Olympiad problems

2000 Tuymaada Olympiad, 6

Let $O$ be the center of the circle circumscribed around the the triangle $ABC$. The centers of the circles circumscribed around the squares $OAB,OBC,OCA$ lie at the vertices of a regular triangle. Prove that the triangle $ABC$ is right.

2005 Estonia National Olympiad, 1

Tags: right triangle , area , geometry

The height drawn on the hypotenuse of a right triangle divides the hypotenuse into two sections with a length ratio of $9: 1$ and two triangles of the starting triangle with a difference of areas of $48$ cm$^2$. Find the original triangle sidelengths.

1998 Argentina National Olympiad, 5

Tags: construction , right triangle , isosceles , inscribed , minimum , area of a triangle , geometry

Let $ABC$ a right isosceles triangle with hypotenuse $AB=\sqrt2$ . Determine the positions of the points $X,Y,Z$ on the sides $BC,CA,AB$ respectively so that the triangle $XYZ$ is isosceles, right, and with minimum area.

2003 District Olympiad, 2

Tags: equal segments , geometry , perpendicular , parallel , right triangle

In the right triangle $ABC$ ( $\angle A = 90^o$), $D$ is the intersection of the bisector of the angle $A$ with the side $(BC)$, and $P$ and $Q$ are the projections of the point $D$ on the sides $(AB),(AC)$ respectively . If $BQ \cap DP=\{M\}$, $CP \cap DQ=\{N\}$, $BQ\cap CP=\{H\}$, show that: a) $PM = DN$ b) $MN \parallel BC$ c) $AH \perp BC$.

Ukrainian TYM Qualifying - geometry, I.17

Tags: right triangle , geometry , 3d geometry , cone

A right triangle when rotating around a large leg forms a cone with a volume of $100\pi$. Calculate the length of the path that passes through each vertex of the triangle at rotation of $180^o$ around the point of intersection of its bisectors, if the sum of the diameters of the circles, inscribed in the triangle and circumscribed around it, are equal to $17$.

2018 Adygea Teachers' Geometry Olympiad, 2

Tags: right triangle , geometry

It is known that in a right triangle: a) The height drawn from the top of the right angle is the geometric mean of the projections of the legs on the hypotenuse; b) the leg is the geometric mean of the hypotenuse and the projection of this leg to the hypotenuse. Are the converse statements true? Formulate them and justify the answer. Is it possible to formulate the criterion of a right triangle based on these statements? If possible, then how? If not, why?

May Olympiad L1 - geometry, 2014.4

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

Let $ABC$ be a right triangle and isosceles, with $\angle C = 90^o$. Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $AC$. Let $ P$ be such that $MNP$ is an equilateral triangle with $ P$ inside the quadrilateral $MBCN$. Calculate the measure of $\angle CAP$

2013 Cuba MO, 9

Tags: right triangle , geometry

Let ABC be a triangle with $\angle A = 90^o$, $\angle B = 75^o$, and $AB = 2$. Points $P$ and $Q$ of the sides $AC$ and $BC$ respectively, are such that $\angle APB = \angle CPQ$ and $\angle BQA = \angle CQP$. Calculate the lenght of $QA$.

2000 Portugal MO, 5

Tags: geometry , right triangle

In the figure, $[ABC]$ and $[DEC]$ are right triangles . Knowing that $EB = 1/2, EC = 1$ and $AD = 1$, calculate $DC$. [img]https://1.bp.blogspot.com/-nAOrVnK5JmI/X4UMb2CNTyI/AAAAAAAAMmk/TtaBESxYyJ0FsBoY2XaCGlCTc6mgmA5TQCLcBGAsYHQ/s0/2000%2Bportugal%2Bp5.png[/img]

1990 Greece National Olympiad, 1

Tags: right triangle , geometry , angle bisector , geometric inequality

Let $ABC$ be a right triangle with $\angle A=90^o$ and $AB<AC$. Let $AH,AD,AM$ be altitude, angle bisector and median respectively. Prove that $\frac{BD}{CD}<\frac{HD}{MD}.$

2020 IMO Shortlist, G1

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

2014 India PRMO, 15

Tags: midpoint , geometry , right triangle

Let $XOY$ be a triangle with $\angle XOY = 90^o$. Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$, respectively. Suppose that $XN = 19$ and $YM =22$. What is $XY$?

