Olympiad problems

2010 Tournament Of Towns, 6

In acute triangle $ABC$, an arbitrary point $P$ is chosen on altitude $AH$. Points $E$ and $F$ are the midpoints of sides $CA$ and $AB$ respectively. The perpendiculars from $E$ to $CP$ and from $F$ to $BP$ meet at point $K$. Prove that $KB = KC$.

1990 Austrian-Polish Competition, 1

Tags: geometric transformation , trapezoid , geometry , perpendicular bisector , reflection

The distinct points $X_1, X_2, X_3, X_4, X_5, X_6$ all lie on the same side of the line $AB$. The six triangles $ABX_i$ are all similar. Show that the $X_i$ lie on a circle.

2006 IMO Shortlist, 5

Tags: circumcircle , perpendicular bisector , angle chasing , geometry

In triangle $ABC$, let $J$ be the center of the excircle tangent to side $BC$ at $A_{1}$ and to the extensions of the sides $AC$ and $AB$ at $B_{1}$ and $C_{1}$ respectively. Suppose that the lines $A_{1}B_{1}$ and $AB$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $DJ$. Determine the angles $\angle{BEA_{1}}$ and $\angle{AEB_{1}}$. [i]Proposed by Dimitris Kontogiannis, Greece[/i]

2013 India IMO Training Camp, 2

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

Let $ABCD$ by a cyclic quadrilateral with circumcenter $O$. Let $P$ be the point of intersection of the diagonals $AC$ and $BD$, and $K, L, M, N$ the circumcenters of triangles $AOP, BOP$, $COP, DOP$, respectively. Prove that $KL = MN$.

2012 JBMO ShortLists, 2

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

Let $ABC$ be an isosceles triangle with $AB=AC$ . Let also $\omega$ be a circle of center $K$ tangent to the line $AC$ at $C$ which intersects the segment $BC$ again at $H$ . Prove that $HK \bot AB $.

2011 All-Russian Olympiad Regional Round, 9.6

Tags: circumcircle , perpendicular bisector , geometry

Initially, there are three different points on the plane. Every minute, three points are chosen, for example $A$, $B$ and $C$, and a new point $D$ is generated which is symmetric to $A$ with respect to the perpendicular bisector of line segment $BC$. 24 hours later, it turns out that among all the points that were generated, there exist three collinear points. Prove that the three initial points were also collinear. (Author: V. Shmarov)

2012 NIMO Problems, 5

Tags: angle bisector , geometry , incenter , perpendicular bisector

In convex hexagon $ABCDEF$, $\angle A \cong \angle B$, $\angle C \cong \angle D$, and $\angle E \cong \angle F$. Prove that the perpendicular bisectors of $\overline{AB}$, $\overline{CD}$, and $\overline{EF}$ pass through a common point. [i]Proposed by Lewis Chen[/i]

2011 All-Russian Olympiad Regional Round, 11.6

Tags: circumcircle , geometry , perpendicular bisector

$\omega$ is the circumcirle of an acute triangle $ABC$. The tangent line passing through $A$ intersects the tangent lines passing through points $B$ and $C$ at points $K$ and $L$, respectively. The line parallel to $AB$ through $K$ and the line parallel to $AC$ through $L$ intersect at point $P$. Prove that $BP=CP$. (Author: P. Kozhevnikov)

2014 India Regional Mathematical Olympiad, 1

Tags: geometry , circumcircle , incenter , perpendicular bisector

In an acute-angled triangle $ABC, \angle ABC$ is the largest angle. The perpendicular bisectors of $BC$ and $BA$ intersect AC at $X$ and $Y$ respectively. Prove that circumcentre of triangle $ABC$ is incentre of triangle $BXY$ .

1996 Taiwan National Olympiad, 3

Tags: perpendicular bisector , angle bisector , geometry

Let be given points $A,B$ on a circle and let $P$ be a variable point on that circle. Let point $M$ be determined by $P$ as the point that is either on segment $PA$ with $AM=MP+PB$ or on segment $PB$ with $AP+MP=PB$. Find the locus of points $M$.

