Olympiad problems

2014 May Olympiad, 4

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 $\angle CAP$

2003 May Olympiad, 2

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

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 $PQR$ .

2019 Denmark MO - Mohr Contest, 5

Tags: equal segments , equilateral , right triangle , geometry

In the figure below the triangles $BCD, CAE$ and $ABF$ are equilateral, and the triangle $ABC$ is right-angled with $\angle A = 90^o$. Prove that $|AD| = |EF|$. [img]https://1.bp.blogspot.com/-QMMhRdej1x8/XzP18QbsXOI/AAAAAAAAMUI/n53OsE8rwZcjB_zpKUXWXq6bg3o8GUfSwCLcBGAsYHQ/s0/2019%2Bmohr%2Bp5.png[/img]

2019 Girls in Mathematics Tournament, 2

Tags: incenter , geometry , right triangle , circumcenter

Let $ABC$ be a right triangle with hypotenuse $BC$ and center $I$. Let bisectors of the angles $\angle B$ and $\angle C$ intersect the sides $AC$ and $AB$ in$ D$ and $E$, respectively. Let $P$ and $Q$ be the feet of the perpendiculars of the points $D$ and $E$ on the side $BC$. Prove that $I$ is the circumcenter of $APQ$.

2010 Saudi Arabia BMO TST, 3

Tags: geometric inequality , right triangle , ratio , geometry

Let $ABC$ be a right angled triangle with $\angle A = 90^o$and $BC = a$, $AC = b$, $AB = c$. Let $d$ be a line passing trough the incenter of triangle and intersecting the sides $AB$ and $AC$ in $P$ and $Q$, respectively. (a) Prove that $$b \cdot \left( \frac{PB}{PA}\right)+ c \cdot \left( \frac{QC}{QA}\right) =a$$ (b) Find the minimum of $$\left( \frac{PB}{PA}\right)^ 2+\left( \frac{QC}{QA}\right)^ 2$$

1974 Chisinau City MO, 76

Tags: isosceles , right triangle , equal angles , geometry

Altitude $AH$ and median $AM$ of the triangle $ABC$ satisfy the relation: $\angle ABM = \angle CBH$. Prove that triangle $ABC$ is isosceles or right-angled.

2020 BMT Fall, Tie 3

Tags: incircle , right triangle , geometry

$\vartriangle ABC$ has $AB = 5$, $BC = 12$, and $AC = 13$. A circle is inscribed in $\vartriangle ABC$, and $MN$ tangent to the circle is drawn such that $M$ is on $\overline{AC}$, $N$ is on $\overline{BC}$, and $\overline{MN} \parallel \overline{AB}$. The area of $\vartriangle MNC$ is $m/n$ , where $m$ and $n $are relatively prime positive integers. Find $m + n$.

1925 Eotvos Mathematical Competition, 3

Tags: geometric inequalities , right triangle , geometry , inradius

Let $r$ be the radius of the inscribed circle of a right triangle $ABC$. Show that $r$ is less than half of either leg and less than one fourth of the hypotenuse.

Novosibirsk Oral Geo Oly VIII, 2021.5

Tags: right triangle , geometry , 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 $BP = PQ$.

2000 Estonia National Olympiad, 2

Tags: right triangle , cyclic quadrilateral , bisects segment

Let $PQRS$ be a cyclic quadrilateral with $\angle PSR = 90^o$, and let $H,K$ be the projections of $Q$ on the lines $PR$ and $PS$, respectively. Prove that the line $HK$ passes through the midpoint of the segment $SQ$.

