Olympiad problems

2011 China Team Selection Test, 1

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

Let $H$ be the orthocenter of an acute trangle $ABC$ with circumcircle $\Gamma$. Let $P$ be a point on the arc $BC$ (not containing $A$) of $\Gamma$, and let $M$ be a point on the arc $CA$ (not containing $B$) of $\Gamma$ such that $H$ lies on the segment $PM$. Let $K$ be another point on $\Gamma$ such that $KM$ is parallel to the Simson line of $P$ with respect to triangle $ABC$. Let $Q$ be another point on $\Gamma$ such that $PQ \parallel BC$. Segments $BC$ and $KQ$ intersect at a point $J$. Prove that $\triangle KJM$ is an isosceles triangle.

2020 Tournament Of Towns, 2

Tags: incenter , geometry , circumcircle , circumradius , altitude

At heights $AA_0, BB_0, CC_0$ of an acute-angled non-equilateral triangle $ABC$, points $A_1, B_1, C_1$ were marked, respectively, so that $AA_1 = BB_1 = CC_1 = R$, where $R$ is the radius of the circumscribed circle of triangle $ABC$. Prove that the center of the circumscribed circle of the triangle $A_1B_1C_1$ coincides with the center of the inscribed circle of triangle $ABC$. E. Bakaev

2009 ISI B.Stat Entrance Exam, 7

Tags: geometry , trigonometry , complex number , circumcircle

Show that the vertices of a regular pentagon are concyclic. If the length of each side of the pentagon is $x$, show that the radius of the circumcircle is $\frac{x}{2\sin 36^\circ}$.

2012 France Team Selection Test, 3

Tags: geometry , parallelogram , circumcircle , perpendicular bisector , power of a point

Let $ABCD$ be a convex quadrilateral whose sides $AD$ and $BC$ are not parallel. Suppose that the circles with diameters $AB$ and $CD$ meet at points $E$ and $F$ inside the quadrilateral. Let $\omega_E$ be the circle through the feet of the perpendiculars from $E$ to the lines $AB,BC$ and $CD$. Let $\omega_F$ be the circle through the feet of the perpendiculars from $F$ to the lines $CD,DA$ and $AB$. Prove that the midpoint of the segment $EF$ lies on the line through the two intersections of $\omega_E$ and $\omega_F$. [i]Proposed by Carlos Yuzo Shine, Brazil[/i]

Kyiv City MO Seniors Round2 2010+ geometry, 2017.10.3

Tags: circumcircle , geometry , diameter , circles

Circles $w_1$ and $w_2$ with centers at points $O_1$ and $O_2$ respectively, intersect at points $A$ and $B$. A line passing through point $B$, intersects the circles $w_1$ and $w_2$ at points $C$ and $D$ other than $B$. Tangents to the circles $w_1$ and $w_2$ at points $C$ and $D$ intersect at point $E$. Line $EA$ intersects the circumscribed circle $w$ of triangle $AO_1O_2$ at point $F$. Prove that the length of the segment is $EF$ is equal to the diameter of the circle $w$. (Vovchenko V., Plotnikov M.)

2006 IMO Shortlist, 3

Tags: circumcircle , geometry , pentagon

Let $ ABCDE$ be a convex pentagon such that \[ \angle BAC \equal{} \angle CAD \equal{} \angle DAE\qquad \text{and}\qquad \angle ABC \equal{} \angle ACD \equal{} \angle ADE. \]The diagonals $BD$ and $CE$ meet at $P$. Prove that the line $AP$ bisects the side $CD$. [i]Proposed by Zuming Feng, USA[/i]

1997 IMO Shortlist, 20

Tags: circumcircle , geometry , cyclic quadrilateral , triangle

A quick solution: Let R be the foot of the perpend. from X to BC. Let's assume Q and R are in the interior of the segms AC and BC (respectively) and P in the ext of AD. P, R, Q are colinear (Simson's thm). PQ tangent to circle XRD iff XRQ=XDR iff Pi-XCA=XDR iff XBA=XDR=XDC=ADB iff XBC+ABC=ADB=DAC+ACB iff XAC+ABC=DAC+ACD iff ABC=ACD=ACB iff AB=AC. It's the same for all the other cases.

