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: 25757

2008 Bundeswettbewerb Mathematik, 3

Prove: In an acute triangle $ ABC$ angle bisector $ w_{\alpha},$ median $ s_b$ and the altitude $ h_c$ intersect in one point if $ w_{\alpha},$ side $ BC$ and the circle around foot of the altitude $ h_c$ have vertex $ A$ as a common point.

2019 Caucasus Mathematical Olympiad, 2

In a triangle $ABC$ let $I$ be the incenter. Prove that the circle passing through $A$ and touching $BI$ at $I$, and the circle passing through $B$ and touching $AI$ at $I$, intersect at a point on the circumcircle of $ABC$.

2010 Mexico National Olympiad, 2

Let $ABC$ be an acute triangle with $AB\neq AC$, $M$ be the median of $BC$, and $H$ be the orthocenter of $\triangle ABC$. The circumcircle of $B$, $H$, and $C$ intersects the median $AM$ at $N$. Show that $\angle ANH=90^\circ$.

2003 Junior Tuymaada Olympiad, 3

In the acute triangle $ ABC $, the point $ I $ is the center of the inscribed the circle, the point $ O $ is the center of the circumscribed circle and the point $ I_a $ is the center the excircle tangent to the side $ BC $ and the extensions of the sides $ AB $ and $ AC $. Point $ A'$ is symmetric to vertex $ A $ with respect to the line $ BC $. Prove that $ \angle IOI_a = \angle IA'I_a $.

2019 Junior Balkan Team Selection Tests - Romania, 3

Let $ABC$ a triangle, $I$ the incenter, $D$ the contact point of the incircle with the side $BC$ and $E$ the foot of the bisector of the angle $A$. If $M$ is the midpoint of the arc $BC$ which contains the point $A$ of the circumcircle of the triangle $ABC$ and $\{F\} = DI \cap AM$, prove that $MI$ passes through the midpoint of $[EF]$.

2006 All-Russian Olympiad Regional Round, 10.8

A convex polyhedron has $2n$ faces ($n\ge 3$), and all faces are triangles. What is the largest number of vertices at which converges exactly $3$ edges at a such a polyhedron ?

2007 Oral Moscow Geometry Olympiad, 5

At the base of the quadrangular pyramid $SABCD$ lies the quadrangle $ABCD$. whose diagonals are perpendicular and intersect at point $P$, and $SP$ is the altitude of the pyramid. Prove that the projections of the point $P$ onto the lateral faces of the pyramid lie on the same circle. (A. Zaslavsky)

1991 USAMO, 1

In triangle $\, ABC, \,$ angle $\,A\,$ is twice angle $\,B,\,$ angle $\,C\,$ is obtuse, and the three side lengths $\,a,b,c\,$ are integers. Determine, with proof, the minimum possible perimeter.

2021 Kosovo National Mathematical Olympiad, 3

Let $ABC$ be a triangle and let $O$ be the centre of its circumscribed circle. Points $X, Y$ which are neither of the points $A, B$ or $C$, lie on the circumscribed circle and are so that the angles $XOY$ and $BAC$ are equal (with the same orientation). Show that the orthocentre of the triangle that is formed by the lines $BY, CX$ and $XY$ is a fixed point.

2020 Ukrainian Geometry Olympiad - April, 4

Inside triangle $ABC$, the point $P$ is chosen such that $\angle PAB = \angle PCB =\frac14 (\angle A+ \angle C)$. Let $BL$ be the bisector of $\vartriangle ABC$. Line $PL$ intersects the circumcircle of $\vartriangle APC$ at point $Q$. Prove that the line $QB$ is the bisector of $\angle AQC$.

2007 South East Mathematical Olympiad, 2

In right-angle triangle $ABC$, $\angle C=90$°, Point $D$ is the midpoint of side $AB$. Points $M$ and $C$ lie on the same side of $AB$ such that $MB\bot AB$, line $MD$ intersects side $AC$ at $N$, line $MC$ intersects side $AB$ at $E$. Show that $\angle DBN=\angle BCE$.

2007 Nicolae Păun, 2

Tags: geometry
The bisector of $ \angle BAC $ of a triangle $ ABC $ meet the segment $ BC $ at $ D. $ Through the midpoint of $ AD $ passes aline that intersects $ AB,AC $ at $ M,N, $ respectively. Show that: $$ \frac{1}{MA}+\frac{1}{NA} =2\left( \frac{1}{AB} +\frac{1}{AC} \right) $$ [i]Toni Mihalcea[/i]

2021 Polish Junior MO First Round, 2

A triangle $ABC$ is given with $AC = BC = 5$. The altitude of this triangle drawn from vertex $A$ has length $4$. Calculate the length of the altitude of $ABC$ drawn from vertex $C$.

