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: 333

2020 Yasinsky Geometry Olympiad, 1
Tags: square , area , right triangle , geometry
The square $ABCD$ is divided into $8$ equal right triangles and the square $KLMN$, as shown in the figure. Find the area of the square $ABCD$ if $KL = 5, PS = 8$. [img]https://1.bp.blogspot.com/-B2QIHvPcIx0/X4BhUTMDhSI/AAAAAAAAMj4/4h0_q1P6drskc5zSvtfTZUskarJjRp5LgCLcBGAsYHQ/s0/Yasinsky%2B2020%2Bp1.png[/img]

1949-56 Chisinau City MO, 32
Tags: midpoint , geometry , locus , right triangle
Determine the locus of points that are the midpoints of segments of equal length, the ends of which lie on the sides of a given right angle.

2008 Hanoi Open Mathematics Competitions, 9
Tags: geometry , right triangle
Consider a right -angle triangle $ABC$ with $A=90^{o}$, $AB=c$ and $AC=b$. Let $P\in AC$ and $Q\in AB$ such that $\angle APQ=\angle ABC$ and $\angle AQP = \angle ACB$. Calculate $PQ+PE+QF$, where $E$ and $F$ are the projections of $B$ and $Q$ onto $BC$, respectively.

1996 Romania National Olympiad, 4
Tags: geometry , right triangle
In the right triangle $ABC$ ($m ( \angle A) = 90^o$) $D$ is the foot of the altitude from $A$. The bisectors of the angles $ABD$ and $ADB$ intersect in $I_1$ and the bisectors of the angles $ACD$ and $ADC$ in $I_2$. Find the angles of the triangle if the sum of distances from $I_1$ and $I_2$ to $AD$ is equal to $\frac14$ of the length of $BC$.

2012 IMO Shortlist, G5
Tags: geometry , circumcircle , right triangle
Let $ABC$ be a triangle with $\angle BCA=90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK=BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL=AC$. Let $M$ be the point of intersection of $AL$ and $BK$. Show that $MK=ML$. [i]Proposed by Josef Tkadlec, Czech Republic[/i]

2010 All-Russian Olympiad Regional Round, 11.1
Tags: geometry , right triangle
Each leg of a right triangle is increased by one. Could its hypotenuse increase by more than $\sqrt2$?

2004 District Olympiad, 4
Tags: geometry , equal angles , isosceles , right triangle , angle
Consider the isosceles right triangle $ABC$ ($AB = AC$) and the points $M, P \in [AB]$ so that $AM = BP$. Let $D$ be the midpoint of the side $BC$ and $R, Q$ the intersections of the perpendicular from $A$ on$ CM$ with $CM$ and $BC$ respectively. Prove that a) $\angle AQC = \angle PQB$ b) $\angle DRQ = 45^o$

1950 Polish MO Finals, 5
Tags: trigonometry , geometry , right triangle , right angle
Prove that if for angles $A,B,C$ of a triangle holds $$\sin^2 A+\sin^2 B +\sin^2 C=2$$ iff the triangle $ABC$ is right.

Estonia Open Senior - geometry, 2002.1.2
Tags: geometry , right triangle , arithmetic progression , inradius
The sidelengths of a triangle and the diameter of its incircle, taken in some order, form an arithmetic progression. Prove that the triangle is right-angled.

2015 Costa Rica - Final Round, G4
Tags: geometry , incenter , right triangle
Consider $\vartriangle ABC$, right at $B$, let $I$ be its incenter and $F,D,E$ the points where the circle inscribed on sides AB, $BC$ and $AC$, respectively. If $M$ is the intersection point of $CI$ and $EF$, and $N$ is the intersection point of $DM$ and $AB$. Prove that $AN = ID$.

2003 District Olympiad, 2
Tags: equal segments , parallel , perpendicular , right triangle , geometry
In the right triangle $ABC$ ( $\angle A = 90^o$), $D$ is the intersection of the bisector of the angle $A$ with the side $(BC)$, and $P$ and $Q$ are the projections of the point $D$ on the sides $(AB),(AC)$ respectively . If $BQ \cap DP=\{M\}$, $CP \cap DQ=\{N\}$, $BQ\cap CP=\{H\}$, show that: a) $PM = DN$ b) $MN \parallel BC$ c) $AH \perp BC$.

2021 Estonia Team Selection Test, 2
Tags: geometry , right triangle , isosceles triangle
Let $ABC$ be an isosceles triangle with $BC=CA$, and let $D$ be a point inside side $AB$ such that $AD< DB$. Let $P$ and $Q$ be two points inside sides $BC$ and $CA$, respectively, such that $\angle DPB = \angle DQA = 90^{\circ}$. Let the perpendicular bisector of $PQ$ meet line segment $CQ$ at $E$, and let the circumcircles of triangles $ABC$ and $CPQ$ meet again at point $F$, different from $C$. Suppose that $P$, $E$, $F$ are collinear. Prove that $\angle ACB = 90^{\circ}$.