2002 Mediterranean Mathematics Olympiad, 3

Tags: circumcircle , geometry , perpendicular bisector

In an acute-angled triangle $ABC$, $M$ and $N$ are points on the sides $AC$ and $BC$ respectively, and $K$ the midpoint of $MN$. The circumcircles of triangles $ACN$ and $BCM$ meet again at a point $D$. Prove that the line $CD$ contains the circumcenter $O$ of $\triangle ABC$ if and only if $K$ is on the perpendicular bisector of $AB.$

1995 Romania Team Selection Test, 1

Tags: incenter , circumcircle , ellipse , perpendicular bisector , conic , geometry

Let AD be the altitude of a triangle ABC and E , F be the incenters of the triangle ABD and ACD , respectively. line EF meets AB and AC at K and L. prove tht AK=AL if and only if AB=AC or A=90

2013 Vietnam National Olympiad, 3

Tags: geometry , geometric transformation , perpendicular bisector , reflection

Let $ABC$ be a triangle such that $ABC$ isn't a isosceles triangle. $(I)$ is incircle of triangle touches $BC,CA,AB$ at $D,E,F$ respectively. The line through $E$ perpendicular to $BI$ cuts $(I)$ again at $K$. The line through $F$ perpendicular to $CI$ cuts $(I)$ again at $L$.$J$ is midpoint of $KL$. [b]a)[/b] Prove that $D,I,J$ collinear. [b]b)[/b] $B,C$ are fixed points,$A$ is moved point such that $\frac{AB}{AC}=k$ with $k$ is constant.$IE,IF$ cut $(I)$ again at $M,N$ respectively.$MN$ cuts $IB,IC$ at $P,Q$ respectively. Prove that bisector perpendicular of $PQ$ through a fixed point.

2014 Contests, 3

Tags: geometry , circumcircle , ellipse , conic , perpendicular bisector , parallelogram

Let $B$ and $C$ be two fixed points on a circle centered at $O$ that are not diametrically opposed. Let $A$ be a variable point on the circle distinct from $B$ and $C$ and not belonging to the perpendicular bisector of $BC$. Let $H$ be the orthocenter of $\triangle ABC$, and $M$ and $N$ be the midpoints of the segments $BC$ and $AH$, respectively. The line $AM$ intersects the circle again at $D$, and finally, $NM$ and $OD$ intersect at $P$. Determine the locus of points $P$ as $A$ moves around the circle.

2015 China Girls Math Olympiad, 6

Tags: geometry , power of a point , radical axis , perpendicular bisector

Let $\Gamma_1$ and $\Gamma_2$ be two non-overlapping circles. $A,C$ are on $\Gamma_1$ and $B,D$ are on $\Gamma_2$ such that $AB$ is an external common tangent to the two circles, and $CD$ is an internal common tangent to the two circles. $AC$ and $BD$ meet at $E$. $F$ is a point on $\Gamma_1$, the tangent line to $\Gamma_1$ at $F$ meets the perpendicular bisector of $EF$ at $M$. $MG$ is a line tangent to $\Gamma_2$ at $G$. Prove that $MF=MG$.

2024 Sharygin Geometry Olympiad, 11

Tags: barycentric coordinates , perpendicular bisector , geometry

Let $M, N$ be the midpoints of sides $AB, AC$ respectively of a triangle $ABC$. The perpendicular bisector to the bisectrix $AL$ meets the bisectrixes of angles $B$ and $C$ at points $P$ and $Q$ respectively. Prove that the common point of lines $PM$ and $QN$ lies on the tangent to the circumcircle of $ABC$ at $A$.

1992 India National Olympiad, 9

Tags: trig identities , geometry , trigonometry , circumcircle , calculus , law of sines , perpendicular bisector

Let $A_1, A_2, \ldots, A_n$ be an $n$ -sided regular polygon. If $\frac{1}{A_1 A_2} = \frac{1}{A_1 A_3} + \frac{1}{A_1A_4}$, find $n$.

