Olympiad problems

Kvant 2024, M2809

Given is a triangle $ABC$ and the points $M, P$ lie on the segments $AB, BC$, respectively, such that $AM=BC$ and $CP=BM$. If $AP$ and $CM$ meet at $O$ and $2\angle AOM=\angle ABC$, find the measure of $\angle ABC$.

2003 Oral Moscow Geometry Olympiad, 2

Tags: ratio , geometry , right angle , equal segments , equal angles

In a convex quadrilateral $ABCD$, $\angle ABC = 90^o$ , $\angle BAC = \angle CAD$, $AC = AD, DH$ is the alltitude of the triangle $ACD$. In what ratio does the line $BH$ divide the segment $CD$?

2013 Sharygin Geometry Olympiad, 1

Tags: geometry , pentagon , right angle , equal segments

Let $ABCDE$ be a pentagon with right angles at vertices $B$ and $E$ and such that $AB = AE$ and $BC = CD = DE$. The diagonals $BD$ and $CE$ meet at point $F$. Prove that $FA = AB$.

2017 Singapore Senior Math Olympiad, 2

Tags: geometry , right angle , perpendicular , cyclic quadrilateral , cyclic

In the cyclic quadrilateral $ABCD$, the sides $AB, DC$ meet at $Q$, the sides $AD,BC$ meet at $P, M$ is the midpoint of $BD$, If $\angle APQ=90^o$, prove that $PM$ is perpendicular to $AB$.

2009 China Northern MO, 2

Tags: geometry , angle bisector , right angle , trigonometry

In an acute triangle $ABC$ , $AB>AC$ , $ \cos B+ \cos C=1$ , $E,F$ are on the extend line of $AB,AC$ such that $\angle ABF = \angle ACE = 90$ . (1) Prove :$BE+CF=EF$ ; (2) Assume the bisector of $\angle EBC$ meet $EF$ at $P$ , prove that $CP$ is the bisector of $\angle BCF$. [img]https://cdn.artofproblemsolving.com/attachments/a/2/c554c2bc0b4e044c45f88138568f5234d544a8.png[/img]

2018 Oral Moscow Geometry Olympiad, 2

Tags: geometry , trapezoid , perpendicular , right angle , diagonal

The diagonals of the trapezoid $ABCD$ are perpendicular ($AD//BC, AD>BC$) . Point $M$ is the midpoint of the side of $AB$, the point $N$ is symmetric of the center of the circumscribed circle of the triangle $ABD$ wrt $AD$. Prove that $\angle CMN = 90^o$. (A. Mudgal, India)

2009 Belarus Team Selection Test, 3

Tags: trapezoid , right angle , parallel , tangential quadrilateral , tangential , geometry

Given trapezoid $ABCD$ ($AD\parallel BC$) with $AD \perp AB$ and $T=AC\cap BD$. A circle centered at point $O$ is inscribed in the trapezoid and touches the side $CD$ at point $Q$. Let $P$ be the intersection point (different from $Q$) of the side $CD$ and the circle passing through $T,Q$ and $O$. Prove that $TP \parallel AD$. I. Voronovich

2016 Saudi Arabia BMO TST, 2

Tags: geometry , circles , right angle

A circle with center $O$ passes through points $A$ and $C$ and intersects the sides $AB$ and $BC$ of triangle $ABC$ at points $K$ and $N$, respectively. The circumcircles of triangles $ABC$ and $KBN$ meet at distinct points $B$ and $M$. Prove that $\angle OMB = 90^o$.

2024 Polish MO Finals, 6

Tags: right angle , parallelogram , circumcircle , geometry

Let $ABCD$ be a parallelogram. Let $X \notin AC $ lie inside $ABCD$ so that $\angle AXB = \angle CXD = 90^ {\circ}$ and $\Omega$ denote the circumcircle of $AXC$. Consider a diameter $EF$ of $\Omega$ and assume neither $E, \ X, \ B$ nor $F, \ X, \ D$ are collinear. Let $K \neq X$ be an intersection point of circumcircles of $BXE$ and $DXF$ and $L \neq X$ be such point on $\Omega$ so that $\angle KXL = 90^{\circ}$. Prove that $AB = KL$.

2018 Iranian Geometry Olympiad, 2

Tags: geometry , right angle , angle

In convex quadrilateral $ABCD$, the diagonals $AC$ and $BD$ meet at the point $P$. We know that $\angle DAC = 90^o$ and $2 \angle ADB = \angle ACB$. If we have $ \angle DBC + 2 \angle ADC = 180^o$ prove that $2AP = BP$. Proposed by Iman Maghsoudi

Ukrainian From Tasks to Tasks - geometry, 2010.13

Tags: geometry , pentagon , right angle , tangential

You can inscribe a circle in the pentagon $ABCDE$. It is also known that $\angle ABC = \angle BAE = \angle CDE = 90^o$. Find the measure of the angle $ADB$.

2023 Euler Olympiad, Round 2, 4

Tags: trapezoid , euler , right angle , geometry , harmonic bundles

Let $ABCD$ be a trapezoid, with $AD \parallel BC$, let $M$ be the midpoint of $AD$, and let $C_1$ be symmetric point to $C$ with respect to line $BD$. Segment $BM$ meets diagonal $AC$ at point $K$, and ray $C_1K$ meets line $BD$ at point $H$. Prove that $\angle{AHD}$ is a right angle. [i]Proposed by Giorgi Arabidze, Georgia[/i]

1979 Austrian-Polish Competition, 1

Tags: geometry , square , right angle , equal segments

On sides $AB$ and $BC$ of a square $ABCD$ the respective points $E$ and $F$ have been chosen so that $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. Prove that $\angle DNF = 90$.

