This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 2023

1988 Romania Team Selection Test, 2

Let $OABC$ be a trihedral angle such that \[ \angle BOC = \alpha, \quad \angle COA = \beta, \quad \angle AOB = \gamma , \quad \alpha + \beta + \gamma = \pi . \] For any interior point $P$ of the trihedral angle let $P_1$, $P_2$ and $P_3$ be the projections of $P$ on the three faces. Prove that $OP \geq PP_1+PP_2+PP_3$. [i]Constantin Cocea[/i]

1998 Turkey MO (2nd round), 2

Variable points $M$ and $N$ are considered on the arms $\left[ OX \right.$ and $\left[ OY \right.$ , respectively, of an angle $XOY$ so that $\left| OM \right|+\left| ON \right|$ is constant. Determine the locus of the midpoint of $\left[ MN \right]$.

2001 India IMO Training Camp, 1

If on $ \triangle ABC$, trinagles $ AEB$ and $ AFC$ are constructed externally such that $ \angle AEB\equal{}2 \alpha$, $ \angle AFB\equal{} 2 \beta$. $ AE\equal{}EB$, $ AF\equal{}FC$. COnstructed externally on $ BC$ is triangle $ BDC$ with $ \angle DBC\equal{} \beta$ , $ \angle BCD\equal{} \alpha$. Prove that 1. $ DA$ is perpendicular to $ EF$. 2. If $ T$ is the projection of $ D$ on $ BC$, then prove that $ \frac{DA}{EF}\equal{} 2 \frac{DT}{BC}$.

2009 Junior Balkan Team Selection Test, 3

Let $ ABCD$ be a convex quadrilateral, such that $ \angle CBD\equal{}2\cdot\angle ADB, \angle ABD\equal{}2\cdot\angle CDB$ and $ AB\equal{}CB$. Prove that quadrilateral $ ABCD$ is a kite.

1996 Baltic Way, 1

Let $\alpha$ be the angle between two lines containing the diagonals of a regular $1996$-gon, and let $\beta\not= 0$ be another such angle. Prove that $\frac{\alpha}{\beta}$ is a rational number.

2003 Tournament Of Towns, 3

Point $M$ is chosen in triangle $ABC$ so that the radii of the circumcircles of triangles $AMC, BMC$, and $BMA$ are no smaller than the radius of the circumcircle of $ABC$. Prove that all four radii are equal.

2014 Contests, 2

Let $ABCD$ be a trapezoid with bases $AB$ and $CD$, inscribed in a circle of center $O$. Let $P$ be the intersection of the lines $BC$ and $AD$. A circle through $O$ and $P$ intersects the segments $BC$ and $AD$ at interior points $F$ and $G$, respectively. Show that $BF=DG$.

2001 Baltic Way, 7

Given a parallelogram $ABCD$. A circle passing through $A$ meets the line segments $AB, AC$ and $AD$ at inner points $M,K,N$, respectively. Prove that \[|AB|\cdot |AM | + |AD|\cdot |AN|=|AK|\cdot |AC|\]

2008 Hungary-Israel Binational, 1

Find the largest value of n, such that there exists a polygon with n sides, 2 adjacent sides of length 1, and all his diagonals have an integer length.

2007 ITAMO, 3

Let ABC be a triangle, G its centroid, M the midpoint of AB, D the point on the line $AG$ such that $AG = GD, A \neq D$, E the point on the line $BG$ such that $BG = GE, B \neq E$. Show that the quadrilateral BDCM is cyclic if and only if $AD = BE$.

2011 All-Russian Olympiad, 3

Let $ABC$ be an equilateral triangle. A point $T$ is chosen on $AC$ and on arcs $AB$ and $BC$ of the circumcircle of $ABC$, $M$ and $N$ are chosen respectively, so that $MT$ is parallel to $BC$ and $NT$ is parallel to $AB$. Segments $AN$ and $MT$ intersect at point $X$, while $CM$ and $NT$ intersect in point $Y$. Prove that the perimeters of the polygons $AXYC$ and $XMBNY$ are the same.

2015 Iran MO (2nd Round), 1

In quadrilateral $ABCD$ , $AC$ is bisector of $\hat{A}$ and $\widehat{ADC}=\widehat{ACB}$. $X$ and $Y$ are feet of perpendicular from $A$ to $BC$ and $CD$,respectively.Prove that orthocenter of triangle $AXY$ is on $BD$.

2000 Turkey MO (2nd round), 1

A circle with center $O$ and a point $A$ in this circle are given. Let $P_{B}$ is the intersection point of $[AB]$ and the internal bisector of $\angle AOB$ where $B$ is a point on the circle such that $B$ doesn't lie on the line $OA$, Find the locus of $P_{B}$ as $B$ varies.

