Olympiad problems

2016 Czech And Slovak Olympiad III A, 2

Let us denote successively $r$ and $r_a$ the radii of the inscribed circle and the exscribed circle wrt to side BC of triangle $ABC$. Prove that if it is true that $r+r_a=|BC|$ , then the triangle $ABC$ is a right one

2014 Contests, 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 $\angle VAU$.

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)

2024 Yasinsky Geometry Olympiad, 1

Tags: geometry , right triangle , concyclic points

Let $ I $ and $ O $ be the incenter and circumcenter of the right triangle $ ABC $ ($ \angle C = 90^\circ $), and let $ K $ be the tangency point of the incircle with $ AC $. Let $ P $ and $ Q $ be the points where the circumcircle of triangle $ AOK $ intersects $ OC $ and the circumcircle of triangle $ ABC $, respectively. Prove that points $ C, I, P, $ and $ Q $ are concyclic. [i]Proposed by Mykhailo Sydorenko[/i]

2013 Costa Rica - Final Round, 3

Tags: geometry , right triangle , concyclic

Let $ABC$ be a triangle, right-angled at point $ A$ and with $AB>AC$. The tangent through $ A$ of the circumcircle $G$ of $ABC$ cuts $BC$ at $D$. $E$ is the reflection of $ A$ over line $BC$. $X$ is the foot of the perpendicular from $ A$ over $BE$. $Y$ is the midpoint of $AX$, $Z$ is the intersection of $BY$ and $G$ other than $ B$, and $F$ is the intersection of $AE$ and $BC$. Prove $D, Z, F, E$ are concyclic.

2021 Novosibirsk Oral Olympiad in Geometry, 5

Tags: geometry , right triangle , isosceles , equal segments

On the legs $AC$ and $BC$ of an isosceles right-angled triangle with a right angle $C$, points $D$ and $E$ are taken, respectively, so that $CD = CE$. Perpendiculars on line $AE$ from points $C$ and $D$ intersect segment $AB$ at points $P$ and $Q$, respectively. Prove that $BP = PQ$.

Denmark (Mohr) - geometry, 2002.4

Tags: geometry , isosceles , right triangle

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 $AB$

2000 Portugal MO, 5

Tags: right triangle , geometry

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]

2021 Thailand TSTST, 1

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

2018 Dutch IMO TST, 2

Tags: geometry , circumcircle , tangent , perpendicular , right triangle

Suppose a triangle $\vartriangle ABC$ with $\angle C = 90^o$ is given. Let $D$ be the midpoint of $AC$, and let $E$ be the foot of the altitude through $C$ on $BD$. Show that the tangent in $C$ of the circumcircle of $\vartriangle AEC$ is perpendicular to $AB$.

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

2007 Thailand Mathematical Olympiad, 3

Tags: geometry , right triangle

A triangle $\vartriangle ABC$ has $\angle B = 90^o$ . A circle is tangent to $AB$ at $B$ and also tangent to $AC$. Another circle is tangent to the first circle as well as the two sides $AB$ and $AC$. Suppose that $AB =\sqrt3$ and $BC = 3$. What is the radius of the second circle?

2014 May Olympiad, 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 $\angle CAP$

2006 Junior Tuymaada Olympiad, 1

Tags: geometry , right triangle , isosceles , equal segments

On the equal $ AC $ and $ BC $ of an isosceles right triangle $ ABC $ , points $ D $ and $ E $ are marked respectively, so that $ CD = CE $. Perpendiculars on the straight line $ AE $, passing through the points $ C $ and $ D $, intersect the side $ AB $ at the points $ P $ and $ Q $.Prove that $ BP = PQ $.

Estonia Open Junior - geometry, 2012.1.3

Tags: ratio , geometry , rectangle , right triangle

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 $| AF | = 2 | AB |$?

2006 Abels Math Contest (Norwegian MO), 4

Tags: right triangle , mixtilinear circle , circumcircle , inradius , incircle , tangent circle

Let $\gamma$ be the circumscribed circle about a right-angled triangle $ABC$ with right angle $C$. Let $\delta$ be the circle tangent to the sides $AC$ and $BC$ and tangent to the circle $\gamma$ internally. (a) Find the radius $i$ of $\delta$ in terms of $a$ when $AC$ and $BC$ both have length $a$. (b) Show that the radius $i$ is twice the radius of the inscribed circle of $ABC$.

2015 BMT Spring, 4

Tags: right triangle , area , geometry

Triangle $ABC$ has side lengths $AB = 3$, $BC = 4$, and $CD = 5$. Draw line $\ell_A$ such that $\ell_A$ is parallel to $BC$ and splits the triangle into two polygons of equal area. Define lines $\ell_B$ and $\ell_C$ analogously. The intersection points of $\ell_A$, $\ell_B$, and $\ell_C$ form a triangle. Determine its area.

Denmark (Mohr) - geometry, 1994.5

Tags: equal area , right triangle , isosceles , area , geometry

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

2007 Austria Beginners' Competition, 4

Tags: geometry , right triangle , parallelogram , right angle

Consider a parallelogram $ABCD$ such that the midpoint $M$ of the side $CD$ lies on the angle bisector of $\angle BAD$. Show that $\angle AMB$ is a right angle.

2002 May Olympiad, 3

Tags: right triangle , isosceles , square , geometry

In a triangle $ABC$, right in $A$ and isosceles, let $D$ be a point on the side $AC$ ($A \ne D \ne C$) and $E$ be the point on the extension of $BA$ such that the triangle $ADE$ is isosceles. Let $P$ be the midpoint of segment $BD$, $R$ be the midpoint of the segment $CE$ and $Q$ the intersection point of $ED$ and $BC$. Prove that the quadrilateral $ARQP$ is a square.

Ukrainian TYM Qualifying - geometry, 2017.4

Tags: integer , right triangle , geometry

Specify at least one right triangle $ABC$ with integer sides, inside which you can specify a point $M$ such that the lengths of the segments $MA, MB, MC$ are integers. Are there many such triangles, none of which are are similar?

2021 Estonia Team Selection Test, 2

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

2021 Taiwan TST Round 2, 4

Tags: geometry , right triangle , 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 BMT Spring, 4

Tags: geometry , right triangle

In a right triangle, the altitude from a vertex to the hypotenuse splits the hypotenuse into two segments of lengths $a$ and $b$. If the right triangle has area $T$ and is inscribed in a circle of area $C$, find $ab$ in terms of $T$ and $C$.

1964 Poland - Second Round, 5

Tags: geometry , 3d geometry , right triangle

Given is a trihedral angle with edges $ SA $, $ SB $, $ SC $, all plane angles of which are acute, and the dihedral angle at edge $ SA $ is right. Prove that the section of this triangle with any plane perpendicular to any edge, at a point different from the vertex $ S $, is a right triangle.