Olympiad problems

Kyiv City MO 1984-93 - geometry, 1993.8.3

In the triangle $ABC$, $\angle .ACB = 60^o$, and the bisectors $AA_1$ and $BB_1$ intersect at the point $M$. Prove that $MB_1 = MA_1$.

2008 Oral Moscow Geometry Olympiad, 3

Tags: midpoint , geometry , parallelogram , equal segments

Given a quadrilateral $ABCD$. $A ', B', C'$ and $D'$ are the midpoints of the sides $BC, CB, BA$ and $AB$, respectively. It is known that $AA'= CC'$, $BB'= DD'$. Is it true that $ABCD$ is a parallelogram? (M. Volchkevich)

2008 Peru MO (ONEM), 3

Tags: perpendicular , equal segments , geometry

$ABC$ is an acute triangle with $\angle ACB = 45^o$. Let $D$ and $E$ be points on the sides $BC$ and $AC$, respectively, such that $AB = AD = BE$. Let $M,N$ and $X$ be the midpoints of $BD, AE$ and $AB$, respectively. Let lines $AM$ and $BN$ intersect at point $P$. Show that lines $XP$ and $DE$ are perpendicular.

2017 Switzerland - Final Round, 5

Tags: geometry , circumcircle , equal segments , tangent

Let $ABC$ be a triangle with $AC> AB$. Let $P$ be the intersection of $BC$ and the tangent through $A$ around the triangle $ABC$. Let $Q$ be the point on the straight line $AC$, so that $AQ = AB$ and $A$ is between $C$ and $Q$. Let $X$ and $Y$ be the center of $BQ$ and $AP$. Let $R$ be the point on $AP$ so that $AR = BP$ and $R$ is between $A$ and $P$. Show that $BR = 2XY$.

2015 Dutch IMO TST, 1

Tags: right angle , equal angles , equal segments , geometry

In a quadrilateral $ABCD$ we have $\angle A = \angle C = 90^o$. Let $E$ be a point in the interior of $ABCD$. Let $M$ be the midpoint of $BE$. Prove that $\angle ADB = \angle EDC$ if and only if $|MA| = |MC|$.

Champions Tournament Seniors - geometry, 2007.3

Tags: circumcircle , locus , equal segments , geometry

Given a triangle $ABC$. Point $M$ moves along the side $BA$ and point $N$ moves along the side $AC$ beyond point $C$ such that $BM=CN$. Find the geometric locus of the centers of the circles circumscribed around the triangle $AMN$.

Denmark (Mohr) - geometry, 2012.5

Tags: geometry , hexagon , equal angles , equilateral , equal segments

In the hexagon $ABCDEF$, all angles are equally large. The side lengths satisfy $AB = CD = EF = 3$ and $BC = DE = F A = 2$. The diagonals $AD$ and $CF$ intersect each other in the point $G$. The point $H$ lies on the side $CD$ so that $DH = 1$. Prove that triangle $EGH$ is equilateral.

2011 Sharygin Geometry Olympiad, 9

Tags: tangent , geometry , orthocenter , equal segments

Let $H$ be the orthocenter of triangle $ABC$. The tangents to the circumcircles of triangles $CHB$ and $AHB$ at point $H$ meet $AC$ at points $A_1$ and $C_1$ respectively. Prove that $A_1H = C_1H$.

2021 New Zealand MO, 1

Tags: geometry , right angle , equal segments

Let $ABCD$ be a convex quadrilateral such that $AB + BC = 2021$ and $AD = CD$. We are also given that $\angle ABC = \angle CDA = 90^o$. Determine the length of the diagonal $BD$.

2013 Junior Balkan Team Selection Tests - Moldova, 7

Tags: geometry , square , equal segments , angle

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 $MDN$.

1993 Tournament Of Towns, (370) 2

Tags: geometry , cyclic quadrilateral , equal segments

Quadrilateral $ABCD$ is inscribed in a circle, $M$ is the intersection point of the lines $AB$ and $CD$ and $N$ is the intersection point of the lines $BC$ and $AD$. It is known that $BM = DN$. Prove that $CM = CN$. (F Nazarov)

2019 Federal Competition For Advanced Students, P1, 2

Tags: geometry , incenter , circles , equal segments

Let $ABC$ be a triangle and $I$ its incenter. The circle passing through $A, C$ and $I$ intersect the line $BC$ for second time at point $X$. The circle passing through $B, C$ and $I$ intersects the line $AC$ for second time at point $Y$. Show that the segments $AY$ and $BX$ have equal length.

2015 Caucasus Mathematical Olympiad, 2

Tags: geometry , equal angles , equal segments , convex quadrilateral

In the convex quadrilateral $ABCD$, point $K$ is the midpoint of $AB$, point $L$ is the midpoint of $BC$, point $M$ is the midpoint of CD, and point $N$ is the midpoint of $DA$. Let $S$ be a point lying inside the quadrilateral $ABCD$ such that $KS = LS$ and $NS = MS$ .Prove that $\angle KSN = \angle MSL$.

