Olympiad problems

2010 Poland - Second Round, 1

In the convex pentagon $ABCDE$ all interior angles have the same measure. Prove that the perpendicular bisector of segment $EA$, the perpendicular bisector of segment $BC$ and the angle bisector of $\angle CDE$ intersect in one point.

2008 Tournament Of Towns, 3

Tags: perpendicular bisector , geometry , perimeter

A $30$-gon $A_1A_2\cdots A_{30}$ is inscribed in a circle of radius $2$. Prove that one can choose a point $B_k$ on the arc $A_kA_{k+1}$ for $1 \leq k \leq 29$ and a point $B_{30}$ on the arc $A_{30}A_1$, such that the numerical value of the area of the $60$-gon $A_1B_1A_2B_2 \dots A_{30}B_{30}$ is equal to the numerical value of the perimeter of the original $30$-gon.

2020 ELMO Problems, P3

Tags: perpendicular bisector , orthocenter

Janabel has a device that, when given two distinct points $U$ and $V$ in the plane, draws the perpendicular bisector of $UV$. Show that if three lines forming a triangle are drawn, Janabel can mark the orthocenter of the triangle using this device, a pencil, and no other tools. [i]Proposed by Fedir Yudin.[/i]

2016 Saudi Arabia GMO TST, 3

Tags: perpendicular bisector , collinear , midpoint , geometry

Let $ABC$ be an acute, non-isosceles triangle with the circumcircle $(O)$. Denote $D, E$ as the midpoints of $AB,AC$ respectively. Two circles $(ABE)$ and $(ACD)$ intersect at $K$ differs from $A$. Suppose that the ray $AK$ intersects $(O)$ at $L$. The line $LB$ meets $(ABE)$ at the second point $M$ and the line $LC$ meets $(ACD)$ at the second point $N$. a) Prove that $M, K, N$ collinear and $MN$ perpendicular to $OL$. b) Prove that $K$ is the midpoint of $MN$

2014 Middle European Mathematical Olympiad, 3

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

Let $ABC$ be a triangle with $AB < AC$ and incentre $I$. Let $E$ be the point on the side $AC$ such that $AE = AB$. Let $G$ be the point on the line $EI$ such that $\angle IBG = \angle CBA$ and such that $E$ and $G$ lie on opposite sides of $I$. Prove that the line $AI$, the line perpendicular to $AE$ at $E$, and the bisector of the angle $\angle BGI$ are concurrent.

2003 Estonia Team Selection Test, 6

Tags: ratio , orthocenter , perpendicular bisector , projection , circumcenter , geometry

Let $ABC$ be an acute-angled triangle, $O$ its circumcenter and $H$ its orthocenter. The orthogonal projection of the vertex $A$ to the line $BC$ lies on the perpendicular bisector of the segment $AC$. Compute $\frac{CH}{BO}$ . (J. Willemson)

2003 China Team Selection Test, 1

Tags: perpendicular bisector , circumcircle , trapezoid , geometry

Triangle $ABC$ is inscribed in circle $O$. Tangent $PD$ is drawn from $A$, $D$ is on ray $BC$, $P$ is on ray $DA$. Line $PU$ ($U \in BD$) intersects circle $O$ at $Q$, $T$, and intersect $AB$ and $AC$ at $R$ and $S$ respectively. Prove that if $QR=ST$, then $PQ=UT$.

2004 Kazakhstan National Olympiad, 8

Tags: perpendicular bisector , circumcircle , convex quadrilateral , geometry

Let $ ABCD$ be a convex quadrilateral. The perpendicular bisectors of its sides $ AB$ and $ CD$ meet at $ Y$. Denote by $ X$ a point inside the quadrilateral $ ABCD$ such that $ \measuredangle ADX \equal{} \measuredangle BCX < 90^{\circ}$ and $ \measuredangle DAX \equal{} \measuredangle CBX < 90^{\circ}$. Show that $ \measuredangle AYB \equal{} 2\cdot\measuredangle ADX$.

2011 IberoAmerican, 1

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

Let $ABC$ be an acute-angled triangle, with $AC \neq BC$ and let $O$ be its circumcenter. Let $P$ and $Q$ be points such that $BOAP$ and $COPQ$ are parallelograms. Show that $Q$ is the orthocenter of $ABC$.

