Olympiad problems

2017 Switzerland - Final Round, 8

Let $ABC$ be an isosceles triangle with vertex $A$ and $AB> BC$. Let $k$ be the circle with center $A$ passsing through $B$ and $C$. Let $H$ be the second intersection of $k$ with the altitude of the triangle $ABC$ through $B$. Further let $G$ be the second intersection of $k$ with the median through $B$ in triangle $ABC$. Let $X$ be the intersection of the lines $AC$ and $GH$. Show that $C$ is the midpoint of $AX$.

2006 Sharygin Geometry Olympiad, 8.3

Tags: parallelogram , equal segments , circles , geometry

A parallelogram $ABCD$ is given. Two circles with centers at the vertices $A$ and $C$ pass through $B$. The straight line $\ell$ that passes through $B$ and crosses the circles at second time at points $X, Y$ respectively. Prove that $DX = DY$.

2000 Belarus Team Selection Test, 8.1

Tags: equal segments , geometry , perpendicular

The diagonals of a convex quadrilateral $ABCD$ with $AB = AC = BD$ intersect at $P$, and $O$ and $I$ are the circumcenter and incenter of $\vartriangle ABP$, respectively. Prove that if $O \ne I$ then $OI$ and $CD$ are perpendicular

2014 Flanders Math Olympiad, 3

Tags: angle , angle chasing , geometry , equal segments

Let $PQRS$ be a quadrilateral with $| P Q | = | QR | = | RS |$, $\angle Q= 110^o$ and $\angle R = 130^o$ . Determine $\angle P$ and $\angle S$ .

1999 Singapore Team Selection Test, 1

Tags: equal segments , geometry

Let $M$ and $N$ be two points on the side BC of a triangle $ABC$ such that $BM =MN = NC$. A line parallel to $AC$ meets the segments $AB, AM$ and $AN$ at the points $D, E$ and $F$ respectively. Prove that $EF = 3DE$

Champions Tournament Seniors - geometry, 2016.3

Tags: geometry , equal segments , equilateral

Let $t$ be a line passing through the vertex $A$ of the equilateral $ABC$, parallel to the side $BC$. On the side $AC$ arbitrarily mark the point $D$. Bisector of the angle $ABD$ intersects the line $t$at the point $E$. Prove that $BD=CD+AE$.

OIFMAT III 2013, 6

Tags: equal segments , geometry

The acute triangle $ABC$ is inscribed in a circle with center $O$. Let $D$ be the intersection of the bisector of angle $BAC$ with segment $BC$ and $ P$ the intersection point of $AB$ with the perpendicular on $OA$ passing through $D$. Show that $AC = AP$.

2003 Estonia National Olympiad, 3

Tags: equal segments , altitude , parallelogram , geometry

In the acute-angled triangle $ABC$ all angles are greater than $45^o$. Let $AM$ and $BN$ be the heights of this triangle and let $X$ and $Y$ be the points on $MA$ and $NB$, respecively, such that $|MX| =|MB|$ and $|NY| =|NA|$. Prove that $MN$ and $XY$ are parallel.

2016 Dutch IMO TST, 3

Tags: equal segments , angle bisector , isosceles , geometry

Let $\vartriangle ABC$ be an isosceles triangle with $|AB| = |AC|$. Let $D, E$ and $F$ be points on line segments $BC, CA$ and $AB$, respectively, such that $|BF| = |BE|$ and such that $ED$ is the internal angle bisector of $\angle BEC$. Prove that $|BD|= |EF|$ if and only if $|AF| = |EC|$.

1993 All-Russian Olympiad Regional Round, 10.1

Tags: geometry , equal segments

Point $D$ is chosen on the side $AC$ of an acute-angled triangle $ABC$. The median $AM$ intersects the altitude $CH$ and the segment $BD$ at points $N$ and $K$ respectively. Prove that if $AK = BK$, then $AN = 2KM$.

2009 Dutch IMO TST, 2

Tags: midpoint , equal segments , geometry

Let $ABC$ be a triangle, $P$ the midpoint of $BC$, and $Q$ a point on segment $CA$ such that $|CQ| = 2|QA|$. Let $S$ be the intersection of $BQ$ and $AP$. Prove that $|AS| = |SP|$.

Novosibirsk Oral Geo Oly VIII, 2021.5

Tags: right triangle , geometry , isosceles , equal segments

On the legs $AC$ and $BC$ of an isosceles right-angled triangle with a right angle $C$, points $D$ and $E$ are taken, respectively, so that $CD = CE$. Perpendiculars on line $AE$ from points $C$ and $D$ intersect segment $AB$ at points $P$ and $Q$, respectively. Prove that $BP = PQ$.

Champions Tournament Seniors - geometry, 2019.2

Tags: cyclic quadrilateral , equal segments , arc midpoint , perpendicular , geometry

The quadrilateral $ABCD$ is inscribed in the circle and the lengths of the sides $BC$ and $DC$ are equal, and the length of the side $AB$ is equal to the length of the diagonal $AC$. Let the point $P$ be the midpoint of the arc $CD$, which does not contain point $A$, and $Q$ is the point of intersection of diagonals $AC$ and $BD$. Prove that the lines $PQ$ and $AB$ are perpendicular.

