Olympiad problems

2012 Balkan MO, 1

Let $A$, $B$ and $C$ be points lying on a circle $\Gamma$ with centre $O$. Assume that $\angle ABC > 90$. Let $D$ be the point of intersection of the line $AB$ with the line perpendicular to $AC$ at $C$. Let $l$ be the line through $D$ which is perpendicular to $AO$. Let $E$ be the point of intersection of $l$ with the line $AC$, and let $F$ be the point of intersection of $\Gamma$ with $l$ that lies between $D$ and $E$. Prove that the circumcircles of triangles $BFE$ and $CFD$ are tangent at $F$.

2011 Czech and Slovak Olympiad III A, 5

Tags: circumcircle , geometry

In acute triangle ABC, which is not equilateral, let $P$ denote the foot of the altitude from $C$ to side $AB$; let $H$ denote the orthocenter; let $O$ denote the circumcenter; let $D$ denote the intersection of line $CO$ with $AB$; and let $E$ denote the midpoint of $CD$. Prove that line $EP$ passes through the midpoint of $OH$.

2015 Korea National Olympiad, 2

Tags: geometry , circumcircle , trapezoid

An isosceles trapezoid $ABCD$, inscribed in $\omega$, satisfies $AB=CD, AD<BC, AD<CD$. A circle with center $D$ and passing $A$ hits $BD, CD, \omega$ at $E, F, P(\not= A)$, respectively. Let $AP \cap EF = Q$, and $\omega$ meet $CQ$ and the circumcircle of $\triangle BEQ$ at $R(\not= C), S(\not= B)$, respectively. Prove that $\angle BER= \angle FSC$.

2001 Tournament Of Towns, 5

Tags: circumcircle , tetrahedron , geometry , 3d geometry

Nine points are drawn on the surface of a regular tetrahedron with an edge of $1$ cm. Prove that among these points there are two located at a distance (in space) no greater than $0.5$ cm.

2010 Contests, 1

Tags: geometry , circumcircle , perpendicular bisector

Let $ABC$ be a triangle with $\angle BAC \neq 90^{\circ}.$ Let $O$ be the circumcenter of the triangle $ABC$ and $\Gamma$ be the circumcircle of the triangle $BOC.$ Suppose that $\Gamma$ intersects the line segment $AB$ at $P$ different from $B$, and the line segment $AC$ at $Q$ different from $C.$ Let $ON$ be the diameter of the circle $\Gamma.$ Prove that the quadrilateral $APNQ$ is a parallelogram.

2021 Israel TST, 3

Tags: geometry , incenter , circumcircle

In an inscribed quadrilateral $ABCD$, we have $BC=CD$ but $AB\neq AD$. Points $I$ and $J$ are the incenters of triangles $ABC$ and $ACD$ respectively. Point $K$ was chosen on segment $AC$ so that $IK=JK$. Points $M$ and $N$ are the incenters of triangles $AIK$ and $AJK$. Prove that the perpendicular to $CD$ at $D$ and the perpendicular to $KI$ at $I$ intersect on the circumcircle of $MAN$.

2000 IMO Shortlist, 8

Tags: incircle , geometry , homothety , circumcircle , triangle

Let $ AH_1, BH_2, CH_3$ be the altitudes of an acute angled triangle $ ABC$. Its incircle touches the sides $ BC, AC$ and $ AB$ at $ T_1, T_2$ and $ T_3$ respectively. Consider the symmetric images of the lines $ H_1H_2, H_2H_3$ and $ H_3H_1$ with respect to the lines $ T_1T_2, T_2T_3$ and $ T_3T_1$. Prove that these images form a triangle whose vertices lie on the incircle of $ ABC$.

Geometry Mathley 2011-12, 8.3

Tags: circumcircle , geometry , concyclic , orthocenter

Let $ABC$ be a scalene triangle, $(O)$ and $H$ be the circumcircle and its orthocenter. A line through $A$ is parallel to $OH$ meets $(O)$ at $K$. A line through $K$ is parallel to $AH$, intersecting $(O)$ again at $L$. A line through $L$ parallel to $OA$ meets $OH$ at $E$. Prove that $B,C,O,E$ are on the same circle. Trần Quang Hùng

2015 Irish Math Olympiad, 8

Tags: perpendicular , circumcircle , geometry

In triangle $\triangle ABC$, the angle $\angle BAC$ is less than $90^o$. The perpendiculars from $C$ on $AB$ and from $B$ on $AC$ intersect the circumcircle of $\triangle ABC$ again at $D$ and $E$ respectively. If $|DE| =|BC|$, find the measure of the angle $\angle BAC$.

