Olympiad problems

Cono Sur Shortlist - geometry, 2005.G5

Let $O$ be the circumcenter of an acute triangle $ABC$ and $A_1$ a point of the minor arc $BC$ of the circle $ABC$ . Let $A_2$ and $A_3$ be points on sides $AB$ and $AC$ respectively such that $\angle BA_1A_2=\angle OAC$ and $\angle CA_1A_3=\angle OAB$ . Points $B_2, B_3, C_2$ and $C_3$ are similarly constructed, with $B_2$ in $BC, B_3$ in $AB, C_2$ in $AC$ and $C_3$ in $BC$. Prove that lines $A_2A_3, B_2B_3$ and $C_2C_3$ are concurrent.

2008 Germany Team Selection Test, 3

Tags: circumcircle , reflection , cyclic quadrilateral , mixtilinear incircle , geometry

Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$. Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$. [i]Author: Farzan Barekat, Canada[/i]

1986 IMO Shortlist, 15

Tags: geometry , circumcircle , ratio , trigonometry

Let $ABCD$ be a convex quadrilateral whose vertices do not lie on a circle. Let $A'B'C'D'$ be a quadrangle such that $A',B', C',D'$ are the centers of the circumcircles of triangles $BCD,ACD,ABD$, and $ABC$. We write $T (ABCD) = A'B'C'D'$. Let us define $A''B''C''D'' = T (A'B'C'D') = T (T (ABCD)).$ [b](a)[/b] Prove that $ABCD$ and $A''B''C''D''$ are similar. [b](b) [/b]The ratio of similitude depends on the size of the angles of $ABCD$. Determine this ratio.

Brazil L2 Finals (OBM) - geometry, 2006.5

Tags: geometry , circumcircle

Let $ABC$ be an acute triangle with orthocenter $H$. Let $M$, $N$ and $R$ be the midpoints of $AB$, $BC$ an $AH$, respectively. If $A\hat{B}C=70^\large\circ$, compute $M\hat{N}R$.

2017 Sharygin Geometry Olympiad, P11

Tags: geometry , circumcircle

A finite number of points is marked on the plane. Each three of them are not collinear. A circle is circumscribed around each triangle with marked vertices. Is it possible that all centers of these circles are also marked? [i]Proposed by A.Tolesnikov[/i]

2012 Romania Team Selection Test, 2

Tags: euler , geometry , circumcircle , parallelogram , geometric transformation , reflection , cyclic quadrilateral

Let $ABCD$ be a cyclic quadrilateral such that the triangles $BCD$ and $CDA$ are not equilateral. Prove that if the Simson line of $A$ with respect to $\triangle BCD$ is perpendicular to the Euler line of $BCD$, then the Simson line of $B$ with respect to $\triangle ACD$ is perpendicular to the Euler line of $\triangle ACD$.

1995 IMO, 1

Tags: geometry , circumcircle , concurrency

Let $ A,B,C,D$ be four distinct points on a line, in that order. The circles with diameters $ AC$ and $ BD$ intersect at $ X$ and $ Y$. The line $ XY$ meets $ BC$ at $ Z$. Let $ P$ be a point on the line $ XY$ other than $ Z$. The line $ CP$ intersects the circle with diameter $ AC$ at $ C$ and $ M$, and the line $ BP$ intersects the circle with diameter $ BD$ at $ B$ and $ N$. Prove that the lines $ AM,DN,XY$ are concurrent.

2014 Turkey EGMO TST, 1

Tags: geometry , circumcircle

Let $D$ be the midpoint of the side $BC$ of a triangle $ABC$ and $AD$ intersect the circumcircle of $ABC$ for the second time at $E$. Let $P$ be the point symmetric to the point $E$ with respect to the point $D$ and $Q$ be the point of intersection of the lines $CP$ and $AB$. Prove that if $A,C,D,Q$ are concyclic, then the lines $BP$ and $AC$ are perpendicular.