1997 Swedish Mathematical Competition, 2

Tags: equal segments , angle , geometry , equal angles

Let $D$ be the point on side $AC$ of a triangle $ABC$ such that $BD$ bisects $\angle B$, and $E$ be the point on side $AB$ such that $3\angle ACE = 2\angle BCE$. Suppose that $BD$ and $CE$ intersect at a point $P$ with $ED = DC = CP$. Determine the angles of the triangle.

2011 Saudi Arabia Pre-TST, 4.4

Tags: equal segments , geometry , orthocenter

In a triangle $ABC$, let $O$ be the circumcenter, $H$ the orthocenter, and $M$ the midpoint of the segment $AH$. The perpendicular at $M$ onto $OM$ intersects lines $AB$ and $AC$ at $P$ and $Q$, respectively. Prove that $MP = MQ$.

2022 Chile National Olympiad, 2

Tags: incenter , equal segments , geometry

Let $ABC$ be a triangle such that $\angle CAB = 60^o$. Consider $D, E$ points on sides $AC$ and $AB$ respectively such that $BD$ bisects angle $\angle ABC$ , $CE$ bisects angle $\angle BCA$ and let $I$ be the intersection of them. Prove that $|ID| =|IE|$.

2020 Dutch BxMO TST, 4

Tags: equal segments , equilateral , geometry

Three different points $A,B$ and $C$ lie on a circle with center $M$ so that $| AB | = | BC |$. Point $D$ is inside the circle in such a way that $\vartriangle BCD$ is equilateral. Let $F$ be the second intersection of $AD$ with the circle . Prove that $| F D | = | FM |$.

2017 Argentina National Math Olympiad Level 2, 5

Tags: circumradius , angle bisector , equal segments , geometry

Let $ABCD$ be a convex quadrilateral with $AB = BD = 8$ and $CD = DA = 6$. Let $P$ be a point on side $AB$ such that $DP$ is bisector of angle $\angle ADB$ and let $Q$ be a point on side $BC$ such that $DQ$ is bisector of angle $\angle CDB$. Calculate the radius of the circumcircle of triangle $DPQ$. Note: The circumcircle of a triangle is the circle that passes through its three vertices.

2013 Junior Balkan Team Selection Tests - Moldova, 7

Tags: equal segments , angle , square , geometry

The points $M$ and $N$ are located respectively on the diagonal $(AC)$ and the side $(BC)$ of the square $ABCD$ such that $MN = MD$. Determine the measure of the angle $MDN$.

1988 All Soviet Union Mathematical Olympiad, 479

Tags: geometry , altitude , equal segments

In the acute-angled triangle $ABC$, the altitudes $BD$ and $CE$ are drawn. Let $F$ and $G$ be the points of the line $ED$ such that $BF$ and $CG$ are perpendicular to $ED$. Prove that $EF = DG$.

1998 Tournament Of Towns, 2

Tags: parallelogram , geometry , equal segments , equal angles

$ABCD$ is a parallelogram. A point $M$ is found on the side $AB$ or its extension such that $\angle MAD = \angle AMO$ where $O$ is the intersection point of the diagonals of the parallelogram. Prove that $MD = MG$. (M Smurov)

2018 Iranian Geometry Olympiad, 1

Tags: circles , geometry , equal segments

Two circles $\omega_1,\omega_2$ intersect each other at points $A,B$. Let $PQ$ be a common tangent line of these two circles with $P \in \omega_1$ and $Q \in \omega_2$. An arbitrary point $X$ lies on $\omega_1$. Line $AX$ intersects $ \omega_2$ for the second time at $Y$ . Point $Y'\ne Y$ lies on $\omega_2$ such that $QY = QY'$. Line $Y'B$ intersects $ \omega_1$ for the second time at $X'$. Prove that $PX = PX'$. Proposed by Morteza Saghafian

Estonia Open Senior - geometry, 2002.1.4

Tags: angle , geometry , equal segments

In a triangle $ABC$ we have $\angle B = 2 \cdot \angle C$ and the angle bisector drawn from $A$ intersects $BC$ in a point $D$ such that $|AB| = |CD|$. Find $\angle A$.

2015 Oral Moscow Geometry Olympiad, 4

Tags: trapezoid , equal segments , angle bisector , geometry , incircle

In trapezoid $ABCD$, the bisectors of angles $A$ and $D$ intersect at point $E$ lying on the side of $BC$. These bisectors divide the trapezoid into three triangles into which the circles are inscribed. One of these circles touches the base $AB$ at the point $K$, and two others touch the bisector $DE$ at points $M$ and $N$. Prove that $BK = MN$.

2013 Sharygin Geometry Olympiad, 1

Tags: geometry , right angle , equal segments , pentagon

Let $ABCDE$ be a pentagon with right angles at vertices $B$ and $E$ and such that $AB = AE$ and $BC = CD = DE$. The diagonals $BD$ and $CE$ meet at point $F$. Prove that $FA = AB$.