2011 China Team Selection Test, 1

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

Let $AA',BB',CC'$ be three diameters of the circumcircle of an acute triangle $ABC$. Let $P$ be an arbitrary point in the interior of $\triangle ABC$, and let $D,E,F$ be the orthogonal projection of $P$ on $BC,CA,AB$, respectively. Let $X$ be the point such that $D$ is the midpoint of $A'X$, let $Y$ be the point such that $E$ is the midpoint of $B'Y$, and similarly let $Z$ be the point such that $F$ is the midpoint of $C'Z$. Prove that triangle $XYZ$ is similar to triangle $ABC$.

2013 Middle European Mathematical Olympiad, 3

Tags: perpendicular bisector , geometry , angle chasing

Let $ABC$ be an isosceles triangle with $AC=BC$. Let $N$ be a point inside the triangle such that $2 \angle ANB = 180 ^\circ + \angle ACB $. Let $ D $ be the intersection of the line $BN$ and the line parallel to $AN$ that passes through $C$. Let $P$ be the intersection of the angle bisectors of the angles $CAN$ and $ABN$. Show that the lines $DP$ and $AN$ are perpendicular.

2008 Tuymaada Olympiad, 6

Tags: perpendicular bisector , trapezoid , analytic geometry , ratio , angle bisector , projective geometry , geometry

Let $ ABCD$ be an isosceles trapezoid with $ AD \parallel BC$. Its diagonals $ AC$ and $ BD$ intersect at point $ M$. Points $ X$ and $ Y$ on the segment $ AB$ are such that $ AX \equal{} AM$, $ BY \equal{} BM$. Let $ Z$ be the midpoint of $ XY$ and $ N$ is the point of intersection of the segments $ XD$ and $ YC$. Prove that the line $ ZN$ is parallel to the bases of the trapezoid. [i]Author: A. Akopyan, A. Myakishev[/i]

2011 CentroAmerican, 6

Tags: perpendicular bisector , trigonometry , geometry

Let $ABC$ be an acute triangle and $D$, $E$, $F$ be the feet of the altitudes through $A$, $B$, $C$ respectively. Call $Y$ and $Z$ the feet of the perpendicular lines from $B$ and $C$ to $FD$ and $DE$, respectively. Let $F_1$ be the symmetric of $F$ with respect to $E$ and $E_1$ be the symmetric of $E$ with respect to $F$. If $3EF=FD+DE$, prove that $\angle BZF_1=\angle CYE_1$.

2011 Czech-Polish-Slovak Match, 3

Tags: perpendicular bisector , reflection , geometry , geometric transformation

Points $A$, $B$, $C$, $D$ lie on a circle (in that order) where $AB$ and $CD$ are not parallel. The length of arc $AB$ (which contains the points $D$ and $C$) is twice the length of arc $CD$ (which does not contain the points $A$ and $B$). Let $E$ be a point where $AC=AE$ and $BD=BE$. Prove that if the perpendicular line from point $E$ to the line $AB$ passes through the center of the arc $CD$ (which does not contain the points $A$ and $B$), then $\angle ACB = 108^\circ$.

2021 OMMock - Mexico National Olympiad Mock Exam, 4

Tags: geometry , perpendicular bisector

Let $ABC$ be an obtuse triangle with $AB = AC$, and let $\Gamma$ be the circle that is tangent to $AB$ at $B$ and to $AC$ at $C$. Let $D$ be the point on $\Gamma$ furthest from $A$ such that $AD$ is perpendicular to $BC$. Point $E$ is the intersection of $AB$ and $DC$, and point $F$ lies on line $AB$ such that $BC = BF$ and $B$ lies on segment $AF$. Finally, let $P$ be the intersection of lines $AC$ and $DB$. Show that $PE = PF$.