1998 China Team Selection Test, 1

Tags: circumcircle , incenter , trigonometry , geometry

In acute-angled $\bigtriangleup ABC$, $H$ is the orthocenter, $O$ is the circumcenter and $I$ is the incenter. Given that $\angle C > \angle B > \angle A$, prove that $I$ lies within $\bigtriangleup BOH$.

2006 MOP Homework, 5

Tags: incenter , circumcircle , trigonometry , geometric transformation , reflection , geometry

Let $ABC$ be an acute triangle with $AC \neq BC$. Points $H$ and $I$ are the orthocenter and incenter of the triangle, respectively. Line $CH$ and $CI$ meet the circumcircle of triangle $ABC$ again at $D$ and $L$ (other than $C$), respectively. Prove that $\angle CIH=90^{\circ}$ if and only if $\angle IDL=90^{\circ}$.

2003 All-Russian Olympiad, 4

Tags: circumcircle , geometry

Let $B$ and $C$ be arbitrary points on sides $AP$ and $PD$ respectively of an acute triangle $APD$. The diagonals of the quadrilateral $ABCD$ meet at $Q$, and $H_1,H_2$ are the orthocenters of triangles $APD$ and $BPC$, respectively. Prove that if the line $H_1H_2$ passes through the intersection point $X \ (X \neq Q)$ of the circumcircles of triangles $ABQ$ and $CDQ$, then it also passes through the intersection point $Y \ (Y \neq Q)$ of the circumcircles of triangles $BCQ$ and $ADQ.$

2022 Singapore MO Open, Q1

Tags: geometry , tangent , symmedian , circumcircle

For $\triangle ABC$ and its circumcircle $\omega$, draw the tangents at $B,C$ to $\omega$ meeting at $D$. Let the line $AD$ meet the circle with center $D$ and radius $DB$ at $E$ inside $\triangle ABC$. Let $F$ be the point on the extension of $EB$ and $G$ be the point on the segment $EC$ such that $\angle AFB=\angle AGE=\angle A$. Prove that the tangent at $A$ to the circumcircle of $\triangle AFG$ is parallel to $BC$. [i]Proposed by 61plus[/i]

2018 Moscow Mathematical Olympiad, 10

Tags: circumcircle , geometry

$ABC$ is acute-angled triangle, $AA_1,CC_1$ are altitudes. $M$ is centroid. $M$ lies on circumcircle of $A_1BC_1$. Find all values of $\angle B$

2016 China Team Selection Test, 1

Tags: geometry , circumcircle , geometric transformation , reflection

$P$ is a point in the interior of acute triangle $ABC$. $D,E,F$ are the reflections of $P$ across $BC,CA,AB$ respectively. Rays $AP,BP,CP$ meet the circumcircle of $\triangle ABC$ at $L,M,N$ respectively. Prove that the circumcircles of $\triangle PDL,\triangle PEM,\triangle PFN$ meet at a point $T$ different from $P$.

2002 Romania National Olympiad, 2

Tags: ratio , geometry , geometric transformation , reflection , circumcircle , inradius , trigonometry

Let $ABC$ be a right triangle where $\measuredangle A = 90^\circ$ and $M\in (AB)$ such that $\frac{AM}{MB}=3\sqrt{3}-4$. It is known that the symmetric point of $M$with respect to the line $GI$ lies on $AC$. Find the measure of $\measuredangle B$.

1989 USAMO, 4

Tags: geometry , circumcircle , incenter , inequalities

Let $ABC$ be an acute-angled triangle whose side lengths satisfy the inequalities $AB < AC < BC$. If point $I$ is the center of the inscribed circle of triangle $ABC$ and point $O$ is the center of the circumscribed circle, prove that line $IO$ intersects segments $AB$ and $BC$.

2004 Germany Team Selection Test, 2

Tags: parallelogram , circumcircle , geometric transformation , reflection , perpendicular bisector , geometry

Let two chords $AC$ and $BD$ of a circle $k$ meet at the point $K$, and let $O$ be the center of $k$. Let $M$ and $N$ be the circumcenters of triangles $AKB$ and $CKD$. Show that the quadrilateral $OMKN$ is a parallelogram.

