Olympiad problems

2014 Contests, 1

$ABCD$ is a cyclic quadrilateral, with diagonals $AC,BD$ perpendicular to each other. Let point $F$ be on side $BC$, the parallel line $EF$ to $AC$ intersect $AB$ at point $E$, line $FG$ parallel to $BD$ intersect $CD$ at $G$. Let the projection of $E$ onto $CD$ be $P$, projection of $F$ onto $DA$ be $Q$, projection of $G$ onto $AB$ be $R$. Prove that $QF$ bisects $\angle PQR$.

2016 Iranian Geometry Olympiad, 4

Tags: perpendicular bisector , right triangle , geometry

In a right-angled triangle $ABC$ ($\angle A = 90^o$), the perpendicular bisector of $BC$ intersects the line $AC$ in $K$ and the perpendicular bisector of $BK$ intersects the line $AB$ in $L$. If the line $CL$ be the internal bisector of angle $C$, find all possible values for angles $B$ and $C$. by Mahdi Etesami Fard

2006 IberoAmerican, 3

Tags: algebra , function , induction , domain , strong induction , perpendicular bisector , geometry

Consider a regular $n$-gon with $n$ odd. Given two adjacent vertices $A_{1}$ and $A_{2},$ define the sequence $(A_{k})$ of vertices of the $n$-gon as follows: For $k\ge 3,\, A_{k}$ is the vertex lying on the perpendicular bisector of $A_{k-2}A_{k-1}.$ Find all $n$ for which each vertex of the $n$-gon occurs in this sequence.

Indonesia MO Shortlist - geometry, g11

Tags: perpendicular bisector , ratio , geometry

Given triangle $ABC$ and point $P$ on the circumcircle of triangle $ABC$. Suppose the line $CP$ intersects line $AB$ at point $E$ and line $BP$ intersect line $AC$ at point $F$. Suppose also the perpendicular bisector of $AB$ intersects $AC$ at point $K$ and the perpendicular bisector of $AC$ intersects $AB$ at point $J$. Prove that $$\left( \frac{CE}{BF}\right)^2= \frac{AJ \cdot JE }{ AK \cdot KF}$$

2015 District Olympiad, 2

Tags: geometry , perpendicular bisector

Let $ ABC $ be an obtuse triangle with $ AB=AC, M $ the symmetric point of $ A $ with respect to $ C, $ and $ P $ the intersection of the line $ AB $ with the perpendicular bisector of the segment $ \overline{AB} . $ Knowing that $ PM $ is perpendicular to $ BC, $ show that $ APM $ is equilateral.

Geometry Mathley 2011-12, 9.4

Tags: geometry , incenter , fixed , fixed line , perpendicular bisector

Let $ABC$ be a triangle inscribed in a circle $(O)$, and $M$ be some point on the perpendicular bisector of $BC$. Let $I_1, I_2$ be the incenters of triangles $MAB,MAC$. Prove that the incenters of triangles $A_II_1I_2$ are on a fixed line when $M$ varies on the perpendicular bisector. Trần Quang Hùng

2014 India Regional Mathematical Olympiad, 5

Tags: circumcircle , incenter , perpendicular bisector , asymptote , geometry , angle bisector

Let $ABC$ be a triangle and let $X$ be on $BC$ such that $AX=AB$. let $AX$ meet circumcircle $\omega$ of triangle $ABC$ again at $D$. prove that circumcentre of triangle $BDX$ lies on $\omega$.

2012 Uzbekistan National Olympiad, 3

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

The inscribed circle $\omega$ of the non-isosceles acute-angled triangle $ABC$ touches the side $BC$ at the point $D$. Suppose that $I$ and $O$ are the centres of inscribed circle and circumcircle of triangle $ABC$ respectively. The circumcircle of triangle $ADI$ intersects $AO$ at the points $A$ and $E$. Prove that $AE$ is equal to the radius $r$ of $\omega$.

