Olympiad problems

2016 Iran Team Selection Test, 5

Let $AD,BF,CE$ be altitudes of triangle $ABC$.$Q$ is a point on $EF$ such that $QF=DE$ and $F$ is between $E,Q$.$P$ is a point on $EF$ such that $EP=DF$ and $E$ is between $P,F$.Perpendicular bisector of $DQ$ intersect with $AB$ at $X$ and perpendicular bisector of $DP$ intersect with $AC$ at $Y$.Prove that midpoint of $BC$ lies on $XY$.

2008 Sharygin Geometry Olympiad, 1

Tags: symmetry , perpendicular bisector , geometry

(B.Frenkin, 8) Does a regular polygon exist such that just half of its diagonals are parallel to its sides?

2014 Saudi Arabia Pre-TST, 2.4

Tags: circumcircle , perpendicular bisector , geometry

Let $ABC$ be an acute triangle with $\angle A < \angle B \le \angle C$, and $O$ its circumcenter. The perpendicular bisector of side $AB$ intersects side $AC$ at $D$. The perpendicular bisector of side $AC$ intersects side $AB$ at $E$. Express the angles of triangle $DEO$ in terms of the angles of triangle $ABC$.

2011 Croatia Team Selection Test, 3

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

Let $K$ and $L$ be the points on the semicircle with diameter $AB$. Denote intersection of $AK$ and $AL$ as $T$ and let $N$ be the point such that $N$ is on segment $AB$ and line $TN$ is perpendicular to $AB$. If $U$ is the intersection of perpendicular bisector of $AB$ an $KL$ and $V$ is a point on $KL$ such that angles $UAV$ and $UBV$ are equal. Prove that $NV$ is perpendicular to $KL$.

2019 Austrian Junior Regional Competition, 2

Tags: geometry , perpendicular bisector

A square $ABCD$ is given. Over the side $BC$ draw an equilateral triangle $BCS$ on the outside. The midpoint of the segment $AS$ is $N$ and the midpoint of the side $CD$ is $H$. Prove that $\angle NHC = 60^o$. . (Karl Czakler)

2001 Canada National Olympiad, 3

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

Let $ABC$ be a triangle with $AC > AB$. Let $P$ be the intersection point of the perpendicular bisector of $BC$ and the internal angle bisector of $\angle{A}$. Construct points $X$ on $AB$ (extended) and $Y$ on $AC$ such that $PX$ is perpendicular to $AB$ and $PY$ is perpendicular to $AC$. Let $Z$ be the intersection point of $XY$ and $BC$. Determine the value of $\frac{BZ}{ZC}$.

2009 Serbia National Math Olympiad, 6

Tags: circumcircle , perpendicular bisector , projective geometry , geometry

Triangle ABC has incircle w centered as S that touches the sides BC,CA and AB at P,Q and R respectively. AB isn't equal AC, the lines QR and BC intersects at point M, the circle that passes through points B and C touches the circle w at point N, circumcircle of triangle MNP intersects with line AP at L (L isn't equal to P). Then prove that S,L and M lie on the same line

2020 Polish Junior MO Second Round, 2.

Tags: geometry , parallelogram , 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 $XE = \frac{1}{2}AD$.

2005 Morocco TST, 4

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

A convex quadrilateral $ABCD$ has an incircle. In each corner a circle is inscribed that also externally touches the two circles inscribed in the adjacent corners. Show that at least two circles have the same size.

2009 Brazil National Olympiad, 2

Tags: geometry , circumcircle , perpendicular bisector

Let $ ABC$ be a triangle and $ O$ its circumcenter. Lines $ AB$ and $ AC$ meet the circumcircle of $ OBC$ again in $ B_1\neq B$ and $ C_1 \neq C$, respectively, lines $ BA$ and $ BC$ meet the circumcircle of $ OAC$ again in $ A_2\neq A$ and $ C_2\neq C$, respectively, and lines $ CA$ and $ CB$ meet the circumcircle of $ OAB$ in $ A_3\neq A$ and $ B_3\neq B$, respectively. Prove that lines $ A_2A_3$, $ B_1B_3$ and $ C_1C_2$ have a common point.

2012 NIMO Problems, 5

Tags: geometry , incenter , perpendicular bisector , angle 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]

2005 China Western Mathematical Olympiad, 5

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

Circles $C(O_1)$ and $C(O_2)$ intersect at points $A$, $B$. $CD$ passing through point $O_1$ intersects $C(O_1)$ at point $D$ and tangents $C(O_2)$ at point $C$. $AC$ tangents $C(O_1)$ at $A$. Draw $AE \bot CD$, and $AE$ intersects $C(O_1)$ at $E$. Draw $AF \bot DE$, and $AF$ intersects $DE$ at $F$. Prove that $BD$ bisects $AF$.

2003 National Olympiad First Round, 25

Tags: geometry , circumcircle , trigonometry , perpendicular bisector

Let $ABC$ be an acute triangle and $O$ be its circumcenter. Let $D$ be the midpoint of $[AB]$. The circumcircle of $\triangle ADO$ meets $[AC]$ at $A$ and $E$. If $|AE|=7$, $|DE|=8$, and $m(\widehat{AOD}) = 45^\circ$, what is the area of $\triangle ABC$? $ \textbf{(A)}\ 56\sqrt 3 \qquad\textbf{(B)}\ 56 \sqrt 2 \qquad\textbf{(C)}\ 50 \sqrt 2 \qquad\textbf{(D)}\ 84 \qquad\textbf{(E)}\ \text{None of the preceding} $

2013 India IMO Training Camp, 3

Tags: circumcircle , angle bisector , perpendicular bisector , geometry

In a triangle $ABC$, with $AB \ne BC$, $E$ is a point on the line $AC$ such that $BE$ is perpendicular to $AC$. A circle passing through $A$ and touching the line $BE$ at a point $P \ne B$ intersects the line $AB$ for the second time at $X$. Let $Q$ be a point on the line $PB$ different from $P$ such that $BQ = BP$. Let $Y$ be the point of intersection of the lines $CP$ and $AQ$. Prove that the points $C, X, Y, A$ are concyclic if and only if $CX$ is perpendicular to $AB$.

