Olympiad problems

Croatia MO (HMO) - geometry, 2012.7

Tags: incircle , right triangle , geometry , concyclic , excircle , cyclic quadrilateral

Let the points $M$ and $N$ be the intersections of the inscribed circle of the right-angled triangle $ABC$, with sides $AB$ and $CA$ respectively , and points $P$ and $Q$ respectively be the intersections of the ex-scribed circles opposite to vertices $B$ and $C$ with direction $BC$. Prove that the quadrilateral $MNPQ$ is a cyclic if and only if the triangle $ABC$ is right-angled with a right angle at the vertex $A$.

2011 Hanoi Open Mathematics Competitions, 10

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.

1982 IMO Shortlist, 17

Tags: geometry , right triangle , triangle , perpendicular , similar triangles

The right triangles $ABC$ and $AB_1C_1$ are similar and have opposite orientation. The right angles are at $C$ and $C_1$, and we also have $ \angle CAB = \angle C_1AB_1$. Let $M$ be the point of intersection of the lines $BC_1$ and $B_1C$. Prove that if the lines $AM$ and $CC_1$ exist, they are perpendicular.

2016 Romania National Olympiad, 4

Tags: geometry , right triangle , angle , perpendicular , isosceles

Consider the isosceles right triangle $ABC$, with $\angle A = 90^o$ and the point $M \in (BC)$ such that $\angle AMB = 75^o$. On the inner bisector of the angle $MAC$ take a point $F$ such that $BF = AB$. Prove that: a) the lines $AM$ and $BF$ are perpendicular; b) the triangle $CFM$ is isosceles.

Denmark (Mohr) - geometry, 2002.4

Tags: right triangle , geometry , isosceles

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 $AB$

Ukraine Correspondence MO - geometry, 2009.3

Tags: geometry , compass , right triangle , inradius

A right triangle is drawn on the plane. How to use only a compass to mark two points, such that the distance between them is equal to the diameter of the circle inscribed in this triangle?

2013 Saudi Arabia BMO TST, 1

Tags: geometry , right triangle , isosceles , cyclic quadrilateral

In triangle $ABC$, $AB = AC = 3$ and $\angle A = 90^o$. Let $M$ be the midpoint of side $BC$. Points $D$ and $E$ lie on sides $AC$ and $AB$ respectively such that $AD > AE$ and $ADME$ is a cyclic quadrilateral. Given that triangle $EMD$ has area $2$, find the length of segment $CD$.

2016 Iranian Geometry Olympiad, 4

Tags: perpendicular bisector , right triangle , geometry

In a right-angled triangle $ABC$ ($\angle A = 90^o$), the perpendicular bisector of $BC$ intersects the line $AC$ in $K$ and the perpendicular bisector of $BK$ intersects the line $AB$ in $L$. If the line $CL$ be the internal bisector of angle $C$, find all possible values for angles $B$ and $C$. by Mahdi Etesami Fard

Denmark (Mohr) - geometry, 1994.4

Tags: right triangle , integer , geometry

In a right-angled triangle in which all side lengths are integers, one has a cathetus length $1994$. Determine the length of the hypotenuse.

2014 Gulf Math Olympiad, 3

Tags: equal segments , geometry , right triangle , semicircle , angle

(i) $ABC$ is a triangle with a right angle at $A$, and $P$ is a point on the hypotenuse $BC$. The line $AP$ produced beyond $P$ meets the line through $B$ which is perpendicular to $BC$ at $U$. Prove that $BU = BA$ if, and only if, $CP = CA$. (ii) $A$ is a point on the semicircle $CB$, and points $X$ and $Y$ are on the line segment $BC$. The line $AX$, produced beyond $X$, meets the line through $B$ which is perpendicular to $BC$ at $U$. Also the line $AY$, produced beyond $Y$, meets the line through $C$ which is perpendicular to $BC$ at $V$. Given that $BY = BA$ and $CX = CA$, determine the angle $\angle VAU$.

Denmark (Mohr) - geometry, 1996.1

Tags: geometry , right triangle , area

In triangle $ABC$, angle $C$ is right and the two catheti are both length $1$. For one given the choice of the point $P$ on the cathetus $BC$, the point $Q$ on the hypotenuse and the point $R$ are plotted on the second cathetus so that $PQ$ is parallel to $AC$ and $QR$ is parallel to $BC$. Thereby the triangle is divided into three parts. Determine the locations of point $P$ for which the rectangular part has a larger area than each of the other two parts.

Estonia Open Junior - geometry, 1995.2.1

Tags: rectangle , right triangle , geometry , inscribed

A rectangle, whose one sidelength is twice the other side, is inscribed inside a triangles with sides $3$ cm, $4$ cm and $5$ cm, such that the long sides lies entirely on the long side of the triangle. The other two remaining vertices of the rectangle lie respectively on the other two sides of the triangle. Find the lengths of the sides of this rectangle.

