Olympiad problems

2022 Oral Moscow Geometry Olympiad, 1

In a circle with center $O$, chords $AB$ and $AC$ are drawn, both equal to the radius. Points $A_1$, $B_1$ and $C_1$ are projections of points $A, B$ and $C$, respectively, onto an arbitrary diameter $XY$. Prove that one of the segments $XB_1$, $OA_1$ and $C_1Y$ is equal to the sum of the other two. (A. Shklover)

2017 Switzerland - Final Round, 1

Tags: angle bisector , geometry , circles , equal segments

Let $A$ and $B$ be points on the circle $k$ with center $O$, so that $AB> AO$. Let $C$ be the intersection of the bisectors of $\angle OAB$ and $k$, different from $A$. Let $D$ be the intersection of the straight line $AB$ with the circumcircle of the triangle $OBC$, different from $B$. Show that $AD = AO$ .

2012 Oral Moscow Geometry Olympiad, 1

Tags: trapezoid , equal segments , angle , geometry

In trapezoid $ABCD$, the sides $AD$ and $BC$ are parallel, and $AB = BC = BD$. The height $BK$ intersects the diagonal $AC$ at $M$. Find $\angle CDM$.

2021 New Zealand MO, 1

Tags: right angle , geometry , 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 $BD$.

2019 Novosibirsk Oral Olympiad in Geometry, 3

Tags: angle , equal segments , geometry

Equal line segments are marked in triangle $ABC$. Find its angles. [img]https://cdn.artofproblemsolving.com/attachments/0/2/bcb756bba15ba57013f1b6c4cbe9cc74171543.png[/img]

2018 Costa Rica - Final Round, G2

Tags: angle bisector , geometry , equal segments

Consider $\vartriangle ABC$, with $AD$ bisecting $\angle BAC$, $D$ on segment $BC$. Let $E$ be a point on $BC$, such that $BD = EC$. Through $E$ we draw the line $\ell$ parallel to $AD$ and consider a point $ P$ on it and inside the $\vartriangle ABC$. Let $G$ be the point where line $BP$ cuts side $AC$ and let F be the point where line $CP$ to side $AB$. Show that $BF = CG$.

OIFMAT II 2012, 4

Tags: geometry , circumcenter , orthocenter , equal segments

Given a $ \vartriangle ABC $ with $ AB> AC $ and $ \angle BAC = 60^o$. Denote the circumcenter and orthocenter as $ O $ and $ H $ respectively. We also have that $ OH $ intersects $ AB $ in $ P $ and $ AC $ in $ Q $. Prove that $ PO = HQ $.

Indonesia Regional MO OSP SMA - geometry, 2015.3

Tags: geometry , equal angles , equal segments , equilateral

Given the isosceles triangle $ABC$, where $AB = AC$. Let $D$ be a point in the segment $BC$ so that $BD = 2DC$. Suppose also that point $P$ lies on the segment $AD$ such that: $\angle BAC = \angle BP D$. Prove that $\angle BAC = 2\angle DP C$.

2015 Thailand TSTST, 2

Tags: geometry , equal segments , projection

In any $\vartriangle ABC, \ell$ is any line through $C$ and points $P, Q$. If $BP, AQ$ are perpendicular to the line $\ell$ and $M$ is the midpoint of the line segment $AB$, then prove that $MP = MQ$

2005 Oral Moscow Geometry Olympiad, 4

Tags: geometry , hexagon , perpendicular , equal segments , right angle

Given a hexagon $ABCDEF$, in which $AB = BC, CD = DE, EF = FA$, and angles $A$ and $C$ are right. Prove that lines $FD$ and $BE$ are perpendicular. (B. Kukushkin)

2020 Kazakhstan National Olympiad, 3

Tags: geometry , equal segments , equal angles

A point $ N $ is marked on the median $ CM $ of the triangle $ ABC $ so that $ MN \cdot MC = AB ^ 2/4 $. Lines $ AN $ and $ BN $ intersect the circumcircle $ \triangle ABC $ for the second time at points $ P $ and $ Q $, respectively. $ R $ is the point of segment $ PQ $, nearest to $ Q $, such that $ \angle NRC = \angle BNC $. $ S $ is the point of the segment $ PQ $ closest to $ P $ such that $ \angle NSC = \angle ANC $. Prove that $ RN = SN $.

2018 Austria Beginners' Competition, 2

Tags: geometry , equal segments , midpoint , altitude

Let $ABC$ be an acute-angled triangle, $M$ the midpoint of the side $AC$ and $F$ the foot on $AB$ of the altitude through the vertex $C$. Prove that $AM = AF$ holds if and only if $\angle BAC = 60^o$. (Karl Czakler)

2023 Yasinsky Geometry Olympiad, 2

Tags: geometry , equal segments

Let $BC$ and $BD$ be the tangent lines to the circle with diameter $AC$. Let $E$ be the second point of intersection of line $CD$ and the circumscribed circle of triangle $ABC$. Prove that $CD= 2DE$. (Matthew Kurskyi)

2011 Tournament of Towns, 3

Tags: geometry , bisects segment , equal segments , altitude , projection

In triangle $ABC$, points $A_1,B_1,C_1$ are bases of altitudes from vertices $A,B,C$, and points $C_A,C_B$ are the projections of $C_1$ to $AC$ and $BC$ respectively. Prove that line $C_AC_B$ bisects the segments $C_1A_1$ and $C_1B_1$.

