Olympiad problems

2003 All-Russian Olympiad Regional Round, 9.1

Prove that the sides of any equilateral triangle you can either increase everything or decrease everything by the same amount so that you get a right triangle.

Estonia Open Senior - geometry, 1999.2.3

Tags: incircle , circumcircle , geometric inequality , area of a triangle , geometry , right triangle , area

Two right triangles are given, of which the incircle of the first triangle is the circumcircle of the second triangle. Let the areas of the triangles be $S$ and $S'$ respectively. Prove that $\frac{S}{S'} \ge 3 +2\sqrt2$

2017 Grand Duchy of Lithuania, 3

Tags: parallel , right triangle , circumcircle , geometry

Let $ABC$ be a triangle with $\angle A = 90^o$ and let $D$ be an orthogonal projection of $A$ onto $BC$. The midpoints of $AD$ and $AC$ are called $E$ and $F$, respectively. Let $M$ be the circumcentre of $\vartriangle BEF$. Prove that $AC\parallel BM$.

Ukrainian From Tasks to Tasks - geometry, 2010.5

Tags: right triangle , geometry

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 $B_1C_1$.

1990 Mexico National Olympiad, 2

Tags: geometry , right triangle , inradius

$ABC$ is a triangle with $\angle B = 90^o$ and altitude $BH$. The inradii of $ABC, ABH, CBH$ are $r, r_1, r_2$. Find a relation between them.

1998 IMO Shortlist, 8

Tags: geometry , harmonic division , inversion , right triangle

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]

2001 German National Olympiad, 6 (12)

Tags: geometry , tangent , circumcircle , right triangle

Let $ABC$ be a triangle with $\angle A = 90^o$ and $\angle B < \angle C$. The tangent at $A$ to the circumcircle $k$ of $\vartriangle ABC$ intersects line $BC$ at $D$. Let $E$ be the reflection of $A$ in $BC$. Also, let $X$ be the feet of the perpendicular from $A$ to $BE$ and let $Y$ be the midpoint of $AX$. Line $BY$ meets $k$ again at $Z$. Prove that line $BD$ is tangent to the circumcircle of $\vartriangle ADZ$.

Ukrainian TYM Qualifying - geometry, 2016.1

Tags: geometry , orthocenter , right triangle

The points $K$ and $N$ lie on the hypotenuse $AB$ of a right triangle $ABC$. Prove that orthocenters the triangles $BCK$ and $ACN$ coincide if and only if $\frac{BN}{AK}=\tan^2 A.$

2009 Moldova National Olympiad, 7.4

Tags: geometry , right triangle

Triangle $ABC$ with $AB = 10$ cm ¸and $\angle C= 15^o$, is right at $B$. Point $D \in (AC)$ is the foot of the altitude taken from $B$. Find the distance from point $D$ to the line $AB$.

2009 Brazil Team Selection Test, 1

Tags: right triangle , orthocenter , geometry , cyclic , pentagon

Let $A, B, C, D, E$ points in circle of radius r, in that order, such that $AC = BD = CE = r$. The points $H_1, H_2, H_3$ are the orthocenters of the triangles $ACD$, $BCD$ and $BCE$, respectively. Prove that $H_1H_2H_3$ is a right triangle .

Kyiv City MO 1984-93 - geometry, 1989.8.5

Tags: geometry , right triangle , equilateral , inscribed , construction

The student drew a right triangle $ABC$ on the board with a right angle at the vertex $B$ and inscribed in it an equilateral triangle $KMP$ such that the points $K, M, P$ lie on the sides $AB, BC, AC$, respectively, and $KM \parallel AC$. Then the picture was erased, leaving only points $A, P$ and $C$. Restore erased points and lines.

2021 Azerbaijan IMO TST, 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 $\angle ACB = 90^{\circ}$.

2010 Saudi Arabia BMO TST, 3

Tags: geometry , ratio , geometric inequality , right triangle

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 $$\left( \frac{PB}{PA}\right)^ 2+\left( \frac{QC}{QA}\right)^ 2$$

2016 India PRMO, 4

Tags: geometry , right triangle

Consider a right-angled triangle $ABC$ with $\angle C = 90^o$. Suppose that the hypotenuse $AB$ is divided into four equal parts by the points $D,E,F$, such that $AD = DE = EF = FB$. If $CD^2 +CE^2 +CF^2 = 350$, find the length of $AB$.