2018 Bosnia And Herzegovina - Regional Olympiad, 4

Tags: perpendicular bisector , geometry , circumcircle

Let $P$ be a point on circumcircle of triangle $ABC$ on arc $\stackrel{\frown}{BC}$ which does not contain point $A$. Let lines $AB$ and $CP$ intersect at point $E$, and lines $AC$ and $BP$ intersect at $F$. If perpendicular bisector of side $AB$ intersects $AC$ in point $K$, and perpendicular bisector of side $AC$ intersects side $AB$ in point $J$, prove that: ${\left(\frac{CE}{BF}\right)}^2=\frac{AJ\cdot JE}{AK \cdot KF}$

2013 Lusophon Mathematical Olympiad, 2

Tags: trigonometry , geometry , circumcircle , perpendicular bisector

Let $ABC$ be an acute triangle. The circumference with diameter $AB$ intersects sides $AC$ and $BC$ at $E$ and $F$ respectively. The tangent lines to the circumference at the points $E$ and $F$ meet at $P$. Show that $P$ belongs to the altitude from $C$ of triangle $ABC$.

1981 Polish MO Finals, 2

Tags: geometry , perpendicular bisector , trigonometry

In a triangle $ABC$, the perpendicular bisectors of sides $AB$ and $AC$ intersect $BC$ at $X$ and $Y$. Prove that $BC = XY$ if and only if $\tan B\tan C = 3$ or $\tan B\tan C = -1$.

May Olympiad L2 - geometry, 2012.3

Tags: projective geometry , ratio , perpendicular bisector , geometry

Given Triangle $ABC$, $\angle B= 2 \angle C$, and $\angle A>90^\circ$. Let $M$ be midpoint of $BC$. Perpendicular of $AC$ at $C$ intersects $AB$ at $D$. Show $\angle AMB = \angle DMC$ [hide]If possible, don't use projective geometry[/hide]

1998 IberoAmerican Olympiad For University Students, 4

Tags: geometry , rhombus , perpendicular bisector

Four circles of radius $1$ with centers $A,B,C,D$ are in the plane in such a way that each circle is tangent to two others. A fifth circle passes through the center of two of the circles and is tangent to the other two. Find the possible values of the area of the quadrilateral $ABCD$.

2002 Belarusian National Olympiad, 6

Tags: search , function , incenter , perpendicular bisector , geometry

The altitude $CH$ of a right triangle $ABC$, with $\angle{C}=90$, cut the angles bisectors $AM$ and $BN$ at $P$ and $Q$, and let $R$ and $S$ be the midpoints of $PM$ and $QN$. Prove that $RS$ is parallel to the hypotenuse of $ABC$

1987 Romania Team Selection Test, 8

Tags: perpendicular bisector , geometry

Let $ABCD$ be a square and $a$ be the length of his edges. The segments $AE$ and $CF$ are perpendicular on the square's plane in the same half-space and they have the length $AE=a$, $CF=b$ where $a<b<a\sqrt 3$. If $K$ denoted the set of the interior points of the square $ABCD$ determine $\min_{M\in K} \left( \max ( EM, FM ) \right) $ and $\max_{M\in K} \left( \min (EM,FM) \right)$. [i]Octavian Stanasila[/i]

1998 Belarus Team Selection Test, 1

Tags: angle bisector , geometry , perpendicular bisector , angle , midpoint

Let $O$ be a point inside an acute angle with the vertex $A$ and $H, N$ be the feet of the perpendiculars drawn from $O$ onto the sides of the angle. Let point $B$ belong to the bisector of the angle, $K$ be the foot of the perpendicular from $B$ onto either side of the angle. Denote by $P,F$ the midpoints of the segments $AK,HN$ respectively. Known that $ON + OH = BK$, prove that $PF$ is perpendicular to $AB$. Ya. Konstantinovski

2002 India IMO Training Camp, 13

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

Let $ABC$ and $PQR$ be two triangles such that [list] [b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$. [b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$ [/list] Prove that $AB+AC=PQ+PR$.

2013 National Olympiad First Round, 33

Tags: angle bisector , perpendicular bisector , geometry

Let $D$ be a point on side $[BC]$ of triangle $ABC$ such that $[AD]$ is an angle bisector, $|BD|=4$, and $|DC|=3$. Let $E$ be a point on side $[AB]$ and different than $A$ such that $m(\widehat{BED})=m(\widehat{DEC})$. If the perpendicular bisector of segment $[AE]$ meets the line $BC$ at $M$, what is $|CM|$? $ \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ \text { None of above} $