1999 Junior Balkan MO, 4

Tags: geometry , circumcircle , trigonometry , perpendicular bisector

Let $ABC$ be a triangle with $AB=AC$. Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$, and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$. Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$). [i]Greece[/i]

2023 China Second Round, 1

Tags: geometry , circumcircle

Let $A,B$ be two fixed points on a plane and $\Omega$ a fixed semicircle arc with diameter $AB$. Let $T$ be another fixed point on $\Omega$, and $\omega$ a fixed circle that passes through $A$ and $T$ and has its center in $\Delta ABT$. Let $P$ be a moving point on the arc $TB$ (endpoints excluded), and $C,D$ be two moving points on $\omega$ such that $C$ lies on segment $AP$, $C,D$ lies on different sides of line $AB$ and $CD\ \bot \ AB$. Denote the circumcenter of $\Delta CDP$ of $K$. Prove that (i) $K$ lies on the circumcircle of $\Delta TDP$. (ii) $K$ is a fixed point.

2018 Rioplatense Mathematical Olympiad, Level 3, 2

Tags: symmetry , equal segments , circumcircle , geometry

Let $P$ be a point outside a circumference $\Gamma$, and let $PA$ be one of the tangents from $P$ to $\Gamma$. Line $l$ passes through $P$ and intersects $\Gamma$ at $B$ and $C$, with $B$ between $P$ and $C$. Let $D$ be the symmetric of $B$ with respect to $P$. Let $\omega_1$ and $\omega_2$ be the circles circumscribed to the triangles $DAC$ and $PAB$ respectively. $\omega_1$ and $\omega _2$ intersect at $E \neq A$. Line $EB$ cuts back to $\omega _1 $ in $F$. Prove that $CF = AB$.

2011 Iran MO (2nd Round), 3

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

The line $l$ intersects the extension of $AB$ in $D$ ($D$ is nearer to $B$ than $A$) and the extension of $AC$ in $E$ ($E$ is nearer to $C$ than $A$) of triangle $ABC$. Suppose that reflection of line $l$ to perpendicular bisector of side $BC$ intersects the mentioned extensions in $D'$ and $E'$ respectively. Prove that if $BD+CE=DE$, then $BD'+CE'=D'E'$.

2011 Turkey MO (2nd round), 2

Tags: geometric transformation , geometry , circumcircle , trigonometry , reflection , cyclic quadrilateral , perpendicular bisector

Let $ABC$ be a triangle $D\in[BC]$ (different than $A$ and $B$).$E$ is the midpoint of $[CD]$. $F\in[AC]$ such that $\widehat{FEC}=90$ and $|AF|.|BC|=|AC|.|EC|.$ Circumcircle of $ADC$ intersect $[AB]$ at $G$ different than $A$.Prove that tangent to circumcircle of $AGF$ at $F$ is touch circumcircle of $BGE$ too.

2017 IMO, 4

Tags: circumcircle , geometry , tangent

Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$. [i]Proposed by Charles Leytem, Luxembourg[/i]

2019 Tuymaada Olympiad, 8

Tags: geometric transformation , circumcircle , geometry , arc midpoint , moving points , inversion , reflection

In $\triangle ABC$ $\angle B$ is obtuse and $AB \ne BC$. Let $O$ is the circumcenter and $\omega$ is the circumcircle of this triangle. $N$ is the midpoint of arc $ABC$. The circumcircle of $\triangle BON$ intersects $AC$ on points $X$ and $Y$. Let $BX \cap \omega = P \ne B$ and $BY \cap \omega = Q \ne B$. Prove that $P, Q$ and reflection of $N$ with respect to line $AC$ are collinear.

2024 Saint Petersburg Mathematical Olympiad, 3

Tags: circumcircle , geometry

In unequal triangle $ABC$ bisector $AK$ was drawn. Diameter $XY$ of its circumcircle is perpendicular to $AK$ (order of points on circumcircle is $B-X-A-Y-C$). A circle, passing on points $X$ and $Y$, intersect segments $BK$ and $CK$ in points $T$ and $Z$ respectively. Prove that if $KZ=KT$, then $XT \perp YZ$.

2018 Korea Junior Math Olympiad, 5

Tags: circumcircle , geometric transformation , reflection , geometry

Let there be an acute scalene triangle $ABC$ with circumcenter $O$. Denote $D,E$ be the reflection of $O$ with respect to $AB,AC$, respectively. The circumcircle of $ADE$ meets $AB$, $AC$, the circumcircle of $ABC$ at points $K,L,M$, respectively, and they are all distinct from $A$. Prove that the lines $BC,KL,AM$ are concurrent.

2021 Iran MO (3rd Round), 2

Tags: geometry , circumcircle

Given an acute triangle $ABC$ let $M$ be the midpoint of $AB$. Point $K$ is given on the other side of line $AC$ from that of point $B$ such that $\angle KMC = 90 ^ \circ $ and $\angle KAC = 180^\circ - \angle ABC$. The tangent to circumcircle of triangle $ABC$ at $A$ intersects line $CK$ at $E$. Prove that the reflection of line $BC$ with respect to $CM$ passes through the midpoint of line segment $ME$.

