In a triangle $ABC$ with $B = 90^\circ$, $D$ is a point on the segment $BC$ such that the inradii of triangles $ABD$ and $ADC$ are equal. If $\widehat{ADB} = \varphi$ then prove that $\tan^2 (\varphi/2) = \tan (C/2)$.

In a triangle $ABC$ with $B = 90^\circ$, $D$ is a point on the segment $BC$ such that the inradii of triangles $ABD$ and $ADC$ are equal. If $\widehat{ADB} = \varphi$ then prove that $\tan^2 (\varphi/2) = \tan (C/2)$.

Let $ ABC$ be a triangle with incenter $ I$ and let $ \Gamma$ be a circle centered at $ I$, whose radius is greater than the inradius and does not pass through any vertex. Let $ X_{1}$ be the intersection point of $ \Gamma$ and line $ AB$, closer to $ B$; $ X_{2}$, $ X_{3}$ the points of intersection of $ \Gamma$ and line $ BC$, with $ X_{2}$ closer to $ B$; and let $ X_{4}$ be the point of intersection of $ \Gamma$ with line $ CA$ closer to $ C$. Let $ K$ be the intersection point of lines $ X_{1}X_{2}$ and $ X_{3}X_{4}$. Prove that $ AK$ bisects segment $ X_{2}X_{3}$.

In the figure, $\triangle ABC$ has $\angle A =45^{\circ}$ and $\angle B =30^{\circ}$. A line $DE$, with $D$ on $AB$ and $\angle ADE =60^{\circ}$, divides $\triangle ABC$ into two pieces of equal area. (Note: the figure may not be accurate; perhaps $E$ is on $CB$ instead of $AC$.) The ratio $\frac{AD}{AB}$ is [asy] size((220)); draw((0,0)--(20,0)--(7,6)--cycle); draw((6,6)--(10,-1)); label("A", (0,0), W); label("B", (20,0), E); label("C", (7,6), NE); label("D", (9.5,-1), W); label("E", (5.9, 6.1), SW); label("$45^{\circ}$", (2.5,.5)); label("$60^{\circ}$", (7.8,.5)); label("$30^{\circ}$", (16.5,.5)); [/asy] $ \textbf{(A)}\ \frac{1}{\sqrt{2}} \qquad\textbf{(B)}\ \frac{2}{2+\sqrt{2}} \qquad\textbf{(C)}\ \frac{1}{\sqrt{3}} \qquad\textbf{(D)}\ \frac{1}{\sqrt[3]{6}} \qquad\textbf{(E)}\ \frac{1}{\sqrt[4]{12}} $

Points $D$ and $E$ divides the base $BC$ of an isosceles triangle $ABC$ into three equal parts and $D$ is between $B$ and $E.$ Show that $\angle BAD<\angle DAE.$

Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP>CP$. Let $O_1$ and $O_2$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB=12$ and $\angle O_1 P O_2 = 120^\circ$, then $AP=\sqrt{a}+\sqrt{b}$ where $a$ and $b$ are positive integers. Find $a+b$.

Let $ABC$ and $A_{1}B_{1}C_{1}$ be two triangles. Prove that $\frac{a}{a_{1}}+\frac{b}{b_{1}}+\frac{c}{c_{1}}\leq\frac{3R}{2r_{1}}$, where $a = BC$, $b = CA$, $c = AB$ are the sidelengths of triangle $ABC$, where $a_{1}=B_{1}C_{1}$, $b_{1}=C_{1}A_{1}$, $c_{1}=A_{1}B_{1}$ are the sidelengths of triangle $A_{1}B_{1}C_{1}$, where $R$ is the circumradius of triangle $ABC$ and $r_{1}$ is the inradius of triangle $A_{1}B_{1}C_{1}$.

Two equilateral triangles are contained in a square whose side length is $2\sqrt3$. The bases of these triangles are the opposite sides of the square, and their intersection is a rhombus. What is the area of the rhombus? $ \textbf{(A)}\ \frac{3}{2}\qquad\textbf{(B)}\ \sqrt3\qquad\textbf{(C)}\ 2\sqrt2-1\qquad\textbf{(D)}\ 8\sqrt3-12\qquad\textbf{(E)}\ \frac{4\sqrt3}{3}$

The isosceles right triangle $ABC$ has right angle at $C$ and area $12.5$. The rays trisecting $\angle{ACB}$ intersect $AB$ at $D$ and $E$. What is the area of $\triangle{CDE}$? $\textbf{(A) }\frac{5\sqrt{2}}{3}\qquad\textbf{(B) }\frac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\frac{15\sqrt{3}}{8}\qquad\textbf{(D) }\frac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\frac{25}{6}$

Let $ABC$ be a triangle such that $|AC|=1$ and $|AB|=\sqrt 2$. Let $M$ be a point such that $|MA|=|AB|$, $m(\widehat{MAB}) = 90^\circ$, and $C$ and $M$ are on the opposite sides of $AB$. Let $N$ be a point such that $|NA|=|AX|$, $m(\widehat{NAC}) = 90^\circ$, and $B$ and $N$ are on the opposite sides of $AC$. If the line passing throung $A$ and the circumcenter of triangle $MAN$ meets $[BC]$ at $F$, what is $\dfrac {|BF|}{|FC|}$? $ \textbf{(A)}\ 2\sqrt 2 \qquad\textbf{(B)}\ 2\sqrt 3 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 3\sqrt 2 $

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]

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$.