2015 Sharygin Geometry Olympiad, 6

Tags: geometry , median , perpendicular bisector , perpendicular

Lines $b$ and $c$ passing through vertices $B$ and $C$ of triangle $ABC$ are perpendicular to sideline $BC$. The perpendicular bisectors to $AC$ and $AB$ meet $b$ and $c$ at points $P$ and $Q$ respectively. Prove that line $PQ$ is perpendicular to median $AM$ of triangle $ABC$. (D. Prokopenko)

2005 All-Russian Olympiad, 1

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

Given a parallelogram $ABCD$ with $AB<BC$, show that the circumcircles of the triangles $APQ$ share a second common point (apart from $A$) as $P,Q$ move on the sides $BC,CD$ respectively s.t. $CP=CQ$.

2013 USA Team Selection Test, 2

Tags: circumcircle , geometry , perpendicular bisector , similar triangles

Let $ABC$ to be an acute triangle. Also, let $K$ and $L$ to be the two intersections of the perpendicular from $B$ with respect to side $AC$ with the circle of diameter $AC$, with $K$ closer to $B$ than $L$. Analogously, $X$ and $Y$ are the two intersections of the perpendicular from $C$ with respect to side $AB$ with the circle of diamter $AB$, with $X$ closer to $C$ than $Y$. Prove that the intersection of $XL$ and $KY$ lies on $BC$.

2025 Thailand Mathematical Olympiad, 7

Tags: geometry , circumcircle , perpendicular bisector , angle bisector

Let $ABC$ be a triangle with $AB < AC$. The tangent to the circumcircle of $\triangle ABC$ at $A$ intersects $BC$ at $D$. The angle bisector of $\angle BAC$ intersect $BC$ at $E$. Suppose that the perpendicular bisector of $AE$ intersect $AB, AC$ at $P,Q$, respectively. Show that $$\sqrt{\frac{BP}{CQ}} = \frac{AC \cdot BD}{AB \cdot CD}$$

2014 Oral Moscow Geometry Olympiad, 4

Tags: circumcircle , concurrency , angle bisector , geometry , perpendicular bisector

In triangle $ABC$, the perpendicular bisectors of sides $AB$ and $BC$ intersect side $AC$ at points $P$ and $Q$, respectively, with point $P$ lying on the segment $AQ$. Prove that the circumscribed circles of the triangles $PBC$ and $QBA$ intersect on the bisector of the angle $PBQ$.

2013 ITAMO, 5

Tags: rhombus , geometry , perpendicular bisector , parallelogram , circumcircle

$ABC$ is an isosceles triangle with $AB=AC$ and the angle in $A$ is less than $60^{\circ}$. Let $D$ be a point on $AC$ such that $\angle{DBC}=\angle{BAC}$. $E$ is the intersection between the perpendicular bisector of $BD$ and the line parallel to $BC$ passing through $A$. $F$ is a point on the line $AC$ such that $FA=2AC$ ($A$ is between $F$ and $C$). Show that $EB$ and $AC$ are parallel and that the perpendicular from $F$ to $AB$, the perpendicular from $E$ to $AC$ and $BD$ are concurrent.

2002 IMO Shortlist, 3

Tags: incenter , circumcircle , geometry , angle chasing , perpendicular bisector

The circle $S$ has centre $O$, and $BC$ is a diameter of $S$. Let $A$ be a point of $S$ such that $\angle AOB<120{{}^\circ}$. Let $D$ be the midpoint of the arc $AB$ which does not contain $C$. The line through $O$ parallel to $DA$ meets the line $AC$ at $I$. The perpendicular bisector of $OA$ meets $S$ at $E$ and at $F$. Prove that $I$ is the incentre of the triangle $CEF.$

2006 MOP Homework, 1

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

$ ABC$ is an acute triangle. The points $ B'$ and $ C'$are the reflections of $ B$ and $ C$ across the lines $ AC$ and $ AB$ respectively. Suppose that the circumcircles of triangles$ ABB$' and $ ACC'$ meet at $ A$ and $ P$. Prove that the line $ PA$ passes through the circumcenter of triangle$ ABC.$