Brazil L2 Finals (OBM) - geometry, 2004.5

Tags: geometry , circumcenter , right triangle , 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 $O_1O_2$ .

Denmark (Mohr) - geometry, 2016.3

Tags: geometry , right angle , equal segments , area

Prove that all quadrilaterals $ABCD$ where $\angle B = \angle D = 90^o$, $|AB| = |BC|$ and $|AD| + |DC| = 1$, have the same area. [img]https://1.bp.blogspot.com/-55lHuAKYEtI/XzRzDdRGDPI/AAAAAAAAMUk/n8lYt3fzFaAB410PQI4nMEz7cSSrfHEgQCLcBGAsYHQ/s0/2016%2Bmohr%2Bp3.png[/img]

2016 Thailand TSTST, 3

Tags: geometry , orthocenter , right angle , parallel

Let $H$ be the orthocenter of acute-angled $\vartriangle ABC$, and $X, Y$ points on the ray $AB, AC$. ($B$ lies between $X, A$, and $C$ lies between $Y, A$.) Lines $HX, HY$ intersect $BC$ at $D, E$ respectively. Let the line through $D$ parallel to $AC$ intersect $XY$ at $Z$. Prove that $\angle XHY = 90^o$ if and only if $ZE \parallel AB$.

2006 Oral Moscow Geometry Olympiad, 3

Tags: geometry , geometric inequality , inradius , right angle

On the sides $AB, BC$ and $AC$ of the triangle $ABC$, points $C', A'$ and $B'$ are selected, respectively, so that the angle $A'C'B'$ is right. Prove that the segment $A'B'$ is longer than the diameter of the inscribed circle of the triangle $ABC$. (M. Volchkevich)

2019 Adygea Teachers' Geometry Olympiad, 4

Tags: geometry , trapezoid , tangential , right angle , trapezium

From which two statements about the trapezoid follows the third: 1) the trapezoid is tangential, 2) the trapezoid is right, 3) its area is equal to the product of the bases?

Kyiv City MO Juniors Round2 2010+ geometry, 2021.9.2

Tags: geometry , circumcircle , right angle

In an acute triangle $AB$ the heights $ BE$ and $CF$ intersect at the orthocenter $H$, and $M$ is the midpoint of $BC$. The line $EF$ intersects the lines $MH$ and $BC$ at the points $P$ and $T$ , respectively. $AP$ intersects the cirumcscribed circle of $\vartriangle ABC$ for second time at the point $Q$ . Prove that $\angle AQT= 90^o$. (Fedir Yudin)

2018 Dutch IMO TST, 3

Tags: geometry , right angle , equal segments

Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude through $A$. On $AD$, there are distinct points $E$ and $F$ such that $|AE| = |BE|$ and $|AF| =|CF|$. A point$ T \ne D$ satises $\angle BTE = \angle CTF = 90^o$. Show that $|TA|^2 =|TB| \cdot |TC|$.

2004 District Olympiad, 4

Tags: trigonometry , 3d geometry , palnes , trapezoid , right angle , angle , geometry

In the right trapezoid $ABCD$ with $AB \parallel CD, \angle B = 90^o$ and $AB = 2DC$. At points $A$ and $D$ there is therefore a part of the plane $(ABC)$ perpendicular to the plane of the trapezoid, on which the points $N$ and $P$ are taken, ($AP$ and $PD$ are perpendicular to the plane) such that $DN = a$ and $AP = \frac{a}{2}$ . Knowing that $M$ is the midpoint of the side $BC$ and the triangle $MNP$ is equilateral, determine: a) the cosine of the angle between the planes $MNP$ and $ABC$. b) the distance from $D$ to the plane $MNP$

2018 Grand Duchy of Lithuania, 3

Tags: geometry , right angle , circumcenter

The altitudes $AD$ and $BE$ of an acute triangle $ABC$ intersect at point $H$. Let $F$ be the intersection of the line $AB$ and the line that is parallel to the side BC and goes through the circumcenter of $ABC$. Let $M$ be the midpoint of the segment $AH$. Prove that $\angle CMF = 90^o$

2012 Dutch IMO TST, 4

Tags: geometry , right angle , equal angles , angle chasing

Let $\vartriangle ABC$ be a triangle. The angle bisector of $\angle CAB$ intersects$ BC$ at $L$. On the interior of line segments $AC$ and $AB$, two points, $M$ and $N$, respectively, are chosen in such a way that the lines $AL, BM$ and $CN$ are concurrent, and such that $\angle AMN = \angle ALB$. Prove that $\angle NML = 90^o$.

2020 Thailand TST, 1

Tags: geometry , circumcircle , right angle , right triangle , tangent circle

Let $ABC$ be a triangle with circumcircle $\Gamma$. Let $\omega_0$ be a circle tangent to chord $AB$ and arc $ACB$. For each $i = 1, 2$, let $\omega_i$ be a circle tangent to $AB$ at $T_i$ , to $\omega_0$ at $S_i$ , and to arc $ACB$. Suppose $\omega_1 \ne \omega_2$. Prove that there is a circle passing through $S_1, S_2, T_1$, and $T_2$, and tangent to $\Gamma$ if and only if $\angle ACB = 90^o$. .

2024 Bulgaria National Olympiad, 2

Tags: conditional geometry , angle measure , construction , right angle , beautiful geometry

Given is a triangle $ABC$ and the points $M, P$ lie on the segments $AB, BC$, respectively, such that $AM=BC$ and $CP=BM$. If $AP$ and $CM$ meet at $O$ and $2\angle AOM=\angle ABC$, find the measure of $\angle ABC$.