Consider a triangle $ABC$ with incenter $I$. Let $D$ be the intersection of $BI,AC$ and $CI$ intersects the circumcircle of $ABC$ at $M$. Point $K$ lies on the line $MD$ and $\angle KIA=90^\circ$. Let $F$ be the reflection of $B$ about $C$. Prove that $BIKF$ is cyclic.

In a triangle $ABC,$ incircle touches the sides $BC, CA, AB$ at $D, E, F,$ respectively. A circle $\omega$ passing through $A$ and tangent to line $BC$ at $D$ intersects the line segments $BF$ and $CE$ at $K$ and $L,$ respectively. The line passing through $E$ and parallel to $DL$ intersects the line passing through $F$ and parallel to $DK$ at $P.$ If $R_1, R_2, R_3, R_4$ denotes the circumradius of the triangles $AFD, AED, FPD, EPD,$ respectively, prove that $R_1R_4=R_2R_3.$

Let $\triangle ABC$ be a triangle. Let $M$ be the midpoint of $BC$ and let $D$ be a point on the interior of side $AB$. The intersection of $AM$ and $CD$ is called $E$. Suppose that $|AD|=|DE|$. Prove that $|AB|=|CE|$.

Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ$. Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ$. Find the number of degrees in $\angle CMB$.

Find the area of rhombus $ABCD$ given that the radii of the circles circumscribed around triangles $ABD$ and $ACD$ are $12.5$ and $25$, respectively.

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

In a triangle $ ABC,$ let $ D$ and $ E$ be the intersections of the bisectors of $ \angle ABC$ and $ \angle ACB$ with the sides $ AC,AB,$ respectively. Determine the angles $ \angle A,\angle B, \angle C$ if $ \angle BDE \equal{} 24 ^{\circ},$ $ \angle CED \equal{} 18 ^{\circ}.$

Let $ABC$ be a triangle with $AB = AC$. The incircle touches $BC$, $AC$ and $AB$ at $D$, $E$ and $F$ respectively. Let $P$ be a point on the arc $\overarc{EF}$ that does not contain $D$. Let $Q$ be the second point of intersection of $BP$ and the incircle of $ABC$. The lines $EP$ and $EQ$ meet the line $BC$ at $M$ and $N$, respectively. Prove that the four points $P, F, B, M$ lie on a circle and $\frac{EM}{EN} = \frac{BF}{BP}$.

In the figure below, $\triangle ABC$ is an equilateral triangle. [asy] import graph; unitsize(60); axes("$x$", "$y$", (0, 0), (1.5, 1.5), EndArrow); real w = sqrt(3) - 1; pair A = (1, 1); pair B = (0, w); pair C = (w, 0); draw(A -- B -- C -- cycle); dot(Label("$A(1, 1)$", A, NE), A); dot(Label("$B$", B, W), B); dot(Label("$C$", C, S), C); [/asy] Point $A$ has coordinates $(1, 1)$, point $B$ is on the positive $y$-axis, and point $C$ is on the positive $x$-axis. What is the area of $\triangle ABC$?

Let $ ABC$ be a triangle with $ \angle A \equal{} 45^\circ$. Let $ P$ be a point on side $ BC$ with $ PB \equal{} 3$ and $ PC \equal{} 5$. Let $ O$ be the circumcenter of $ ABC$. Determine the length $ OP$.

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths $3$, $4$, and $5$. What is the area of the triangle? $\textbf{(A)}\ 6 \qquad \textbf{(B)}\ \frac{18}{\pi^2} \qquad \textbf{(C)}\ \frac{9}{\pi^2}\left(\sqrt{3}-1\right) \qquad \textbf{(D)}\ \frac{9}{\pi^2}\left(\sqrt{3}+1\right) \qquad \textbf{(E)}\ \frac{9}{\pi^2}\left(\sqrt{3}+3\right)$

Given a triangle $ABC$ and external points $X, Y$ , and $Z$ such that $\angle BAZ = \angle CAY , \angle CBX = \angle ABZ$, and $\angle ACY = \angle BCX$, prove that $AX,BY$ , and $CZ$ are concurrent.

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