2007 Romania Team Selection Test, 2

Tags: geometry , trigonometry , perpendicular bisector

Let $ ABC$ be a triangle, let $ E, F$ be the tangency points of the incircle $ \Gamma(I)$ to the sides $ AC$, respectively $ AB$, and let $ M$ be the midpoint of the side $ BC$. Let $ N \equal{} AM \cap EF$, let $ \gamma(M)$ be the circle of diameter $ BC$, and let $ X, Y$ be the other (than $ B, C$) intersection points of $ BI$, respectively $ CI$, with $ \gamma$. Prove that \[ \frac {NX} {NY} \equal{} \frac {AC} {AB}. \] [i]Cosmin Pohoata[/i]

2004 JBMO Shortlist, 5

Tags: geometry , angle bisector , circumcircle , symmetry , perpendicular bisector

Let $ABC$ be an isosceles triangle with $AC=BC$, let $M$ be the midpoint of its side $AC$, and let $Z$ be the line through $C$ perpendicular to $AB$. The circle through the points $B$, $C$, and $M$ intersects the line $Z$ at the points $C$ and $Q$. Find the radius of the circumcircle of the triangle $ABC$ in terms of $m = CQ$.

2018 Korea - Final Round, 4

Tags: geometry , perpendicular bisector

Triangle $ABC$ satisfies $\angle C=90^{\circ}$. A circle passing $A,B$ meets segment $AC$ at $G(\neq A,C)$ and it meets segment $BC$ at point $D(\neq B)$. Segment $AD$ cuts segment $BG$ at $H$, and let $l$, the perpendicular bisector of segment $AD$, cuts the perpendicular bisector of segment $AB$ at point $E$. A line passing $D$ is perpendicular to $DE$ and cuts $l$ at point $F$. If the circumcircle of triangle $CFH$ cuts $AC$, $BC$ at $P(\neq C),Q(\neq C)$ respectively, then prove that $PQ$ is perpendicular to $FH$.

2007 All-Russian Olympiad, 6

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

Let $ABC$ be an acute triangle. The points $M$ and $N$ are midpoints of $AB$ and $BC$ respectively, and $BH$ is an altitude of $ABC$. The circumcircles of $AHN$ and $CHM$ meet in $P$ where $P\ne H$. Prove that $PH$ passes through the midpoint of $MN$. [i]V. Filimonov[/i]

1993 ITAMO, 4

Tags: geometry , projection , similar triangles , perpendicular bisector

Let $P$ be a point in the plane of a triangle $ABC$, different from its circumcenter. Prove that the triangle whose vertices are the projections of $P$ on the perpendicular bisectors of the sides of $ABC$, is similar to $ABC$.

1967 IMO Shortlist, 2

Tags: tetrahedron , 3d geometry , perpendicular bisector , geometry

Let $ABCD$ be a regular tetrahedron. To an arbitrary point $M$ on one edge, say $CD$, corresponds the point $P = P(M)$ which is the intersection of two lines $AH$ and $BK$, drawn from $A$ orthogonally to $BM$ and from $B$ orthogonally to $AM$. What is the locus of $P$ when $M$ varies ?

2014 PUMaC Geometry A, 3

Tags: circumcircle , perpendicular bisector , geometry , trigonometry

Let $O$ be the circumcenter of triangle $ABC$ with circumradius $15$. Let $G$ be the centroid of $ABC$ and let $M$ be the midpoint of $BC$. If $BC=18$ and $\angle MOA=150^\circ$, find the area of $OMG$.

2013 Iran Team Selection Test, 17

Tags: circumcircle , angle bisector , asymptote , geometry , perpendicular bisector

In triangle $ABC$, $AD$ and $AH$ are the angle bisector and the altitude of vertex $A$, respectively. The perpendicular bisector of $AD$, intersects the semicircles with diameters $AB$ and $AC$ which are drawn outside triangle $ABC$ in $X$ and $Y$, respectively. Prove that the quadrilateral $XYDH$ is concyclic. [i]Proposed by Mahan Malihi[/i]

2011 Sharygin Geometry Olympiad, 2

Tags: circumcircle , geometry , angle chasing , perpendicular bisector

In triangle $ABC, \angle B = 2\angle C$. Points $P$ and $Q$ on the medial perpendicular to $CB$ are such that $\angle CAP = \angle PAQ = \angle QAB = \frac{\angle A}{3}$ . Prove that $Q$ is the circumcenter of triangle $CPB$.