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 $ABCD$ a convex quadrilateral with $AB=BC=CD$, with $AC$ not equal to $BD$ and $E$ be the intersection point of it's diagonals. Prove that $AE=DE$ if and only if $\angle BAD+\angle ADC = 120$.

In $\triangle PQR$, $PQ=8$, $QR=13$, and $RP=15$. Prove that there is a point $S$ on line segment $\overline{PR}$, but not at its endpoints, such that $PS$ and $QS$ are also integers. [asy] size(200); defaultpen(linewidth(0.8)); pair P=origin,Q=(8,0),R=(7,10),S=(3/2,15/7); draw(P--Q--R--cycle); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,NE); draw(Q--S,linetype("4 4")); label("$S$",S,NW); [/asy]

In a triangle $ABC$, let $D$ be the point on the segment $BC$ such that $AB+BD=AC+CD$. Suppose that the points $B$, $C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB=AC$.

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}$

Points $A$ and $B$ lie on circle $\omega$. Point $P$ lies on the extension of segment $AB$ past $B$. Line $\ell$ passes through $P$ and is tangent to $\omega$. The tangents to $\omega$ at points $A$ and $B$ intersect $\ell$ at points $D$ and $C$ respectively. Given that $AB=7$, $BC=2$, and $AD=3$, compute $BP$.

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]

Given a triangle with $120^\circ$. Let $x,\ y,\ z$ be the side lengths of the triangle such that $x<y<z$. (1) Find all triplets $(x,\ y,\ z)$ of positive integers $x,\ y,\ z$ such that $x+y-z=2$. (2) Find all triplets $(x,\ y,\ z)$ of positive integers $x,\ y,\ z$ such that $x+y-z=3$. (3) Let $a,\ b$ be non-negative integers. Express the number of $(x,\ y,\ z)$ such that $x+y-z=2^a3^b$ in terms of $a,\ b$. 2012 Hitotsubashi University entrance exam, problem 1

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 $\triangle{CDE}$? $\textbf{(A) }\frac{5\sqrt{2}}{3}\qquad\textbf{(B) }\frac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\frac{15\sqrt{3}}{8}\qquad\textbf{(D) }\frac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\frac{25}{6}$

A hexagon inscribed in a circle has three consecutive sides each of length $3$ and three consecutive sides each of length $5$. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length $3$ and the other with three sides each of length $5$, has length equal to $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. $\text{(A)}\ 309 \qquad \text{(B)}\ 349 \qquad \text{(C)}\ 369 \qquad \text{(D)}\ 389\qquad \text{(E)}\ 409$

Let $ABC$ be equilateral, and $D, E,$ and $F$ be the midpoints of $\overline{BC}, \overline{CA},$ and $\overline{AB},$ respectively. There exist points $P, Q,$ and $R$ on $\overline{DE}, \overline{EF},$ and $\overline{FD},$ respectively, with the property that $P$ is on $\overline{CQ}, Q$ is on $\overline{AR},$ and $R$ is on $\overline{BP}.$ The ratio of the area of triangle $ABC$ to the area of triangle $PQR$ is $a+b\sqrt{c},$ where $a, b$ and $c$ are integers, and $c$ is not divisible by the square of any prime. What is $a^{2}+b^{2}+c^{2}$?

Circles $ C_1,C_2$ intersect at $ P$. A line $ \Delta$ is drawn arbitrarily from $ P$ and intersects with $ C_1,C_2$ at $ B,C$. What is locus of $ A$ such that the median of $ AM$ of triangle $ ABC$ has fixed length $ k$.

In the adjoining figure of a rectangular solid, $\angle DHG=45^\circ$ and $\angle FHB=60^\circ$. Find the cosine of $\angle BHD$. [asy] size(200); import three;defaultpen(linewidth(0.7)+fontsize(10)); currentprojection=orthographic(1/3+1/10,1-1/10,1/3); real r=sqrt(3); triple A=(0,0,r), B=(0,r,r), C=(1,r,r), D=(1,0,r), E=O, F=(0,r,0), G=(1,0,0), H=(1,r,0); draw(D--G--H--D--A--B--C--D--B--F--H--B^^C--H); draw(A--E^^G--E^^F--E, linetype("4 4")); label("$A$", A, N); label("$B$", B, dir(0)); label("$C$", C, N); label("$D$", D, W); label("$E$", E, NW); label("$F$", F, S); label("$G$", G, W); label("$H$", H, S); triple H45=(1,r-0.15,0.1), H60=(1-0.05, r, 0.07); label("$45^\circ$", H45, dir(125), fontsize(8)); label("$60^\circ$", H60, dir(25), fontsize(8));[/asy] $\textbf {(A) } \frac{\sqrt{3}}{6} \qquad \textbf {(B) } \frac{\sqrt{2}}{6} \qquad \textbf {(C) } \frac{\sqrt{6}}{3} \qquad \textbf {(D) } \frac{\sqrt{6}}{4} \qquad \textbf {(E) } \frac{\sqrt{6}-\sqrt{2}}{4}$

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]

Eight spheres of radius 100 are placed on a flat surface so that each sphere is tangent to two others and their centers are the vertices of a regular octagon. A ninth sphere is placed on the flat surface so that it is tangent to each of the other eight spheres. The radius of this last sphere is $a+b\sqrt{c},$ where $a, b,$ and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c.$

In $\triangle BAC$, $\angle BAC=40^\circ$, $AB=10$, and $AC=6$. Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$? $\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sqrt 3+9\qquad$

In quadrilateral $ABCD$, $BC=8$, $CD=12$, $AD=10$, and $m\angle A= m\angle B = 60^\circ$. Given that $AB=p + \sqrt{q}$, where $p$ and $q$ are positive integers, find $p+q$.

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}$