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: 131

2005 Oral Moscow Geometry Olympiad, 4
Tags: geometry , perpendicular , hexagon , equal segments , right angle
Given a hexagon $ABCDEF$, in which $AB = BC, CD = DE, EF = FA$, and angles $A$ and $C$ are right. Prove that lines $FD$ and $BE$ are perpendicular. (B. Kukushkin)

2000 Junior Balkan Team Selection Tests - Moldova, 7
Tags: geometry , right angle , concurrency
Let a triangle $ABC, A_1$ be the midpoint of the segment $[BC], B_1 \in (AC)$ ¸and $C_1 \in (AB)$ such that $[A_1B_1$ is the bisector of the angle $AA_1C$ and $A_1C_1$ is perpendicular to $AB$. Show that the lines $AA_1, BB_1$ and $CC_1$ are concurrent if and only if $ \angle BAC = 90^o$

2021 Abels Math Contest (Norwegian MO) Final, 4a
Tags: right angle , geometry , 3d geometry , tetrahedron
A tetrahedron $ABCD$ satisfies $\angle BAC=\angle CAD=\angle DAB=90^o$. Show that the areas of its faces satisfy the equation $area(BAC)^2 + area(CAD)^2 + area(DAB)^2 = area(BCD)^2$. .

2021 Dutch IMO TST, 2
Tags: right angle , centroid , right triangle , geometry
Let $ABC $be a right triangle with $\angle C = 90^o$ and let $D$ be the foot of the altitude from $C$. Let $E$ be the centroid of triangle $ACD$ and let $F$ be the centroid of triangle $BCD$. The point $P$ satisfies $\angle CEP = 90^o$ and $|CP| = |AP|$, while point $Q$ satisfies $\angle CFQ = 90^o$ and $|CQ| = |BQ|$. Prove that $PQ$ passes through the centroid of triangle $ABC$.

2021 Bosnia and Herzegovina Junior BMO TST, 3
Tags: geometry , right angle , equal angles
In the convex quadrilateral $ABCD$, $AD = BD$ and $\angle ACD = 3 \angle BAC$. Let $M$ be the midpoint of side $AD$. If the lines $CM$ and $AB$ are parallel, prove that the angle $\angle ACB$ is right.

Denmark (Mohr) - geometry, 2016.3
Tags: geometry , right angle , equal segments , area
Prove that all quadrilaterals $ABCD$ where $\angle B = \angle D = 90^o$, $|AB| = |BC|$ and $|AD| + |DC| = 1$, have the same area. [img]https://1.bp.blogspot.com/-55lHuAKYEtI/XzRzDdRGDPI/AAAAAAAAMUk/n8lYt3fzFaAB410PQI4nMEz7cSSrfHEgQCLcBGAsYHQ/s0/2016%2Bmohr%2Bp3.png[/img]