \textbf{(B) }\dfrac{\sqrt3}3\qquad \textbf{(C) }\dfrac23\qquad \textbf{(D) }\dfrac{\sqrt2}2\qquad \textbf{(E) }\dfrac{\sqrt3}2$

In triangle $ABC$, $\angle ABC$ is obtuse. Point $D$ lies on side $AC$ such that $\angle ABD$ is right, and point $E$ lies on side $AC$ between $A$ and $D$ such that $BD$ bisects $\angle EBC$. Find $CE$ given that $AC=35$, $BC=7$, and $BE=5$.

$ABCD$ is a convex quadrilateral with \[\angle BAC = 30^\circ \]\[\angle CAD = 20^\circ\]\[\angle ABD = 50^\circ\]\[\angle DBC = 30^\circ\] If the diagonals intersect at $P$, show that $PC = PD$.

Let $ AB$ be the diameter of semicircle $O$ , $C, D $ be points on the arc $AB$, $P, Q$ be respectively the circumcenter of $\triangle OAC $ and $\triangle OBD $ . Prove that:$CP\cdot CQ=DP \cdot DQ$.[asy] import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black; real h=sqrt(55/64); pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B); D(arc(O,1,0,180),darkgreen); D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue); D(O); [/asy]

In the triangle $ ABC,$ with $ \angle A \equal{} 60 ^{\circ},$ a parallel $ IF$ to $ AC$ is drawn through the incenter $ I$ of the triangle, where $ F$ lies on the side $ AB.$ The point $ P$ on the side $ BC$ is such that $ 3BP \equal{} BC.$ Show that $ \angle BFP \equal{} \frac{\angle B}{2}.$

Triangle $ ABC$ has $ AB \equal{} 2 \cdot AC$. Let $ D$ and $ E$ be on $ \overline{AB}$ and $ \overline{BC}$, respectively, such that $ \angle{BAE} \equal{} \angle{ACD}.$ Let $ F$ be the intersection of segments $ AE$ and $ CD$, and suppose that $ \triangle{CFE}$ is equilateral. What is $ \angle{ACB}$? $ \textbf{(A)}\ 60^{\circ}\qquad \textbf{(B)}\ 75^{\circ}\qquad \textbf{(C)}\ 90^{\circ}\qquad \textbf{(D)}\ 105^{\circ}\qquad \textbf{(E)}\ 120^{\circ}$

The base of an isosceles triangle is $ 6$ inches and one of the equal sides is $ 12$ inches. The radius of the circle through the vertices of the triangle is: $ \textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad\textbf{(B)}\ 4\sqrt{3} \qquad\textbf{(C)}\ 3\sqrt{5} \qquad\textbf{(D)}\ 6\sqrt{3} \qquad\textbf{(E)}\ \text{none of these}$