2010 AIME Problems, 14

Tags: trigonometry , ratio , geometry , quadratic , circumcircle

In right triangle $ ABC$ with right angle at $ C$, $ \angle BAC < 45$ degrees and $ AB \equal{} 4$. Point $ P$ on $ AB$ is chosen such that $ \angle APC \equal{} 2\angle ACP$ and $ CP \equal{} 1$. The ratio $ \frac{AP}{BP}$ can be represented in the form $ p \plus{} q\sqrt{r}$, where $ p,q,r$ are positive integers and $ r$ is not divisible by the square of any prime. Find $ p\plus{}q\plus{}r$.

2009 Greece Team Selection Test, 2

Tags: circumcircle , geometry , perpendicular bisector

Given is a triangle $ABC$ with barycenter $G$ and circumcenter $O$.The perpendicular bisectors of $GA,GB,GC$ intersect at $A_1,B_1,C_1$.Show that $O$ is the barycenter of $\triangle{A_1B_1C_1}$.

2019 Turkey Junior National Olympiad, 3

Tags: incenter , geometry , circumcircle

In $ABC$ triangle $I$ is incenter and incircle of $ABC$ tangents to $BC,AC,AB$ at $D,E,F$, respectively. If $AI$ intersects $DE$ and $DF$ at $P$ and $Q$, prove that the circumcenter of $DPQ$ triangle is the midpoint of $BC$.

2021 ITAMO, 2

Tags: circumcircle , geometry

Let $ABC$ a triangle and let $I$ be the center of its inscribed circle. Let $D$ be the symmetric point of $I$ with respect to $AB$ and $E$ be the symmetric point of $I$ with respect to $AC$. Show that the circumcircles of the triangles $BID$ and $CIE$ are eachother tangent.

2002 Poland - Second Round, 2

Tags: geometry , circumcircle

In a convex quadrilateral $ABCD$, both $\angle ADB=2\angle ACB$ and $\angle BDC=2\angle BAC$. Prove that $AD=CD$.

2024 German National Olympiad, 5

Tags: circumcircle , circles , geometry , concyclic

Let $\triangle ABC$ be a triangle and let $X$ be a point in the interior of the triangle. The second intersection points of the lines $XA,XB$ and $XC$ with the circumcircle of $\triangle ABC$ are $P,Q$ and $R$. Let $U$ be a point on the ray $XP$ (these are the points on the line $XP$ such that $P$ and $U$ lie on the same side of $X$). The line through $U$ parallel to $AB$ intersects $BQ$ in $V$ . The line through $U$ parallel to $AC$ intersects $CR$ in $W$. Prove that $Q, R, V$ , and $W$ lie on a circle.

2023 4th Memorial "Aleksandar Blazhevski-Cane", P4

Tags: cyclic quadrilateral , geometry , angle bisector , segment sum , circumcenter , circumcircle

Let $ABCD$ be a cyclic quadrilateral such that $AB = AD + BC$ and $CD < AB$. The diagonals $AC$ and $BD$ intersect at $P$, while the lines $AD$ and $BC$ intersect at $Q$. The angle bisector of $\angle APB$ meets $AB$ at $T$. Show that the circumcenter of the triangle $CTD$ lies on the circumcircle of the triangle $CQD$. [i]Proposed by Nikola Velov[/i]

2022 Taiwan TST Round 2, G

Tags: circumcircle , incenter , geometry

Let $I$, $O$, $H$, and $\Omega$ be the incenter, circumcenter, orthocenter, and the circumcircle of the triangle $ABC$, respectively. Assume that line $AI$ intersects with $\Omega$ again at point $M\neq A$, line $IH$ and $BC$ meets at point $D$, and line $MD$ intersects with $\Omega$ again at point $E\neq M$. Prove that line $OI$ is tangent to the circumcircle of triangle $IHE$. [i]Proposed by Li4 and Leo Chang.[/i]

