Olympiad problems

1966 Spain Mathematical Olympiad, 5

The length of the hypotenuse $BC$ of a right triangle $ABC$ is $a$, and on it the points $M$ and $N$ are taken such that $BM = NC = k$, with $k < a/2$. Assuming that (only) the data $a$ and $k$ are known, calculate: a) The value of the sum of the squares of the lengths $AM$ and $AN$. b) The ratio of the areas of triangles $ABC$ and $AMN$. c) The area enclosed by the circle that passes through the points $A, M' , N'$ , where $M'$ is the orthogonal projection of $M$ onto $AC$ and $N'$ that of $N$ onto $AB$.

1992 Austrian-Polish Competition, 2

Tags: coloring , combinatorics , right triangle , square

Each point on the boundary of a square has to be colored in one color. Consider all right triangles with the vertices on the boundary of the square. Determine the least number of colors for which there is a coloring such that no such triangle has all its vertices of the same color.

2006 Junior Balkan Team Selection Tests - Romania, 1

Tags: equal ratio , geometry , angle , ratio , right triangle

Let $ABC$ be a triangle right in $C$ and the points $D, E$ on the sides $BC$ and $CA$ respectively, such that $\frac{BD}{AC} =\frac{AE}{CD} = k$. Lines $BE$ and $AD$ intersect at $O$. Show that the angle $\angle BOD = 60^o$ if and only if $k =\sqrt3$.

Kyiv City MO Juniors 2003+ geometry, 2004.7.3

Tags: geometry , right triangle , right angle

Given a right triangle $ABC$ ($\angle A <45^o$,$ \angle C = 90^o$), on the sides $AC$ and $AB$ which are selected points $D,E$ respectively, such that $BD = AD$ and $CB = CE$. Let the segments $BD$ and $CE$ intersect at the point $O$. Prove that $\angle DOE = 90^o$.

2014 Contests, 2

Tags: geometry , rhombus , square , perpendicular , isosceles , right triangle

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 .

2010 Cuba MO, 3

Tags: ratio , geometry , right triangle

Let $ABC$ be a right triangle at $B$. Let $D$ be a point such that $BD\perp AC$ and $DC = AC$. Find the ratio $\frac{AD}{AB}$.

2010 Sharygin Geometry Olympiad, 7

Tags: geometry , right triangle , symmetry , tangent circle

Given triangle $ABC$. Lines $AL_a$ and $AM_a$ are the internal and the external bisectrix of angle $A$. Let $\omega_a$ be the reflection of the circumcircle of $\triangle AL_aM_a$ in the midpoint of $BC$. Circle $\omega_b$ is defined similarly. Prove that $\omega_a$ and $\omega_b$ touch if and only if $\triangle ABC$ is right-angled.

2020-21 KVS IOQM India, 22

Tags: geometry , right triangle , inradius

Let $ABC$ be a triangle with $\angle BAC = 90^o$ and $D$ be the point on the side $BC$ such that $AD \perp BC$. Let$ r, r_1$, and $r_2$ be the inradii of triangles $ABC, ABD$, and $ACD$, respectively. If $r, r_1$, and $r_2$ are positive integers and one of them is $5$, find the largest possible value of $r+r_1+ r_2$.

2016 Iranian Geometry Olympiad, 4

Tags: right triangle , perpendicular bisector , 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

2021 Argentina National Olympiad, 3

Tags: geometry , right triangle , angle chasing

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 $\angle MAN.$

Estonia Open Junior - geometry, 2016.1.5

Tags: geometry , isosceles , right triangle

A right triangle $ABC$ has the right angle at vertex $A$. Circle $c$ passes through vertices $A$ and $B$ of the triangle $ABC$ and intersects the sides $AC$ and $BC$ correspondingly at points $D$ and $E$. The line segment $CD$ has the same length as the diameter of the circle $c$. Prove that the triangle $ABE$ is isosceles.

2022 Novosibirsk Oral Olympiad in Geometry, 6

Tags: partition , geometry , combinatorial geometry , right triangle , isosceles

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]

1994 Denmark MO - Mohr Contest, 4

Tags: integer , geometry , right triangle

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 Hanoi Open Mathematics Competitions, 8

Tags: geometry , right triangle , isosceles , isosceles triangle

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.

Ukrainian From Tasks to Tasks - geometry, 2013.4

Tags: geometry , trapezoid , right triangle

The trapezoid is composed of three conguent right isosceles triangles as shown in the figure. It is necessary to cut it into $4$ equal parts. How to do it? [img]https://cdn.artofproblemsolving.com/attachments/f/e/87b07ae823190f26b70bfa22824679a829e649.png[/img]

2021 Dutch IMO TST, 2

Tags: right angle , centroid , right triangle , geometry

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 $ABC$.

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.

1997 Argentina National Olympiad, 2

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

Let $ABC$ be a triangle and $M$ be the midpoint of $AB$. If it is known that $\angle CAM + \angle MCB = 90^o$, show that triangle $ABC$ is isosceles or right.

Kyiv City MO Juniors Round2 2010+ geometry, 2016.8.1

Tags: ratio , radius , right triangle , geometry

In a right triangle, the point $O$ is the center of the circumcircle. Another circle of smaller radius centered at the point $O$ touches the larger leg and the altitude drawn from the top of the right angle. Find the acute angles of a right triangle and the ratio of the radii of the circumscribed and smaller circles.

1992 Denmark MO - Mohr Contest, 2

Tags: geometry , right triangle , tangent circle

In a right-angled triangle, $a$ and $b$ denote the lengths of the two catheti. A circle with radius $r$ has the center on the hypotenuse and touches both catheti. Show that $\frac{1}{a}+\frac{1}{b}=\frac{1}{r}$.

2021 Puerto Rico Team Selection Test, 2

Tags: geometry , incenter , right triangle , angle

Let $ABC$ be a right triangle with right angle at $ B$ and $\angle C=30^o$. If $M$ is midpoint of the hypotenuse and $I$ the incenter of the triangle, show that $ \angle IMB=15^o$.

1982 IMO Longlists, 54

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

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 Dutch BxMO TST, 3

Tags: geometry , incircle , collinear , right triangle

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 $AC$.

2014 Gulf Math Olympiad, 3

Tags: geometry , equal segments , 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 $\angle VAU$.

1984 Brazil National Olympiad, 4

Tags: geometry , right triangle , minimum , projection , segment

$ABC$ is a triangle with $\angle A = 90^o$. For a point $D$ on the side $BC$, the feet of the perpendiculars to $AB$ and $AC$ are $E$ and$ F$. For which point $D$ is $ EF$ a minimum?