2017 Germany, Landesrunde - Grade 11/12, 5

Tags: inequalities , geometric inequality , right triangle , inradius , geometry

In a right-angled triangle let $r$ be the inradius and $s_a,s_b$ be the lengths of the medians of the legs $a,b$. Prove the inequality \[ \frac{r^2}{s_a^2+s_b^2} \leq \frac{3-2 \sqrt2}{5}. \]

2012 Sharygin Geometry Olympiad, 4

Tags: right triangle , locus , square , geometry

Consider a square. Find the locus of midpoints of the hypothenuses of rightangled triangles with the vertices lying on three different sides of the square and not coinciding with its vertices. (B.Frenkin)

2013 Costa Rica - Final Round, 3

Tags: right triangle , concyclic , geometry

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 Harvard-MIT Mathematics Tournament., 2

Tags: right triangle , geometry

Let $ABC$ be a right triangle with $\angle A= 90^{\circ}$. A circle $\omega$ centered on $BC$ is tangent to $AB$ at $D$ and $AC$ at $E$. Let $F$ and $G$ be the intersections of $\omega$ and $BC$ so that $F$ lies between $B$ and $G$. If lines $DG$ and $EF$ intersect at $X$, show that $AX=AD.$

Brazil L2 Finals (OBM) - geometry, 2004.5

Tags: geometry , right triangle , circumcenter , right angle , tangent

Let $D$ be the midpoint of the hypotenuse $AB$ of a right triangle $ABC$. Let $O_1$ and $O_2$ be the circumcenters of the $ADC$ and $DBC$ triangles, respectively. a) Prove that $\angle O_1DO_2$ is right. b) Prove that $AB$ is tangent to the circle of diameter $O_1O_2$ .

2011 Peru MO (ONEM), 3

Tags: parallelogram , geometry , right triangle , perpendicular , midpoint

Let $ABC$ be a right triangle, right in $B$. Inner bisectors are drawn $CM$ and $AN$ that intersect in $I$. Then, the $AMIP$ and $CNIQ$ parallelograms are constructed. Let $U$ and $V$ are the midpoints of the segments $AC$ and $PQ$, respectively. Prove that $UV$ is perpendicular to $AC$.

Ukrainian From Tasks to Tasks - geometry, 2014.15

Tags: construction , right triangle , geometry

Construct a right triangle given the hypotenuse and the median drawn to the leg.

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]

1996 Tournament Of Towns, (485) 3

Tags: right triangle , geometry , angle

The two tangents to the incircle of a right-angled triangle $ABC$ that are perpendicular to the hypotenuse $AB$ intersect it at the points $P$ and $Q$. Find $\angle PCQ$. (M Evdokimov,)

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 $C$.

2021 Sharygin Geometry Olympiad, 10-11.7

Tags: similar triangles , humpty point , concurrency , right triangle , geometry

Let $I$ be the incenter of a right-angled triangle $ABC$, and $M$ be the midpoint of hypothenuse $AB$. The tangent to the circumcircle of $ABC$ at $C$ meets the line passing through $I$ and parallel to $AB$ at point $P$. Let $H$ be the orthocenter of triangle $PAB$. Prove that lines $CH$ and $PM$ meet at the incircle of triangle $ABC$.

1951 Poland - Second Round, 1

Tags: geometry , right triangle , incircle

In a right triangle $ ABC $, the altitude $ CD $ is drawn from the vertex of the right angle $ C $ and a circle is inscribed in each of the triangles $ ABC $, $ ACD $ and $ BCD $. Prove that the sum of the radii of these circles equals the height $ CD $.

1999 Yugoslav Team Selection Test, Problem 2

Tags: right triangle , harmonic division , inversion , geometry

Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$. The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ in the line $BC$, let $X$ be the foot of the perpendicular from $A$ to $BE$, and let $Y$ be the midpoint of the segment $AX$. Let the line $BY$ intersect the circle $\omega$ again at $Z$. Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$. [hide="comment"] [i]Edited by Orl.[/i] [/hide]

2000 May Olympiad, 2

Tags: geometry , midpoint , right triangle , angle bisector

Let $ABC$ be a right triangle in $A$ , whose leg measures $1$ cm. The bisector of the angle $BAC$ cuts the hypotenuse in $R$, the perpendicular to $AR$ on $R$ , cuts the side $AB$ at its midpoint. Find the measurement of the side $AB$ .

2010 Grand Duchy of Lithuania, 4

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

In the triangle $ABC$ angle $C$ is a right angle. On the side $AC$ point $D$ has been found, and on the segment $BD$ point K has been found such that $\angle ABC = \angle KAD = \angle AKD$. Prove that $BK = 2DC$.