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

2021 Argentina National Olympiad, 3
Tags: right triangle , angle chasing , geometry
Let $ABC$ be an isosceles right triangle at $A$ with $AB=AC$. Let $M$ and $N$ be on side $BC$, with $M$ between $B$ and $N,$ such that $$BM^2+ NC^2= MN^2.$$ Determine the measure of the angle $\angle MAN.$

2021 Azerbaijan IMO TST, 1
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 $\angle ACB = 90^{\circ}$.

1993 Rioplatense Mathematical Olympiad, Level 3, 6
Tags: equal segments , geometry , pentagon , right triangle , angle
Let $ABCDE$ be pentagon such that $AE = ED$ and $BC = CD$. It is known that $\angle BAE + \angle EDC + \angle CB A = 360^o$ and that $P$ is the midpoint of $AB$. Show that the triangle $ECP$ is right.

2009 Junior Balkan Team Selection Tests - Romania, 1
Tags: right triangle , geometric inequality , inequalities , geometry
Show that in any triangle $ABC$ with $A = 90^0$ the following inequality holds: $$(AB -AC)^2(BC^2 + 4AB \cdot AC)^2 \le 2BC^6$$

Kyiv City MO Seniors 2003+ geometry, 2005.10.4
Tags: right triangle , equal segments , bisects segment , equal angles , geometry
In a right triangle $ABC $ with a right angle $\angle C $, n the sides $AC$ and $AB$, the points $M$ and $N$ are selected, respectively, that $CM = MN$ and $\angle MNB = \angle CBM$. Let the point $K$ be the projection of the point $C $ on the segment $MB $. Prove that the line $NK$ passes through the midpoint of the segment $BC$. (Alex Klurman)

2003 Estonia National Olympiad, 3
Tags: angle , right triangle , geometry
Let $ABC$ be a triangle with $\angle C = 90^o$ and $D$ a point on the ray $CB$ such that $|AC| \cdot |CD| = |BC|^2$. A parallel line to $AB$ through $D$ intersects the ray $CA$ at $E$. Find $\angle BEC$.

2005 Portugal MO, 2
Tags: right triangle , geometry , geometric inequality
Consider the triangles $[ABC]$ and $[EDC]$, right at $A$ and $D$, respectively. Show that, if $E$ is the midpoint of $[AC]$, then $AB <BD$. [img]https://cdn.artofproblemsolving.com/attachments/c/3/75bc1bda1a22bcf00d4fe7680c80a81a9aaa4c.png[/img]

1949-56 Chisinau City MO, 22
Tags: right triangle , angle bisector , geometry
Show that in a right-angled triangle the bisector of the right angle divides into equal parts the angle between the altitude and the median, drawn from the same vertex.

Ukrainian TYM Qualifying - geometry, 2016.1
Tags: right triangle , orthocenter , geometry
The points $K$ and $N$ lie on the hypotenuse $AB$ of a right triangle $ABC$. Prove that orthocenters the triangles $BCK$ and $ACN$ coincide if and only if $\frac{BN}{AK}=\tan^2 A.$

2014 Junior Balkan Team Selection Tests - Moldova, 3
Tags: right triangle , area , geometry , area of a triangle
Let $ABC$ be a right triangle with $\angle ABC = 90^o$ . Points $D$ and $E$, located on the legs $(AC)$ and $(AB)$ respectively, are the legs of the inner bisectors taken from the vertices $B$ and $C$, respectively. Let $I$ be the center of the circle inscribed in the triangle $ABC$. If $BD \cdot CE = m^2 \sqrt2$ , find the area of the triangle $BIC$ (in terms of parameter $m$)

1987 Mexico National Olympiad, 5
Tags: geometry , area , rectangle , right triangle
In a right triangle $ABC$, M is a point on the hypotenuse $BC$ and $P$ and $Q$ the projections of $M$ on $AB$ and $AC$ respectively. Prove that for no such point $M$ do the triangles $BPM, MQC$ and the rectangle $AQMP$ have the same area.

1983 IMO Shortlist, 3
Tags: right triangle , combinatorics , ramsey theory , extremal combinatorics , geometry
Let $ABC$ be an equilateral triangle and $\mathcal{E}$ the set of all points contained in the three segments $AB$, $BC$, and $CA$ (including $A$, $B$, and $C$). Determine whether, for every partition of $\mathcal{E}$ into two disjoint subsets, at least one of the two subsets contains the vertices of a right-angled triangle.