2013 Tournament of Towns, 3

Tags: angle bisector , geometry , right triangle , semicircle

Assume that $C$ is a right angle of triangle $ABC$ and $N$ is a midpoint of the semicircle, constructed on $CB$ as on diameter externally. Prove that $AN$ divides the bisector of angle $C$ in half.

Estonia Open Junior - geometry, 2008.2.2

Tags: equal segments , right triangle , geometry , equal angles

In a right triangle $ABC$, $K$ is the midpoint of the hypotenuse $AB$ and $M$ such a point on the $BC$ that $| B M | = 2 | MC |$. Prove that $\angle MAB = \angle MKC$.

Denmark (Mohr) - geometry, 1991.3

Tags: right triangle , geometry , perimeter

A right-angled triangle has perimeter $60$ and the altitude of the hypotenuse has a length $12$. Determine the lengths of the sides.

Ukrainian TYM Qualifying - geometry, 2011.2

Tags: right triangle , equal circles , geometry

Eight circles of radius $r$ located in a right triangle $ABC$ (angle $C$ is right) as shown in figure (each of the circles touches the respactive sides of the triangle and the other circles). Find the radius of the inscribed circle of triangle $ABC$. [img]https://cdn.artofproblemsolving.com/attachments/4/7/1b1cd7d6bc7f5004b8e94468d723ed16e9a039.png[/img]

Kyiv City MO Juniors 2003+ geometry, 2006.8.3

Tags: geometry , right triangle , equal segments , perpendicular , llllllllllllllllllllllllllllll

On the legs $AC, BC$ of a right triangle $\vartriangle ABC$ select points $M$ and $N$, respectively, so that $\angle MBC = \angle NAC$. The perpendiculars from points $M$ and $C$ on the line $AN$ intersect $AB$ at points $K$ and $L$, respectively. Prove that $KL=LB$. (O. Clurman)

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 $\frac{AD}{AB}$.

2018 Czech-Polish-Slovak Junior Match, 2

Tags: area , ratio , geometry , symmetry , right triangle

Given a right triangle $ABC$ with the hypotenuse $AB$. Let $K$ be any interior point of triangle $ABC$ and points $L, M$ are symmetric of point $K$ wrt lines $BC, AC$ respectively. Specify all possible values for $S_{ABLM} / S_{ABC}$, where $S_{XY ... Z}$ indicates the area of the polygon $XY...Z$ .

1988 IMO, 2

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

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 \[ \frac {E}{E_1} \geq 2. \]

2015 Saudi Arabia GMO TST, 3

Tags: geometry , right triangle , perpendicular , symmetric

Let $ABC$ be a triangle, with $AB < AC$, $D$ the foot of the altitude from $A, M$ the midpoint of $BC$, and $B'$ the symmetric of $B$ with respect to $D$. The perpendicular line to $BC$ at $B'$ intersects $AC$ at point $P$ . Prove that if $BP$ and $AM$ are perpendicular then triangle $ABC$ is right-angled. Liana Topan

2015 Hanoi Open Mathematics Competitions, 10

Tags: geometry , square , concyclic , right triangle

A right-angled triangle has property that, when a square is drawn externally on each side of the triangle, the six vertices of the squares that are not vertices of the triangle are concyclic. Assume that the area of the triangle is $9$ cm$^2$. Determine the length of sides of the triangle.

2012 Romania National Olympiad, 2

Tags: geometry , equal angles , right triangle

Let $ABC$ be a triangle with right $\angle A$. Consider points $D \in (AC)$ and $E \in (BD)$ such that $\angle ABC = \angle ECD = \angle CED$. Prove that $BE = 2 \cdot AD$

Ukrainian From Tasks to Tasks - geometry, 2010.5

Tags: geometry , right triangle

In a right triangle $ABC$ ($\angle C = 90^o$) it is known that $AC = 4$ cm, $BC = 3$ cm. The points $A_1, B_1$ and $C_1$ are such that $AA_1 \parallel BC$, $BB_1\parallel A_1C$, $CC_1\parallel A_1B_1$, $A_1B_1C_1= 90^o$, $A_1B_1= 1$ cm. Find $B_1C_1$.

2021 Dutch IMO TST, 2

Tags: geometry , right angle , centroid , right triangle

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 $ABC$.