2019 IFYM, Sozopol, 2

Tags: geometry
In $\Delta ABC$ with $\angle ACB=135^\circ$, are chosen points $M$ and $N$ on side $AB$, so that $\angle MCN=90^\circ$. Segments $MD$ and $NQ$ are angle bisectors of $\Delta AMC$ and $\Delta NBC$ respectively. Prove that the reflection of $C$ in line $PQ$ lies on the line $AB$.

2008 CentroAmerican, 6

Let $ ABC$ be an acute triangle. Take points $ P$ and $ Q$ inside $ AB$ and $ AC$, respectively, such that $ BPQC$ is cyclic. The circumcircle of $ ABQ$ intersects $ BC$ again in $ S$ and the circumcircle of $ APC$ intersects $ BC$ again in $ R$, $ PR$ and $ QS$ intersect again in $ L$. Prove that the intersection of $ AL$ and $ BC$ does not depend on the selection of $ P$ and $ Q$.

1969 IMO Longlists, 27

$(GBR 4)$ The segment $AB$ perpendicularly bisects $CD$ at $X$. Show that, subject to restrictions, there is a right circular cone whose axis passes through $X$ and on whose surface lie the points $A,B,C,D.$ What are the restrictions?

2013 Finnish National High School Mathematics Competition, 3

The points $A,B,$ and $C$ lies on the circumference of the unit circle. Furthermore, it is known that $AB$ is a diameter of the circle and \[\frac{|AC|}{|CB|}=\frac{3}{4}.\] The bisector of $ABC$ intersects the circumference at the point $D$. Determine the length of the $AD$.

2008 JBMO Shortlist, 5

Tags: geometry
Is it possible to cover a given square with a few congruent right-angled triangles with acute angle equal to ${{30}^{o}}$? (The triangles may not overlap and may not exceed the margins of the square.)

2022 Princeton University Math Competition, A3 / B5

Tags: geometry
Daeun draws a unit circle centered at the origin and inscribes within it a regular hexagon $ABCDEF$. Then Dylan chooses a point $P$ within the circle of radius $2$ centered at the origin. Let $M$ be the maximum possible value of $|PA| \cdot |PB| \cdot |PC| \cdot |PD| \cdot |PE| \cdot |PF|$, and let $N$ be the number of possible points $P$ for which this maximal value is obtained. Find $M + N^2$.

2023 VN Math Olympiad For High School Students, Problem 9

Tags: geometry
Given a quadrilateral $ABCD$ inscribed in $(O)$. Let $L, J$ be the [i]Lemoine[/i] point of $\triangle ABC$ and $\triangle ACD$. Prove that: $AC, BD, LJ$ are concurrent.

2018 Romanian Master of Mathematics, 6

Tags: geometry
Fix a circle $\Gamma$, a line $\ell$ to tangent $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ meet $\ell$ at $Y$ and $Z$. Prove that, as $X$ varies over $\Omega$, the circumcircle of $XYZ$ is tangent to two fixed circles.

1993 IMO Shortlist, 1

Let $ABC$ be a triangle, and $I$ its incenter. Consider a circle which lies inside the circumcircle of triangle $ABC$ and touches it, and which also touches the sides $CA$ and $BC$ of triangle $ABC$ at the points $D$ and $E$, respectively. Show that the point $I$ is the midpoint of the segment $DE$.

1981 Austrian-Polish Competition, 3

Given is a triangle $ABC$, the inscribed circle $G$ of which has radius $r$. Let $r_a$ be the radius of the circle touching $AB$, $AC$ and $G$. [This circle lies inside triangle $ABC$.] Define $r_b$ and $r_c$ similarly. Prove that $r_a + r_b + r_c \geq r$ and find all cases in which equality occurs. [i]Bosnia - Herzegovina Mathematical Olympiad 2002[/i]

2001 Baltic Way, 6

The points $A, B, C, D, E$ lie on the circle $c$ in this order and satisfy $AB\parallel EC$ and $AC\parallel ED$. The line tangent to the circle $c$ at $E$ meets the line $AB$ at $P$. The lines $BD$ and $EC$ meet at $Q$. Prove that $|AC|=|PQ|$.

2019 Estonia Team Selection Test, 7

An acute-angled triangle $ABC$ has two altitudes $BE$ and $CF$. The circle with diameter $AC$ intersects the segment $BE$ at point $P$. A circle with diameter $AB$ intersects the segment $CF$ at point $Q$ and the extension of this altitude at point $Q'$. Prove that $\angle PQ'Q = \angle PQB$.