2023 European Mathematical Cup, 2
Tags: right triangle , geometry , circumcircle
Let $ABC$ be a triangle such that $\angle BAC = 90^{\circ}$. The incircle of triangle $ABC$ is tangent to the sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D,E,F$ respectively. Let $M$ be the midpoint of $\overline{EF}$. Let $P$ be the projection of $A$ onto $BC$ and let $K$ be the intersection of $MP$ and $AD$. Prove that the circumcircles of triangles $AFE$ and $PDK$ have equal radius. [i]Kyprianos-Iason Prodromidis[/i]

2010 Oral Moscow Geometry Olympiad, 4
Tags: right triangle , triangle inequality , geometry , 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 $MK$.

Novosibirsk Oral Geo Oly VIII, 2022.6
Tags: right triangle , isosceles , partition , combinatorial geometry , geometry
Anton has an isosceles right triangle, which he wants to cut into $9$ triangular parts in the way shown in the picture. What is the largest number of the resulting $9$ parts that can be equilateral triangles? A more formal description of partitioning. Let triangle $ABC$ be given. We choose two points on its sides so that they go in the order $AC_1C_2BA_1A_2CB_1B_2$, and no two coincide. In addition, the segments $C_1A_2$, $A_1B_2$ and $B_1C_2$ must intersect at one point. Then the partition is given by segments $C_1A_2$, $A_1B_2$, $B_1C_2$, $A_1C_2$, $B_1A_2$ and $C_1B_2$. [img]https://cdn.artofproblemsolving.com/attachments/0/5/5dd914b987983216342e23460954d46755d351.png[/img]

1990 Mexico National Olympiad, 6
Tags: similar triangles , right triangle , geometry
$ABC$ is a triangle with $\angle C = 90^o$. $E$ is a point on $AC$, and $F$ is the midpoint of $EC$. $CH$ is an altitude. $I$ is the circumcenter of $AHE$, and $G$ is the midpoint of $BC$. Show that $ABC$ and $IGF$ are similar.

Ukrainian TYM Qualifying - geometry, 2015.21
Tags: right triangle , geometry , construction
Let $CH$ be the altitude of the triangle $ABC$ drawn on the board, in which $\angle C = 90^o$, $CA \ne CB$. The mathematics teacher drew the perpendicular bisectors of segments$ CA$ and $CB$, which cut the line CH at points $K$ and $M$, respectively, and then erased the drawing, leaving only the points $C, K$ and $M$ on the board. Restore triangle $ABC$, using only a compass and a ruler.

2014 Contests, 2
Tags: right triangle , rhombus , square , perpendicular , geometry , isosceles
Outside the square $ABCD$, the rhombus $BCMN$ is constructed with angle $BCM$ obtuse . Let $P$ be the intersection point of the lines $BM$ and $AN$ . Prove that $DM \perp CP$ and the triangle $DPM$ is right isosceles .

1949-56 Chisinau City MO, 27
Tags: similar triangles , geometry , similar , right triangle
The areas of two right-angled triangles have ratio equal to the squares of their hypotenuses. Show that these triangles are similar.

2011 Tournament of Towns, 2
Tags: right triangle , geometry
On side $AB$ of triangle $ABC$ a point $P$ is taken such that $AP = 2PB$. It is known that $CP = 2PQ$ where $Q$ is the midpoint of $AC$. Prove that $ABC$ is a right triangle.

2010 Argentina National Olympiad, 2
Tags: incircle , right triangle , geometry
Let $ABC$ be a triangle with $\angle C = 90^o$ and $AC = 1$. The median $AM$ intersects the incircle at the points $P$ and $Q$, with $P$ between $A$ and $Q$, such that $AP = QM$. Find the length of $PQ$.

2021 Germany Team Selection Test, 2
Tags: right triangle , geometry , 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 $\angle ACB = 90^{\circ}$.

2020 Puerto Rico Team Selection Test, 3
Tags: geometry , equal segments , right triangle
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.

2015 Czech-Polish-Slovak Junior Match, 4
Tags: midpoint , right triangle , angle bisector , incenter , geometry
Let $ABC$ ne a right triangle with $\angle ACB=90^o$. Let $E, F$ be respecitvely the midpoints of the $BC, AC$ and $CD$ be it's altitude. Next, let $P$ be the intersection of the internal angle bisector from $A$ and the line $EF$. Prove that $P$ is the center of the circle inscribed in the triangle $CDE$ .

2013 Argentina National Olympiad Level 2, 2
Tags: equal segments , right triangle , ratio , geometry
Let $ABC$ be a right triangle. It is known that there are points $D$ on the side $AC$ and $E$ on the side $BC$ such that $AB = AD = BE$ and $BD$ is perpendicular to $DE$. Calculate the ratios $\frac{AB}{BC}$ and $\frac{BC}{CA}$.