Estonia Open Junior - geometry, 2008.2.2
Tags: equal angles , geometry , equal segments , right triangle
In a right triangle $ABC$, $K$ is the midpoint of the hypotenuse $AB$ and $M$ such a point on the $BC$ that $| B M | = 2 | MC |$. Prove that $\angle MAB = \angle MKC$.

2014 Federal Competition For Advanced Students, 4
Tags: geometry , angle bisector , right triangle
We are given a right-angled triangle $MNP$ with right angle in $P$. Let $k_M$ be the circle with center $M$ and radius $MP$, and let $k_N$ be the circle with center $N$ and radius $NP$. Let $A$ and $B$ be the common points of $k_M$ and the line $MN$, and let $C$ and $D$ be the common points of $k_N$ and the line $MN$ with with $C$ between $A$ and $B$. Prove that the line $PC$ bisects the angle $\angle APB$.

2019 Singapore Junior Math Olympiad, 1
Tags: right triangle , geometry , isosceles triangle , isosceles
In the triangle $ABC, AC=BC, \angle C=90^o, D$ is the midpoint of $BC, E$ is the point on $AB$ such that $AD$ is perpendicular to $CE$. Prove that $AE=2EB$.

1993 Mexico National Olympiad, 1
Tags: right triangle , geometry , area of a triangle
$ABC$ is a triangle with $\angle A = 90^o$. Take $E$ such that the triangle $AEC$ is outside $ABC$ and $AE = CE$ and $\angle AEC = 90^o$. Similarly, take $D$ so that $ADB$ is outside $ABC$ and similar to $AEC$. $O$ is the midpoint of $BC$. Let the lines $OD$ and $EC$ meet at $D'$, and the lines $OE$ and $BD$ meet at $E'$. Find area $DED'E'$ in terms of the sides of $ABC$.

2012 Romania National Olympiad, 2
Tags: geometry , equal angles , right triangle
Let $ABC$ be a triangle with right $\angle A$. Consider points $D \in (AC)$ and $E \in (BD)$ such that $\angle ABC = \angle ECD = \angle CED$. Prove that $BE = 2 \cdot AD$

1992 Tournament Of Towns, (325) 2
Tags: geometry , right triangle , ratio
Consider a right triangle $ABC$, where $A$ is the right angle, and $AC > AB$. Points $E$ on $AC$ and $D$ on $BC$ are chosen so that$ AB = AE = BD$. Prove that the triangle $ADE$ is right if and only if the ratio $AB : AC : BC$ of sides of the triangle $ABC$ is $3 : 4 : 5$. (A. Parovan)

2010 Oral Moscow Geometry Olympiad, 4
Tags: geometry , triangle inequality , right triangle , isosceles
An isosceles triangle $ABC$ with base $AC$ is given. Point $H$ is the intersection of altitudes. On the sides $AB$ and $BC$, points $M$ and $K$ are selected, respectively, so that the angle $KMH$ is right. Prove that a right-angled triangle can be constructed from the segments $AK, CM$ and $MK$.

1974 Chisinau City MO, 76
Tags: geometry , isosceles , right triangle , equal angles
Altitude $AH$ and median $AM$ of the triangle $ABC$ satisfy the relation: $\angle ABM = \angle CBH$. Prove that triangle $ABC$ is isosceles or right-angled.

2020 Puerto Rico Team Selection Test, 3
Tags: geometry , right triangle , equal segments
The side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$, such that $CD=BC$. The side $CA$ is extended beyond $A$ to $E$, such that $AE=2CA$. Prove that if $AD=BE$, then the triangle $ABC$ is right.

1991 Denmark MO - Mohr Contest, 3
Tags: geometry , perimeter , right triangle
A right-angled triangle has perimeter $60$ and the altitude of the hypotenuse has a length $12$. Determine the lengths of the sides.

2002 Denmark MO - Mohr Contest, 4
Tags: geometry , isosceles , right triangle
In triangle $ABC$ we have $\angle C = 90^o$ and $AC = BC$. Furthermore $M$ is an interior pont in the triangle so that $MC = 1 , MA = 2$ and $MB =\sqrt2$. Determine $AB$

1925 Eotvos Mathematical Competition, 3
Tags: geometry , inradius , geometric inequalities , right triangle
Let $r$ be the radius of the inscribed circle of a right triangle $ABC$. Show that $r$ is less than half of either leg and less than one fourth of the hypotenuse.

2015 Czech-Polish-Slovak Junior Match, 1
Tags: max , area of a triangle , angle , right triangle , geometry
In the right triangle $ABC$ with shorter side $AC$ the hypotenuse $AB$ has length $12$. Denote $T$ its centroid and $D$ the feet of altitude from the vertex $C$. Determine the size of its inner angle at the vertex $B$ for which the triangle $DTC$ has the greatest possible area.