2014 Iran MO (2nd Round), 2

Let $ABCD$ be a square. Let $N,P$ be two points on sides $AB, AD$, respectively such that $NP=NC$, and let $Q$ be a point on $AN$ such that $\angle QPN = \angle NCB$. Prove that \[ \angle BCQ = \dfrac{1}{2} \angle AQP .\]

2007 Iran MO (3rd Round), 1

Let $ ABC$, $ l$ and $ P$ be arbitrary triangle, line and point. $ A',B',C'$ are reflections of $ A,B,C$ in point $ P$. $ A''$ is a point on $ B'C'$ such that $ AA''\parallel l$. $ B'',C''$ are defined similarly. Prove that $ A'',B'',C''$ are collinear.

2012 China Team Selection Test, 1

Given two circles ${\omega _1},{\omega _2}$, $S$ denotes all $\Delta ABC$ satisfies that ${\omega _1}$ is the circumcircle of $\Delta ABC$, ${\omega _2}$ is the $A$- excircle of $\Delta ABC$ , ${\omega _2}$ touches $BC,CA,AB$ at $D,E,F$. $S$ is not empty, prove that the centroid of $\Delta DEF$ is a fixed point.

2014 Turkey MO (2nd round), 3

Let $D, E, F$ be points on the sides $BC, CA, AB$ of a triangle $ABC$, respectively such that the lines $AD, BE, CF$ are concurrent at the point $P$. Let a line $\ell$ through $A$ intersect the rays $[DE$ and $[DF$ at the points $Q$ and $R$, respectively. Let $M$ and $N$ be points on the rays $[DB$ and $[DC$, respectively such that the equation \[ \frac{QN^2}{DN}+\frac{RM^2}{DM}=\frac{(DQ+DR)^2-2\cdot RQ^2+2\cdot DM\cdot DN}{MN} \] holds. Show that the lines $AD$ and $BC$ are perpendicular to each other.

JBMO Geometry Collection, 2009

Let $ ABCDE$ be a convex pentagon such that $ AB\plus{}CD\equal{}BC\plus{}DE$ and $ k$ a circle with center on side $ AE$ that touches the sides $ AB$, $ BC$, $ CD$ and $ DE$ at points $ P$, $ Q$, $ R$ and $ S$ (different from vertices of the pentagon) respectively. Prove that lines $ PS$ and $ AE$ are parallel.

1994 Baltic Way, 12

The inscribed circle of the triangle $A_1A_2A_3$ touches the sides $A_2A_3,A_3A_1,A_1A_2$ at points $S_1,S_2,S_3$, respectively. Let $O_1,O_2,O_3$ be the centres of the inscribed circles of triangles $A_1S_2S_3, A_2S_3S_1,A_3S_1S_2$, respectively. Prove that the straight lines $O_1S_1,O_2S_2,O_3S_3$ intersect at one point.

2001 Romania National Olympiad, 3

We consider the points $A,B,C,D$, not in the same plane, such that $AB\perp CD$ and $AB^2+CD^2=AD^2+BC^2$. a) Prove that $AC\perp BD$. b) Prove that if $CD<BC<BD$, then the angle between the planes $(ABC)$ and $(ADC)$ is greater than $60^{\circ}$.

2010 Middle European Mathematical Olympiad, 3

We are given a cyclic quadrilateral $ABCD$ with a point $E$ on the diagonal $AC$ such that $AD=AE$ and $CB=CE$. Let $M$ be the center of the circumcircle $k$ of the triangle $BDE$. The circle $k$ intersects the line $AC$ in the points $E$ and $F$. Prove that the lines $FM$, $AD$ and $BC$ meet at one point. [i](4th Middle European Mathematical Olympiad, Individual Competition, Problem 3)[/i]

2009 Indonesia TST, 4

Given triangle $ ABC$. Let the tangent lines of the circumcircle of $ AB$ at $ B$ and $ C$ meet at $ A_0$. Define $ B_0$ and $ C_0$ similarly. a) Prove that $ AA_0,BB_0,CC_0$ are concurrent. b) Let $ K$ be the point of concurrency. Prove that $ KG\parallel BC$ if and only if $ 2a^2\equal{}b^2\plus{}c^2$.

1988 India National Olympiad, 7

Given an angle $ \angle QBP$ and a point $ L$ outside the angle $ \angle QBP$. Draw a straight line through $ L$ meeting $ BQ$ in $ A$ and $ BP$ in $ C$ such that the triangle $ \triangle ABC$ has a given perimeter.

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.

2011 Turkey Team Selection Test, 1

Let $D$ be a point different from the vertices on the side $BC$ of a triangle $ABC.$ Let $I, \: I_1$ and $I_2$ be the incenters of the triangles $ABC, \: ABD$ and $ADC,$ respectively. Let $E$ be the second intersection point of the circumcircles of the triangles $AI_1I$ and $ADI_2,$ and $F$ be the second intersection point of the circumcircles of the triangles $AII_2$ and $AI_1D.$ Prove that if $AI_1=AI_2,$ then \[ \frac{EI}{FI} \cdot \frac{ED}{FD}=\frac{{EI_1}^2}{{FI_1}^2}.\]