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

2016 India PRMO, 4

Consider a right-angled triangle $ABC$ with $\angle C = 90^o$. Suppose that the hypotenuse $AB$ is divided into four equal parts by the points $D,E,F$, such that $AD = DE = EF = FB$. If $CD^2 +CE^2 +CF^2 = 350$, find the length of $AB$.

1979 Chisinau City MO, 172

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.

2003 Argentina National Olympiad, 4

The trapezoid $ABCD$ of bases $AB$ and $CD$, has $\angle A = 90^o, AB = 6, CD = 3$ and $AD = 4$. Let $E, G, H$ be the circumcenters of triangles $ABC, ACD, ABD$, respectively. Find the area of the triangle $EGH$.

Kyiv City MO 1984-93 - geometry, 1989.8.5

The student drew a right triangle $ABC$ on the board with a right angle at the vertex $B$ and inscribed in it an equilateral triangle $KMP$ such that the points $K, M, P$ lie on the sides $AB, BC, AC$, respectively, and $KM \parallel AC$. Then the picture was erased, leaving only points $A, P$ and $C$. Restore erased points and lines.

2021 Taiwan TST Round 2, 4

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 $\angle ACB = 90^{\circ}$.

1975 Czech and Slovak Olympiad III A, 1

Let $\mathbf T$ be a triangle with $[\mathbf T]=1.$ Show that there is a right triangle $\mathbf R$ such that $[\mathbf R]\le\sqrt3$ and $\mathbf T\subseteq\mathbf R.$ ($[-]$ denotes area of a triangle.)

2002 Denmark MO - Mohr Contest, 4

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 $AB$

1983 IMO, 1

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.

2014 Junior Balkan Team Selection Tests - Moldova, 7

Let the isosceles right triangle $ABC$ with $\angle A= 90^o$ . The points $E$ and $F$ are taken on the ray $AC$ so that $\angle ABE = 15^o$ and $CE = CF$. Determine the measure of the angle $CBF$.

1962 Dutch Mathematical Olympiad, 1

Given a triangle $ABC$ with $\angle C = 90^o$. a) Construct the circle with center $C$, so that one of the tangents from $A$ to that circle is parallel to one of the tangents from $B$ to that circle. b) A circle with center $C$ has two parallel tangents passing through A and go respectively. If $AC = b$ and $BC = a$, express the radius of the circle in terms of $a$ and $b$.

1996 Romania National Olympiad, 4

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 $BC$.

Ukrainian From Tasks to Tasks - geometry, 2014.4

In the triangle $ABC$ it is known that $AC = 21$ cm, $BC = 28$ cm and $\angle C = 90^o$. On the hypotenuse $AB$, we construct a square $ABMN$ with center $O$ such that the segment $CO$ intersects the hypotenuse $AB$ at the point $K$. Find the lengths of the segments $AK$ and $KB$.

2019 Singapore Junior Math Olympiad, 1

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 $AE=2EB$.

2016 Sharygin Geometry Olympiad, 1

An altitude $AH$ of triangle $ABC$ bisects a median $BM$. Prove that the medians of triangle $ABM$ are sidelengths of a right-angled triangle. by Yu.Blinkov

1994 Denmark MO - Mohr Contest, 5

In a right-angled and isosceles triangle, the two catheti are both length $1$. Find the length of the shortest line segment dividing the triangle into two figures with the same area, and specify the location of this line segment

2013 Estonia Team Selection Test, 4

Let $D$ be the point different from $B$ on the hypotenuse $AB$ of a right triangle $ABC$ such that $|CB| = |CD|$. Let $O$ be the circumcenter of triangle $ACD$. Rays $OD$ and $CB$ intersect at point $P$, and the line through point $O$ perpendicular to side AB and ray $CD$ intersect at point $Q$. Points $A, C, P, Q$ are concyclic. Does this imply that $ACPQ$ is a square?

Estonia Open Senior - geometry, 2012.1.3

Let $ABC$ be a triangle with median AK. Let $O$ be the circumcenter of the triangle $ABK$. a) Prove that if $O$ lies on a midline of the triangle $ABC$, but does not coincide with its endpoints, then $ABC$ is a right triangle. b) Is the statement still true if $O$ can coincide with an endpoint of the midsegment?

2017 Czech-Polish-Slovak Junior Match, 4

Given is a right triangle $ABC$ with perimeter $2$, with $\angle B=90^o$ . Point $S$ is the center of the excircle to the side $AB$ of the triangle and $H$ is the intersection of the heights of the triangle $ABS$ . Determine the smallest possible length of the segment $HS $.

Estonia Open Junior - geometry, 2012.1.3

A rectangle $ABEF$ is drawn on the leg $AB$ of a right triangle $ABC$, whose apex $F$ is on the leg $AC$. Let $X$ be the intersection of the diagonal of the rectangle $AE$ and the hypotenuse $BC$ of the triangle. In what ratio does point $X$ divide the hypotenuse $BC$ if it is known that $| AC | = 3 | AB |$ and $| AF | = 2 | AB |$?

Denmark (Mohr) - geometry, 2019.5

In the figure below the triangles $BCD, CAE$ and $ABF$ are equilateral, and the triangle $ABC$ is right-angled with $\angle A = 90^o$. Prove that $|AD| = |EF|$. [img]https://1.bp.blogspot.com/-QMMhRdej1x8/XzP18QbsXOI/AAAAAAAAMUI/n53OsE8rwZcjB_zpKUXWXq6bg3o8GUfSwCLcBGAsYHQ/s0/2019%2Bmohr%2Bp5.png[/img]

1988 IMO Shortlist, 13

In a right-angled triangle $ ABC$ let $ AD$ be the altitude drawn to the hypotenuse and let the straight line joining the incentres of the triangles $ ABD, ACD$ intersect the sides $ AB, AC$ at the points $ K,L$ respectively. If $ E$ and $ E_1$ dnote the areas of triangles $ ABC$ and $ AKL$ respectively, show that \[ \frac {E}{E_1} \geq 2. \]

2024 Mozambique National Olympiad, P3

Let $ACE$ be a triangle with $\angle ECA=60^{\circ}, \angle AEC=90^{\circ}$. Let $B$ and $D$ be points on the sides $AC$ and $CE$ respectively such that the $\triangle BCD$ is equilateral. Now suppose $BD \cap AE=F$. Find $\angle EAC+\angle EFD$.

2020 Argentina National Olympiad, 3

Let $ABC$ be a right isosceles triangle with right angle at $A$. Let $E$ and $F$ be points on A$B$ and $AC$ respectively such that $\angle ECB = 30^o$ and $\angle FBC = 15^o$. Lines $CE$ and $BF$ intersect at $P$ and line $AP$ intersects side $BC$ at $D$. Calculate the measure of angle $\angle FDC$.

1955 Moscow Mathematical Olympiad, 288

We are given a right triangle $ABC$ and the median $BD$ drawn from the vertex $B$ of the right angle. Let the circle inscribed in $\vartriangle ABD$ be tangent to side $AD$ at $K$. Find the angles of $\vartriangle ABC$ if $K$ divides $AD$ in halves.

2011 May Olympiad, 3

In a right triangle rectangle $ABC$ such that $AB = AC$, $M$ is the midpoint of $BC$. Let $P$ be a point on the perpendicular bisector of $AC$, lying in the semi-plane determined by $BC$ that does not contain $A$. Lines $CP$ and $AM$ intersect at $Q$. Calculate the angles that form the lines $AP$ and $BQ$.