Found problems: 124
1985 IMO Longlists, 39
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.
1981 IMO Shortlist, 11
On a semicircle with unit radius four consecutive chords $AB,BC, CD,DE$ with lengths $a, b, c, d$, respectively, are given. Prove that
\[a^2 + b^2 + c^2 + d^2 + abc + bcd < 4.\]
2005 Junior Balkan Team Selection Tests - Moldova, 5
Let $ABC$ be an acute-angled triangle, and let $F$ be the foot of its altitude from the vertex $C$. Let $M$ be the midpoint of the segment $CA$. Assume that $CF=BM$. Then the angle $MBC$ is equal to angle $FCA$ if and only if the triangle $ABC$ is equilateral.
2013 India IMO Training Camp, 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 $\tan^2 (\varphi/2) = \tan (C/2)$.
2006 Kyiv Mathematical Festival, 4
See all the problems from 5-th Kyiv math festival
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url]
Let $O$ be the circumcenter and $H$ be the intersection point of the altitudes of acute triangle $ABC.$ The straight lines $BH$ and $CH$ intersect the segments $CO$ and $BO$ at points $D$ and $E$ respectively. Prove that if triangles $ODH$ and $OEH$ are isosceles then triangle $ABC$ is isosceles too.
2005 Postal Coaching, 10
On the sides $AB$ and $BC$ of triangle $ABC$, points $K$ and $M$ are chosen such that the quadrilaterals $AKMC$ and $KBMN$ are cyclic , where $N = AM \cap CK$ . If these quads have the same circumradii, find $\angle ABC$
2001 AIME Problems, 4
In triangle $ABC$, angles $A$ and $B$ measure 60 degrees and 45 degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$, and $AT=24.$ The area of triangle $ABC$ can be written in the form $a+b\sqrt{c},$ where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c.$
1986 China National Olympiad, 2
In $\triangle ABC$, the length of altitude $AD$ is $12$, and the bisector $AE$ of $\angle A$ is $13$. Denote by $m$ the length of median $AF$. Find the range of $m$ when $\angle A$ is acute, orthogonal and obtuse respectively.
2010 Contests, 3
$ABCD$ 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.
2013 Sharygin Geometry Olympiad, 3
Let $ABC$ be a right-angled triangle ($\angle B = 90^\circ$). The excircle inscribed into the angle $A$ touches the extensions of the sides $AB$, $AC$ at points $A_1, A_2$ respectively; points $C_1, C_2$ are defined similarly. Prove that the perpendiculars from $A, B, C$ to $C_1C_2, A_1C_1, A_1A_2$ respectively, concur.
2003 France Team Selection Test, 3
$M$ is an arbitrary point inside $\triangle ABC$. $AM$ intersects the circumcircle of the triangle again at $A_1$. Find the points $M$ that minimise $\frac{MB\cdot MC}{MA_1}$.
2003 AMC 10, 17
The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?
$ \textbf{(A)}\ \frac{3\sqrt2}{\pi} \qquad
\textbf{(B)}\ \frac{3\sqrt3}{\pi} \qquad
\textbf{(C)}\ \sqrt3 \qquad
\textbf{(D)}\ \frac{6}{\pi} \qquad
\textbf{(E)}\ \sqrt3\pi$
2008 IberoAmerican, 2
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.
1993 IMO Shortlist, 4
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
\[\frac{1}{MB}+\frac{1}{MD}= \frac{1}{NC}+\frac{1}{NE}. \]
1996 USAMO, 5
Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles.
1995 AMC 12/AHSME, 18
Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is
$\textbf{(A)}\ 1 \qquad
\textbf{(B)}\ \dfrac{1+\sqrt{3}}{\sqrt{2}} \qquad
\textbf{(C)}\ \sqrt{3} \qquad
\textbf{(D)}\ 2 \qquad
\textbf{(E)}\ \dfrac{4}{\sqrt{3}}$
2000 Brazil Team Selection Test, Problem 1
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]
2014 Dutch IMO TST, 2
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 $|AB|=|CE|$.
2004 China Team Selection Test, 2
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$.
2007 Harvard-MIT Mathematics Tournament, 23
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 $BE=5$.
2010 AMC 10, 14
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}$
2014 NIMO Problems, 4
Points $A$, $B$, $C$, and $D$ lie on a circle such that chords $\overline{AC}$ and $\overline{BD}$ intersect at a point $E$ inside the circle. Suppose that $\angle ADE =\angle CBE = 75^\circ$, $BE=4$, and $DE=8$. The value of $AB^2$ can be written in the form $a+b\sqrt{c}$ for positive integers $a$, $b$, and $c$ such that $c$ is not divisible by the square of any prime. Find $a+b+c$.
[i]Proposed by Tony Kim[/i]
1995 AIME Problems, 9
Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$
[asy] import graph; size(7cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(0.75,1.63),SE*lsf); dot((1,3),ds); label("$A$",(0.96,3.14),NE*lsf); dot((0,0),ds); label("$B$",(-0.15,-0.18),NE*lsf); dot((2,0),ds); label("$C$",(2.06,-0.18),NE*lsf); dot((1,0),ds); label("$M$",(0.97,-0.27),NE*lsf); dot((1,0.7),ds); label("$D$",(1.05,0.77),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]
2012 Tuymaada Olympiad, 2
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]
2011 AMC 10, 18
Rectangle $ABCD$ has $AB=6$ and $BC=3$. Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$. What is the degree measure of $\angle AMD$?
$ \textbf{(A)}\ 15 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 45 \qquad
\textbf{(D)}\ 60 \qquad
\textbf{(E)}\ 75 $