2012 Bosnia And Herzegovina - Regional Olympiad, 4

Tags: geometry , circumcircle

In triangle $ABC$ point $O$ is circumcenter. Point $T$ is centroid of $ABC$, and points $D$, $E$ and $F$ are circumcenters of triangles $TBC$, $TCA$ and $TAB$. Prove that $O$ is centroid of $DEF$

2006 Peru IMO TST, 4

Tags: circumcircle , parallelogram , geometry , similar triangles , perpendicular bisector

[color=blue][size=150]PERU TST IMO - 2006[/size] Saturday, may 20.[/color] [b]Question 04[/b] In an actue-angled triangle $ABC$ draws up: its circumcircle $w$ with center $O$, the circumcircle $w_1$ of the triangle $AOC$ and the diameter $OQ$ of $w_1$. The points are chosen $M$ and $N$ on the straight lines $AQ$ and $AC$, respectively, in such a way that the quadrilateral $AMBN$ is a parallelogram. Prove that the intersection point of the straight lines $MN$ and $BQ$ belongs the circumference $w_1.$ --- [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=88513]Spanish version[/url] $\text{\LaTeX}{}$ed by carlosbr

1998 Korea Junior Math Olympiad, 7

Tags: geometry , angle bisector , circumcircle

$O$ is a circumcircle of non-isosceles triangle $ABC$ and the angle bisector of $A$ meets $BC$ at $D$. If the line perpendicular to $BC$ passing through $D$ meets $AO$ at $E$, show that $ADE$ is an isosceles triangle.

2017 Poland - Second Round, 4

Tags: geometry , circumcircle

Incircle of a triangle $ABC$ touches $AB$ and $AC$ at $D$ and $E$, respectively. Point $J$ is the excenter of $A$. Points $M$ and $N$ are midpoints of $JD$ and $JE$. Lines $BM$ and $CN$ cross at point $P$. Prove that $P$ lies on the circumcircle of $ABC$.

2009 Czech and Slovak Olympiad III A, 6

Tags: circumcircle , geometry , perpendicular bisector

Given two fixed points $O$ and $G$ in the plane. Find the locus of the vertices of triangles whose circumcenters and centroids are $O$ and $G$ respectively.

2014 Vietnam National Olympiad, 3

Tags: geometry , circumcircle , combinatorics

Given a regular 103-sided polygon. 79 vertices are colored red and the remaining vertices are colored blue. Let $A$ be the number of pairs of adjacent red vertices and $B$ be the number of pairs of adjacent blue vertices. a) Find all possible values of pair $(A,B).$ b) Determine the number of pairwise non-similar colorings of the polygon satisfying $B=14.$ 2 colorings are called similar if they can be obtained from each other by rotating the circumcircle of the polygon.

2006 Romania Team Selection Test, 3

Tags: circumcircle , incenter , geometry , power of a point , radical axis

Let $\gamma$ be the incircle in the triangle $A_0A_1A_2$. For all $i\in\{0,1,2\}$ we make the following constructions (all indices are considered modulo 3): $\gamma_i$ is the circle tangent to $\gamma$ which passes through the points $A_{i+1}$ and $A_{i+2}$; $T_i$ is the point of tangency between $\gamma_i$ and $\gamma$; finally, the common tangent in $T_i$ of $\gamma_i$ and $\gamma$ intersects the line $A_{i+1}A_{i+2}$ in the point $P_i$. Prove that a) the points $P_0$, $P_1$ and $P_2$ are collinear; b) the lines $A_0T_0$, $A_1T_1$ and $A_2T_2$ are concurrent.

2005 Germany Team Selection Test, 2

Tags: geometry , circumcircle , homothety , triangle

Let $O$ be the circumcenter of an acute-angled triangle $ABC$ with ${\angle B<\angle C}$. The line $AO$ meets the side $BC$ at $D$. The circumcenters of the triangles $ABD$ and $ACD$ are $E$ and $F$, respectively. Extend the sides $BA$ and $CA$ beyond $A$, and choose on the respective extensions points $G$ and $H$ such that ${AG=AC}$ and ${AH=AB}$. Prove that the quadrilateral $EFGH$ is a rectangle if and only if ${\angle ACB-\angle ABC=60^{\circ }}$. [i]Proposed by Hojoo Lee, Korea[/i]