2012 China Northern MO, 1

Tags: geometry , right triangle , perpendicular

As shown in figure, given right $\vartriangle ABC$ with $\angle C=90^o$. $I$ is the incenter. The line $BI$ intersects segment $AC$ at the point $D$ . The line passing through $D$ parallel to $AI$ intersects $BC$ at point $E$. The line $EI$ intersects segment $AB$ at point $F$. Prove that $DF \perp AI$. [img]https://cdn.artofproblemsolving.com/attachments/2/4/6fc94adb4ce12c3bf07948b8c57170ca01b256.png[/img]

1993 Tournament Of Towns, (390) 2

Tags: right triangle , angle , geometry

Points $M$ and $N$ are taken on the hypotenuse $AB$ of a right triangle $ABC$ so that $BC = BM$ and $AC = AN$. Prove that the angle $MCN$ is equal to $45$ degrees. (Folklore)

2020 OMpD, 4

Tags: geometry , incenter , right triangle , right angle

Let $ABC$ be a triangle and $P$ be any point on the side $BC$. Let $I_1$,$I_2$ be the incenters of triangles $ABP$ and $ACP$, respectively. If $D$ is the point of tangency of the incircle of $ABC$ with the side $BC$, prove that $\angle I_1DI_2 = 90^o$.

2014 Hanoi Open Mathematics Competitions, 8

Tags: right triangle , isosceles triangle , geometry , isosceles

Let $ABC$ be a triangle. Let $D,E$ be the points outside of the triangle so that $AD=AB,AC=AE$ and $\angle DAB =\angle EAC =90^o$. Let $F$ be at the same side of the line $BC$ as $A$ such that $FB = FC$ and $\angle BFC=90^o$. Prove that the triangle $DEF$ is a right- isosceles triangle.

2021 Dutch IMO TST, 2

Tags: right triangle , geometry , right angle , centroid

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 $ABC$.

2018 Irish Math Olympiad, 2

Tags: geometry , right triangle , angle chasing

The triangle $ABC$ is right-angled at $A$. Its incentre is $I$, and $H$ is the foot of the perpendicular from $I$ on $AB$. The perpendicular from $H$ on $BC$ meets $BC$ at $E$, and it meets the bisector of $\angle ABC$ at $D$. The perpendicular from $A$ on $BC$ meets $BC$ at $F$. Prove that $\angle EFD = 45^o$

India EGMO 2021 TST, 5

Tags: right triangle , isosceles triangle , geometry

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

1990 Mexico National Olympiad, 2

Tags: right triangle , inradius , geometry

$ABC$ is a triangle with $\angle B = 90^o$ and altitude $BH$. The inradii of $ABC, ABH, CBH$ are $r, r_1, r_2$. Find a relation between them.

2015 India PRMO, 9

Tags: right triangle , geometry

$9.$ What is the greatest possible perimeter of a right-angled triangle with integer side lengths if one of the sides has length $12 ?$

2005 Denmark MO - Mohr Contest, 3

Tags: geometry , right triangle , isosceles

The point $P$ lies inside $\vartriangle ABC$ so that $\vartriangle BPC$ is isosceles, and angle $P$ is a right angle. Furthermore both $\vartriangle BAN$ and $\vartriangle CAM$ are isosceles with a right angle at $A$, and both are outside $\vartriangle ABC$. Show that $\vartriangle MNP$ is isosceles and right-angled. [img]https://1.bp.blogspot.com/-i9twOChu774/XzcBLP-RIXI/AAAAAAAAMXA/n5TJCOJypeMVW28-9GDG4st5C47yhvTCgCLcBGAsYHQ/s0/2005%2BMohr%2Bp3.png[/img]

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.

2019 Oral Moscow Geometry Olympiad, 4

Tags: right triangle , geometry , angle , excircle

Given a right triangle $ABC$ ($\angle C=90^o$). The $C$-excircle touches the hypotenuse $AB$ at point $C_1, A_1$ is the touchpoint of $B$-excircle with line $BC, B_1$ is the touchpoint of $A$-excircle with line $AC$. Find the angle $\angle A_1C_1B_1$.

1965 Czech and Slovak Olympiad III A, 2

Tags: construction , right triangle , geometric construction , geometry , ratio

Line segment $AM=d>0$ is given in the plane. Furthermore, a positive number $v$ is given. Construct a right triangle $ABC$ with hypotenuse $AB$, altitude to the hypotenuse of the length $v$ and the leg $BC$ being divided by $M$ in ration $MB/MC=2/3$. Discuss conditions of solvability in terms of $d, v$.