1995 IMO Shortlist, 1

Tags: geometry , circumcircle , concurrency

Let $ A,B,C,D$ be four distinct points on a line, in that order. The circles with diameters $ AC$ and $ BD$ intersect at $ X$ and $ Y$. The line $ XY$ meets $ BC$ at $ Z$. Let $ P$ be a point on the line $ XY$ other than $ Z$. The line $ CP$ intersects the circle with diameter $ AC$ at $ C$ and $ M$, and the line $ BP$ intersects the circle with diameter $ BD$ at $ B$ and $ N$. Prove that the lines $ AM,DN,XY$ are concurrent.

2002 Italy TST, 1

Tags: circumcircle , geometry

Given that in a triangle $ABC$, $AB=3$, $BC=4$ and the midpoints of the altitudes of the triangle are collinear, find all possible values of the length of $AC$.

2008 Rioplatense Mathematical Olympiad, Level 3, 2

Tags: circumcircle , trigonometry , cyclic quadrilateral , angle bisector , geometry

In triangle $ABC$, where $AB<AC$, let $X$, $Y$, $Z$ denote the points where the incircle is tangent to $BC$, $CA$, $AB$, respectively. On the circumcircle of $ABC$, let $U$ denote the midpoint of the arc $BC$ that contains the point $A$. The line $UX$ meets the circumcircle again at the point $K$. Let $T$ denote the point of intersection of $AK$ and $YZ$. Prove that $XT$ is perpendicular to $YZ$.

2020 Silk Road, 2

Tags: circumcircle , circles

The triangle $ ABC $ is inscribed in the circle $ \omega $. Points $ K, L, M $ are marked on the sides $ AB, BC, CA $, respectively, and $ CM \cdot CL = AM \cdot BL $. Ray $ LK $ intersects line $ AC $ at point $ P $. The common chord of the circle $ \omega $ and the circumscribed circle of the triangle $ KMP $ meets the segment $ AM $ at the point $ S $. Prove that $ SK \parallel BC $.

2018 Azerbaijan BMO TST, 4

Tags: circumcircle , projective geometry , concurrency , geometry

Let $ABC$ be an acute angled triangle with orthocenter $H$. centroid $G$ and circumcircle $\omega$. Let $D$ and $M$ respectively be the intersection of lines $AH$ and $AG$ with side $BC$. Rays $MH$ and $DG$ interect $ \omega$ again at $P$ and $Q$ respectively. Prove that $PD$ and $QM$ intersect on $\omega$.

2008 May Olympiad, 4

Tags: geometry , circumcircle

Let $ABF$ be a right-angled triangle with $\angle AFB = 90$, a square $ABCD$ is externally to the triangle. If $FA = 6$, $FB = 8$ and $E$ is the circumcenter of the square $ABCD$, determine the value of $EF$

2010 Tuymaada Olympiad, 2

Tags: geometry , circumcircle , geometric transformation , reflection , parallelogram , calculus , trigonometry

Let $ABC$ be an acute triangle, $H$ its orthocentre, $D$ a point on the side $[BC]$, and $P$ a point such that $ADPH$ is a parallelogram. Show that $\angle BPC > \angle BAC$.

2016 Saudi Arabia IMO TST, 1

Tags: orthocenter , circumcircle , geometry

Let $ABC$ be a triangle inscribed in the circle $(O)$. The bisector of $\angle BAC$ cuts the circle $(O)$ again at $D$. Let $DE$ be the diameter of $(O)$. Let $G$ be a point on arc $AB$ which does not contain $C$. The lines $GD$ and $BC$ intersect at $F$. Let $H$ be a point on the line $AG$ such that $FH \parallel AE$. Prove that the circumcircle of triangle $HAB$ passes through the orthocenter of triangle $HAC$.