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 $\angle CAP$

1999 Junior Balkan Team Selection Tests - Moldova, 2

Tags: geometry , right triangle , isosceles , angle

Let $ABC$ be an isosceles right triangle with $\angle A=90^o$. Point $D$ is the midpoint of the side $[AC]$, and point $E \in [AC]$ is so that $EC = 2AE$. Calculate $\angle AEB + \angle ADB$ .

2010 Grand Duchy of Lithuania, 4

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

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 $BK = 2DC$.

1955 Moscow Mathematical Olympiad, 288

Tags: geometry , area , right triangle , bisects segment , angle

We are given a right triangle $ABC$ and the median $BD$ drawn from the vertex $B$ of the right angle. Let the circle inscribed in $\vartriangle ABD$ be tangent to side $AD$ at $K$. Find the angles of $\vartriangle ABC$ if $K$ divides $AD$ in halves.

2019 District Olympiad, 4

Tags: geometry , equilateral , right triangle , isosceles

Consider the isosceles right triangle$ ABC, \angle A = 90^o$, and point $D \in (AB)$ such that $AD = \frac13 AB$. In the half-plane determined by the line $AB$ and point $C$ , consider a point $E$ such that $\angle BDE = 60^o$ and $\angle DBE = 75^o$. Lines $BC$ and $DE$ intersect at point $G$, and the line passing through point $G$ parallel to the line $AC$ intersects the line $BE$ at point $H$. Prove that the triangle $CEH$ is equilateral.

1951 Poland - Second Round, 1

Tags: geometry , incircle , right triangle

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 $.

2010 Austria Beginners' Competition, 4

Tags: geometry , right triangle , angle

In the right-angled triangle $ABC$ with a right angle at $C$, the side $BC$ is longer than the side $AC$. The perpendicular bisector of $AB$ intersects the line $BC$ at point $D$ and the line $AC$ at point $E$. The segments $DE$ has the same length as the side $AB$. Find the measures of the angles of the triangle $ABC$. (R. Henner, Vienna)

1998 Romania National Olympiad, 3

Tags: geometry , isosceles , right triangle

In the exterior of the triangle $ABC$ with $m(\angle B) > 45^o$, $m(\angle C) >45°^o$ one constructs the right isosceles triangles $ACM$ and $ABN$ such that $m(\angle CAM) = m(\angle BAN) = 90^o$ and, in the interior of $ABC$, the right isosceles triangle $BCP$, with $m(\angle P) = 90^o$. Show that the triangle $MNP$ is a right isosceles triangle.

Denmark (Mohr) - geometry, 1992.2

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 $\frac{1}{a}+\frac{1}{b}=\frac{1}{r}$.

2016 Bulgaria EGMO TST, 2

Tags: geometry , similar triangles , right triangle , circumcircle

Let $ABC$ be a right triangle with $\angle ACB = 90^{\circ}$ and centroid $G$. The circumcircle $k_1$ of triangle $AGC$ and the circumcircle $k_2$ of triangle $BGC$ intersect $AB$ at $P$ and $Q$, respectively. The perpendiculars from $P$ and $Q$ respectively to $AC$ and $BC$ intersect $k_1$ and $k_2$ at $X$ and $Y$. Determine the value of $\frac{CX \cdot CY}{AB^2}$.

1959 AMC 12/AHSME, 7

Tags: ratio , right triangle

The sides of a right triangle are $a, a+d,$ and $a+2d$, with $a$ and $d$ both positive. The ratio of $a$ to $d$ is: $ \textbf{(A)}\ 1:3 \qquad\textbf{(B)}\ 1:4 \qquad\textbf{(C)}\ 2:1\qquad\textbf{(D)}\ 3:1\qquad\textbf{(E)}\ 3:4 $