Cono Sur Shortlist - geometry, 1993.11

Tags: geometry , equal segments , semicircle

Let $\Gamma$ be a semicircle with center $O$ and diameter $AB$. $D$ is the midpoint of arc $AB$. On the ray $OD$, we take $E$ such that $OE = BD$. $BE$ intersects the semicircle at $F$ and $ P$ is the point on $AB$ such that $FP$ is perpendicular to $AB$. Prove that $BP=\frac13 AB$.

2002 Kazakhstan National Olympiad, 1

Tags: geometry , incenter , equal segments , angle

Let $ O $ be the center of the inscribed circle of the triangle $ ABC $, tangent to the side of $ BC $. Let $ M $ be the midpoint of $ AC $, and $ P $ be the intersection point of $ MO $ and $ BC $. Prove that $ AB = BP $ if $ \angle BAC = 2 \angle ACB $.

1988 Greece National Olympiad, 2

Tags: geometry , isosceles , equal segments

In isosceles triangle $ABC$ with $AB=AC$, consider point $D$ on the base $BC$ and point $E$ on side $AC$ such that $ \angle BAD = 2 \angle CDE$. Prove that $AD=AE$.

1999 All-Russian Olympiad Regional Round, 10.6

Tags: geometry , incircle , equal segments

Triangle $ABC$ has an inscribed circle tangent to sides $AB$, $AC$ and $BC$ at points $C_1$, $B_1$ and $A_1 $ respectively. Let $K$ be a point on the circle diametrically opposite to point $C_1$, $D$ be the intersection point of lines $B_1C_1$ and $A_1K$. Prove that $CD = CB_1$.

2005 Peru MO (ONEM), 3

Tags: geometry , angle , equilateral triangle , equilateral , equal segments

Let $A,B,C,D$, be four different points on a line $\ell$, so that $AB=BC=CD$. In one of the semiplanes determined by the line $\ell$, the points $P$ and $Q$ are chosen in such a way that the triangle $CPQ$ is equilateral with its vertices named clockwise. Let $M$ and $N$ be two points of the plane be such that the triangles $MAP$ and $NQD$ are equilateral (the vertices are also named clockwise). Find the angle $\angle MBN$.

2020 Estonia Team Selection Test, 2

Tags: excircle , equal segments , circumcircle , geometry

Let $M$ be the midpoint of side BC of an acute-angled triangle $ABC$. Let $D$ and $E$ be the center of the excircle of triangle $AMB$ tangent to side $AB$ and the center of the excircle of triangle $AMC$ tangent to side $AC$, respectively. The circumscribed circle of triangle $ABD$ intersects line$ BC$ for the second time at point $F$, and the circumcircle of triangle $ACE$ is at point $G$. Prove that $| BF | = | CG|$.

Kyiv City MO Juniors Round2 2010+ geometry, 2018.8.31

Tags: geometry , isosceles , equal angles , equal segments

On the sides $AB$, $BC$ and $CA$ of the isosceles triangle $ABC$ with the vertex at the point $B$ marked the points $M$, $D$ and $K$ respectively so that $AM = 2DC$ and $\angle AMD = \angle KDC$. Prove that $MD = KD$.

2010 Abels Math Contest (Norwegian MO) Final, 1a

Tags: geometry , right angle , equal segments , angle

The point $P$ lies on the edge $AB$ of a quadrilateral $ABCD$. The angles $BAD, ABC$ and $CPD$ are right, and $AB = BC + AD$. Show that $BC = BP$ or $AD = BP$.

2012 Korea Junior Math Olympiad, 5

Tags: geometry , equal angles , equal segments , tangent , cyclic quadrilateral

Let $ABCD$ be a cyclic quadrilateral inscirbed in a circle $O$ ($AB> AD$), and let $E$ be a point on segment $AB$ such that $AE = AD$. Let $AC \cap DE = F$, and $DE \cap O = K(\ne D)$. The tangent to the circle passing through $C,F,E$ at $E$ hits $AK$ at $L$. Prove that $AL = AD$ if and only if $\angle KCE = \angle ALE$.

2020 Ukrainian Geometry Olympiad - December, 4

Tags: geometry , isosceles , equal segments , angle

In an isosceles triangle $ABC$ with an angle $\angle A= 20^o$ and base $BC=12$ point $E$ on the side $AC$ is chosen such that $\angle ABE= 30^o$ , and point $F$ on the side $AB$ such that $EF = FC$ . Find the length of $FC$.

1993 All-Russian Olympiad Regional Round, 10.1

Tags: equal segments , geometry

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 $AN = 2KM$.

2020 Regional Olympiad of Mexico Center Zone, 3

Tags: geometry , equal segments

In an acute triangle $ABC$, an arbitrary point $P$ is chosen on the altitude $AH$. The points $E$ and $F$ are the midpoints of $AC$ and $AB$, respectively. The perpendiculars from $E$ on $CP$ and from $F$ on $BP$ intersect at the point $K$. Show that $KB = KC$.