2012 ISI Entrance Examination, 6

Tags: geometry , perimeter , calculus , ellipse , inequalities , perpendicular bisector , conic

[b]i)[/b] Let $0<a<b$.Prove that amongst all triangles having base $a$ and perimeter $a+b$ the triangle having two sides(other than the base) equal to $\frac {b}{2}$ has the maximum area. [b]ii)[/b]Using $i)$ or otherwise, prove that amongst all quadrilateral having give perimeter the square has the maximum area.

2010 Sharygin Geometry Olympiad, 6

Tags: circumcircle , perpendicular bisector , geometry

Points $M$ and $N$ lie on the side $BC$ of the regular triangle $ABC$ ($M$ is between $B$ and $N$), and $\angle MAN=30^\circ.$ The circumcircles of triangles $AMC$ and $ANB$ meet at a point $K.$ Prove that the line $AK$ passes through the circumcenter of triangle $AMN.$

1995 Turkey Team Selection Test, 1

Tags: perpendicular bisector , angle bisector , geometry

In a convex quadrilateral $ABCD$ it is given that $\angle{CAB} = 40^{\circ}, \angle{CAD} = 30^{\circ}, \angle{DBA} = 75^{\circ}$, and $\angle{DBC}=25^{\circ}$. Find $\angle{BDC}$.

2013 Sharygin Geometry Olympiad, 6

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

Diagonals $AC$ and $BD$ of a trapezoid $ABCD$ meet at $P$. The circumcircles of triangles $ABP$ and $CDP$ intersect the line $AD$ for the second time at points $X$ and $Y$ respectively. Let $M$ be the midpoint of segment $XY$. Prove that $BM = CM$.

2013 Olympic Revenge, 2

Tags: geometry , circumcircle , similar triangles , perpendicular bisector

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 $BC$.

1995 IMO Shortlist, 2

Tags: geometry , circumcircle , reflection , complex number , perpendicular bisector

Let $ A, B$ and $ C$ be non-collinear points. Prove that there is a unique point $ X$ in the plane of $ ABC$ such that \[ XA^2 \plus{} XB^2 \plus{} AB^2 \equal{} XB^2 \plus{} XC^2 \plus{} BC^2 \equal{} XC^2 \plus{} XA^2 \plus{} CA^2.\]

2022 VN Math Olympiad For High School Students, Problem 6

Tags: geometry , perpendicular bisector

Let $ABC$ be a triangle with $\angle A,\angle B,\angle C <120^{\circ}$, $T$ is its [i]Fermat-Torricelli[/i] point. Let $G$ be the centroid of $\triangle ABC$. Prove that: the distances from $G$ to the perpendicular bisectors of $TA, TB, TC$ are the same.

2003 Junior Balkan Team Selection Tests - Romania, 1

Tags: geometry , perpendicular bisector , rhombus , rectangle

Consider a rhombus $ABCD$ with center $O$. A point $P$ is given inside the rhombus, but not situated on the diagonals. Let $M,N,Q,R$ be the projections of $P$ on the sides $(AB), (BC), (CD), (DA)$, respectively. The perpendicular bisectors of the segments $MN$ and $RQ$ meet at $S$ and the perpendicular bisectors of the segments $NQ$ and $MR$ meet at $T$. Prove that $P, S, T$ and $O$ are the vertices of a rectangle.

1983 IMO Longlists, 74

Tags: hyperbola , asymptote , geometry , circumcircle , analytic geometry , perpendicular bisector , conic

In a plane we are given two distinct points $A,B$ and two lines $a, b$ passing through $B$ and $A$ respectively $(a \ni B, b \ni A)$ such that the line $AB$ is equally inclined to a and b. Find the locus of points $M$ in the plane such that the product of distances from $M$ to $A$ and a equals the product of distances from $M$ to $B$ and $b$ (i.e., $MA \cdot MA' = MB \cdot MB'$, where $A'$ and $B'$ are the feet of the perpendiculars from $M$ to $a$ and $b$ respectively).

2018 Danube Mathematical Competition, 3

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

Let $ABC$ be an acute non isosceles triangle. The angle bisector of angle $A$ meets again the circumcircle of the triangle $ABC$ in $D$. Let $O$ be the circumcenter of the triangle $ABC$. The angle bisectors of $\angle AOB$, and $\angle AOC$ meet the circle $\gamma$ of diameter $AD$ in $P$ and $Q$ respectively. The line $PQ$ meets the perpendicular bisector of $AD$ in $R$. Prove that $AR // BC$.

Cono Sur Shortlist - geometry, 2005.G3.4

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

Let $ABC$ be a isosceles triangle, with $AB=AC$. A line $r$ that pass through the incenter $I$ of $ABC$ touches the sides $AB$ and $AC$ at the points $D$ and $E$, respectively. Let $F$ and $G$ be points on $BC$ such that $BF=CE$ and $CG=BD$. Show that the angle $\angle FIG$ is constant when we vary the line $r$.