2020 Polish Junior MO Second Round, 2.

Tags: parallelogram , geometry , perpendicular bisector

Let $ABCD$ be the parallelogram, such that angle at vertex $A$ is acute. Perpendicular bisector of the segment $AB$ intersects the segment $CD$ in the point $X$. Let $E$ be the intersection point of the diagonals of the parallelogram $ABCD$. Prove that $XE = \frac{1}{2}AD$.

2006 Bulgaria National Olympiad, 2

Tags: circumcircle , cyclic quadrilateral , geometry , perpendicular bisector

The triangle $ABC$ is such that $\angle BAC=30^{\circ},\angle ABC=45^{\circ}$. Prove that if $X$ lies on the ray $AC$, $Y$ lies on the ray $BC$ and $OX=BY$, where $O$ is the circumcentre of triangle $ABC$, then $S_{XY}$ passes through a fixed point. [i]Emil Kolev [/i]

2021 Korea Junior Math Olympiad, 3

Tags: geometry , circumcircle , tangent , cyclic quadrilateral , perpendicular bisector

Let $ABCD$ be a cyclic quadrilateral with circumcircle $\Omega$ and let diagonals $AC$ and $BD$ intersect at $X$. Suppose that $AEFB$ is inscribed in a circumcircle of triangle $ABX$ such that $EF$ and $AB$ are parallel. $FX$ meets the circumcircle of triangle $CDX$ again at $G$. Let $EX$ meets $AB$ at $P$, and $XG$ meets $CD$ at $Q$. Denote by $S$ the intersection of the perpendicular bisector of $\overline{EG}$ and $\Omega$ such that $S$ is closer to $A$ than $B$. Prove that line through $S$ parallel to $PQ$ is tangent to $\Omega$.

2002 AMC 12/AHSME, 23

Tags: angle bisector , geometry , perpendicular bisector , trigonometry , trig identities , law of sines , area of a triangle

In triangle $ ABC$, side $ AC$ and the perpendicular bisector of $ BC$ meet in point $ D$, and $ BD$ bisects $ \angle ABC$. If $ AD \equal{} 9$ and $ DC \equal{} 7$, what is the area of triangle $ ABD$? $ \textbf{(A)}\ 14 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 14\sqrt5 \qquad \textbf{(E)}\ 28\sqrt5$

2012 Sharygin Geometry Olympiad, 3

Tags: angle bisector , perpendicular bisector , incenter , geometry

A circle with center $I$ touches sides $AB,BC,CA$ of triangle $ABC$ in points $C_{1},A_{1},B_{1}$. Lines $AI, CI, B_{1}I$ meet $A_{1}C_{1}$ in points $X, Y, Z$ respectively. Prove that $\angle Y B_{1}Z = \angle XB_{1}Z$.

2010 India National Olympiad, 5

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

Let $ ABC$ be an acute-angled triangle with altitude $ AK$. Let $ H$ be its ortho-centre and $ O$ be its circum-centre. Suppose $ KOH$ is an acute-angled triangle and $ P$ its circum-centre. Let $ Q$ be the reflection of $ P$ in the line $ HO$. Show that $ Q$ lies on the line joining the mid-points of $ AB$ and $ AC$.

1989 India National Olympiad, 7

Tags: circumcircle , perpendicular bisector , geometry

Let $ A$ be one of the two points of intersection of two circles with centers $ X, Y$ respectively.The tangents at $ A$ to the two circles meet the circles again at $ B, C$. Let a point $ P$ be located so that $ PXAY$ is a parallelogram. Show that $ P$ is also the circumcenter of triangle $ ABC$.

2014 Saudi Arabia Pre-TST, 3.3

Tags: incenter , geometry , perpendicular bisector , isosceles

Let $ABC$ be a triangle and $I$ its incenter. The line $AI$ intersects the side $BC$ at $D$ and the perpendicular bisector of $BC$ at $E$. Let $J$ be the incenter of triangle $CDE$. Prove that triangle $CIJ$ is isosceles.