2003 China Team Selection Test, 1

Tags: circumcircle , trigonometry , geometry , angle bisector

$ABC$ is an acute-angled triangle. Let $D$ be the point on $BC$ such that $AD$ is the bisector of $\angle A$. Let $E, F$ be the feet of perpendiculars from $D$ to $AC,AB$ respectively. Suppose the lines $BE$ and $CF$ meet at $H$. The circumcircle of triangle $AFH$ meets $BE$ at $G$ (apart from $H$). Prove that the triangle constructed from $BG$, $GE$ and $BF$ is right-angled.

2010 All-Russian Olympiad, 2

Tags: angle bisector , geometry , geometric transformation , circumcircle

Into triangle $ABC$ gives point $K$ lies on bisector of $ \angle BAC$. Line $CK$ intersect circumcircle $ \omega$ of triangle $ABC$ at $M \neq C$. Circle $ \Omega$ passes through $A$, touch $CM$ at $K$ and intersect segment $AB$ at $P \neq A$ and $\omega $ at $Q \neq A$. Prove, that $P$, $Q$, $M$ lies at one line.

2015 Iran MO (3rd round), 5

Tags: geometry , incenter , circumcircle

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $R$ be the radius of circumcircle of $\triangle ABC$. Let $A',B',C'$ be the points on $\overrightarrow{AH},\overrightarrow{BH},\overrightarrow{CH}$ respectively such that $AH.AA'=R^2,BH.BB'=R^2,CH.CC'=R^2$. Prove that $O$ is incenter of $\triangle A'B'C'$.

2012 Indonesia MO, 3

Tags: geometry , circumcircle , angle bisector , perpendicular

Given an acute triangle $ABC$ with $AB>AC$ that has circumcenter $O$. Line $BO$ and $CO$ meet the bisector of $\angle BAC$ at $P$ and $Q$, respectively. Moreover, line $BQ$ and $CP$ meet at $R$. Show that $AR$ is perpendicular to $BC$. [i]Proposer: Soewono and Fajar Yuliawan[/i]

2011 AMC 12/AHSME, 25

Tags: inequalities , calculus , trigonometry , function , geometry , incenter , circumcircle

Triangle $ABC$ has $\angle BAC=60^\circ$, $\angle CBA \le 90^\circ$, $BC=1$, and $AC \ge AB$. Let $H$, $I$, and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$, respectively. Assume that the area of the pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$? $\textbf{(A)}\ 60 ^\circ \qquad \textbf{(B)}\ 72 ^\circ\qquad \textbf{(C)}\ 75 ^\circ \qquad \textbf{(D)}\ 80 ^\circ\qquad \textbf{(E)}\ 90 ^\circ$

2007 Romania Team Selection Test, 4

Tags: homothety , circumcircle , cyclic quadrilateral , geometry , geometric transformation , euler , inequalities

The points $M, N, P$ are chosen on the sides $BC, CA, AB$ of a triangle $\Delta ABC$, such that the triangle $\Delta MNP$ is acute-angled. We denote with $x$ the length of the shortest altitude of the triangle $\Delta ABC$, and with $X$ the length of the longest altitudes of the triangle $\Delta MNP$. Prove that $x \leq 2X$.

2005 Rioplatense Mathematical Olympiad, Level 3, 2

Tags: trapezoid , circumcircle , greatest common divisor , parallelogram , geometry , geometric transformation , reflection

In trapezoid $ABCD$, the sum of the lengths of the bases $AB$ and $CD$ is equal to the length of the diagonal $BD$. Let $M$ denote the midpoint of $BC$, and let $E$ denote the reflection of $C$ about the line $DM$. Prove that $\angle AEB=\angle ACD$.

2010 Kazakhstan National Olympiad, 5

Tags: circumcircle , geometry

Arbitrary triangle $ABC$ is given (with $AB<BC$). Let $M$ - midpoint of $AC$, $N$- midpoint of arc $AC$ of circumcircle $ABC$, which is contains point $B$. Let $I$ - in-center of $ABC$. Proved, that $ \angle IMA = \angle INB$