Olympiad problems

Estonia Open Junior - geometry, 1995.1.4

The midpoint of the hypotenuse $AB$ of the right triangle $ABC$ is $K$. The point $M$ on the side $BC$ is taken such that $BM = 2 \cdot MC$. Prove that $\angle BAM = \angle CKM$.

Ukraine Correspondence MO - geometry, 2009.3

Tags: compass , inradius , right triangle , geometry

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?

2022 Yasinsky Geometry Olympiad, 5

Tags: geometry , geometric inequality , right triangle

Let $ABC$ be a right triangle with leg $CB = 2$ and hypotenuse $AB= 4$. Point $K$ is chosen on the hypotenuse $AB$, and point $L$ is chosen on the leg $AC$. a) Describe and justify how to construct such points $K$ and $ L$ so that the sum of the distances $CK+KL$ is the smallest possible. b) Find the smallest possible value of $CK+KL$. (Olexii Panasenko)

2008 Tournament Of Towns, 3

Tags: geometry , circumcircle , equal segments , right triangle

In triangle $ABC, \angle A = 90^o$. $M$ is the midpoint of $BC$ and $H$ is the foot of the altitude from $A$ to $BC$. The line passing through $M$ and perpendicular to $AC$ meets the circumcircle of triangle $AMC$ again at $P$. If $BP$ intersects $AH$ at $K$, prove that $AK = KH$.

1999 Tournament Of Towns, 2

Tags: ratio , square , geometry , right triangle

$ABC$ is a right-angled triangle. A square $ABDE$ is constructed on the opposite side of the hypothenuse $AB$ from $C$. The bisector of $\angle C$ cuts $DE$ at $F$. If $AC = 1$ and $BC = 3$, compute $\frac{DF}{EF}$. (A Blinkov)

2019 Yasinsky Geometry Olympiad, p5

Tags: geometry , right triangle , construction

In a right triangle $ABC$ with a hypotenuse $AB$, the angle $A$ is greater than the angle $B$. Point $N$ lies on the hypotenuse $AB$ , such that $BN = AC$. Construct this triangle $ABC$ given the point $N$, point $F$ on the side $AC$ and a straight line $\ell$ containing the bisector of the angle $A$ of the triangle $ABC$. (Grigory Filippovsky)

May Olympiad L1 - geometry, 2003.2

Tags: geometry , area of a triangle , right triangle , area

The triangle $ABC$ is right in $A$ and $R$ is the midpoint of the hypotenuse $BC$ . On the major leg $AB$ the point $P$ is marked such that $CP = BP$ and on the segment $BP$ the point $Q$ is marked such that the triangle $PQR$ is equilateral. If the area of triangle $ABC$ is $27$, calculate the area of triangle $PQR$ .

2016 Costa Rica - Final Round, G2

Tags: right triangle , geometry

Consider $\vartriangle ABC$ right at $B, F$ a point such that $B - F - C$ and $AF$ bisects $\angle BAC$, $I$ a point such that $A - I - F$ and CI bisect $\angle ACB$, and $E$ a point such that $A- E - C$ and $AF \perp EI$. If $AB = 4$ and $\frac{AI}{IF}={4}{3}$ , determine $AE$. Notation: $A-B-C$ means than points $A,B,C$ are collinear in that order i.e. $ B$ lies between $ A$ and $C$.

Denmark (Mohr) - geometry, 2010.1

Tags: geometry , circles , right triangle , area

Four right triangles, each with the sides $1$ and $2$, are assembled to a figure as shown. How large a fraction does the area of the small circle make up of that of the big one? [img]https://1.bp.blogspot.com/-XODK1XKCS0Q/XzXDtcA-xAI/AAAAAAAAMWA/zSLPpf3IcX0rgaRtOxm_F2begnVdUargACLcBGAsYHQ/s0/2010%2BMohr%2Bp1.png[/img]

2016 Dutch BxMO TST, 3

Tags: incircle , collinear , right triangle , geometry

Let $\vartriangle ABC$ be a right-angled triangle with $\angle A = 90^o$ and circumcircle $\Gamma$. The inscribed circle is tangent to $BC$ in point $D$. Let $E$ be the midpoint of the arc $AB$ of $\Gamma$ not containing $C$ and let $F$ be the midpoint of the arc $AC$ of $\Gamma$ not containing $B$. (a) Prove that $\vartriangle ABC \sim \vartriangle DEF$. (b) Prove that $EF$ goes through the points of tangency of the incircle to $AB$ and $AC$.

2015 India PRMO, 9

Tags: geometry , right triangle

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

2017 Ukrainian Geometry Olympiad, 3

Tags: geometry , circumcircle , tangent , right triangle

On the hypotenuse $AB$ of a right triangle $ABC$, we denote a point $K$ such that $BK = BC$. Let $P$ be a point on the perpendicular from the point $K$ to line $CK$, equidistant from the points $K$ and $B$. Let $L$ be the midpoint of $CK$. Prove that line $AP$ is tangent to the circumcircle of $\Delta BLP$.

2013 India PRMO, 19

Tags: ratio , right triangle , area

In a triangle $ABC$ with $\angle BC A = 90^o$, the perpendicular bisector of $AB$ intersects segments $AB$ and $AC$ at $X$ and $Y$, respectively. If the ratio of the area of quadrilateral $BXYC$ to the area of triangle $ABC$ is $13 : 18$ and $BC = 12$ then what is the length of $AC$?

2016 Oral Moscow Geometry Olympiad, 3

Tags: right triangle , distance , geometry

A circle with center $O$ passes through the ends of the hypotenuse of a right-angled triangle and intersects its legs at points $M$ and $K$. Prove that the distance from point $O$ to line $MK$ is half the hypotenuse.

Denmark (Mohr) - geometry, 2005.3

Tags: geometry , isosceles , right triangle

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]

Ukrainian TYM Qualifying - geometry, 2015.21

Tags: geometry , construction , right triangle

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.

Croatia MO (HMO) - geometry, 2012.7

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

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 $A$.

Ukrainian From Tasks to Tasks - geometry, 2014.4

Tags: geometry , right triangle , square

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 $KB$.

2003 Argentina National Olympiad, 4

Tags: geometry , trapezoid , right triangle , circumcenter , area of a triangle

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 $EGH$.

2018 Oral Moscow Geometry Olympiad, 1

Tags: projection , right triangle , geometry , inradius

In a right triangle $ABC$ with a right angle $C$, let $AK$ and $BN$ be the angle bisectors. Let $D,E$ be the projections of $C$ on $AK, BN$ respectively. Prove that the length of the segment $DE$ is equal to the radius of the inscribed circle.

2022 New Zealand MO, 4

Tags: geometry , incircle , right triangle

Triangle $ABC$ is right-angled at $B$ and has incentre $I$. Points $D$, $E$ and $F$ are the points where the incircle of the triangle touches the sides $BC$, $AC$ and AB respectively. Lines $CI$ and $EF$ intersect at point $P$. Lines $DP$ and $AB$ intersect at point $Q$. Prove that $AQ = BF$.