2010 Dutch IMO TST, 1

Tags: geometry , perpendicular , equal segments

Let $ABC$ be an acute triangle such that $\angle BAC = 45^o$. Let $D$ a point on $AB$ such that $CD \perp AB$. Let $P$ be an internal point of the segment $CD$. Prove that $AP\perp BC$ if and only if $|AP| = |BC|$.

Ukrainian TYM Qualifying - geometry, XI.6

Tags: geometry , concurrency , equal segments , centroid

Prove that there exists a point $K$ in the plane of $\vartriangle ABC$ such that $$AK^2 - BC^2 = BK^2 - AC^2 = CK^2 - AB^2.$$ Let $Q, N, T$ be the points of intersection of the medians of the triangles $BKC, CKA, AKB$, respectively. Prove that the segments $AQ, BN$ and $CT$ are equal and have a common point.

Kyiv City MO Seniors 2003+ geometry, 2016.11.4.1

Tags: geometry , circumcircle , equal segments , incircle

In the triangle $ABC$ the angle bisector $AD$ is drawn, $E$ is the point of tangency of the inscribed circle to the side $BC$, $I$ is the center of the inscribed circle $\Delta ABC$. The point ${{A} _ {1}}$ on the circumscribed circle $\Delta ABC$ is such that $A {{A} _ {1}} || BC$. Denote by $T$ - the second point of intersection of the line $E {{A} _ {1}}$ and the circumscribed circle $\Delta AED$. Prove that $IT = IA$.

1939 Eotvos Mathematical Competition, 3

Tags: construction , semicircle , equal segments , geometry

$ABC$ is an acute triangle. Three semicircles are constructed outwardly on the sides $BC$, $CA$ and $AB$ respectively. Construct points $A'$ , $B'$ and $C' $ on these semicìrcles respectively so that $AB' = AC'$, $BC' = BA'$ and $CA'= CB'$.

2015 Caucasus Mathematical Olympiad, 3

Tags: equal segments , angle bisector , circumcircle , geometry , equal angles

Let $AL$ be the angle bisector of the acute-angled triangle $ABC$. and $\omega$ be the circle circumscribed about it. Denote by $P$ the intersection point of the extension of the altitude $BH$ of the triangle $ABC$ with the circle $\omega$ . Prove that if $\angle BLA= \angle BAC$, then $BP = CP$.

2018 Ecuador NMO (OMEC), 5

Tags: cyclic quadrilateral , concyclic , equal segments

Let $ABC$ be an acute triangle and let $M$, $N$, and $ P$ be on $CB$, $AC$, and $AB$, respectively, such that $AB = AN + PB$, $BC = BP + MC$, $CA = CM + AN$. Let $\ell$ be a line in a different half plane than $C$ with respect to to the line $AB$ such that if $X, Y$ are the projections of $A, B$ on $\ell$ respectively, $AX = AP$ and $BY = BP$. Prove that $NXYM$ is a cyclic quadrilateral.

2021 Austrian Junior Regional Competition, 2

Tags: geometry , equal segments , circumcenter

A triangle $ABC$ with circumcenter $U$ is given, so that $\angle CBA = 60^o$ and $\angle CBU = 45^o$ apply. The straight lines $BU$ and $AC$ intersect at point $D$. Prove that $AD = DU$. (Karl Czakler)

2018 Thailand TSTST, 2

Tags: geometry , circumcircle , equal segments

In triangle $\vartriangle ABC$, $\angle BAC = 135^o$. $M$ is the midpoint of $BC$, and $N \ne M$ is on $BC$ such that $AN = AM$. The line $AM$ meets the circumcircle of $\vartriangle ABC$ at $D$. Point $E$ is chosen on segment $AN$ such that $AE = MD$. Show that $ME = BC$.

2020 Ukrainian Geometry Olympiad - April, 3

Tags: geometry , circumcircle , parallel , perpendicular , equal segments

Triangle $ABC$. Let $B_1$ and $C_1$ be such points, that $AB= BB_1, AC=CC_1$ and $B_1, C_1$ lie on the circumscribed circle $\Gamma$ of $\vartriangle ABC$. Perpendiculars drawn from from points $B_1$ and $C_1$ on the lines $AB$ and $AC$ intersect $\Gamma$ at points $B_2$ and $C_2$ respectively, these points lie on smaller arcs $AB$ and $AC$ of circle $\Gamma$ respectively, Prove that $BB_2 \parallel CC_2$.

1979 Austrian-Polish Competition, 1

Tags: square , right angle , equal segments , geometry

On sides $AB$ and $BC$ of a square $ABCD$ the respective points $E$ and $F$ have been chosen so that $BE = BF$. Let $BN$ be the altitude in triangle $BCE$. Prove that $\angle DNF = 90$.

2001 Greece Junior Math Olympiad, 4

Tags: geometry , equal segments , angle chasing

Let $ABC$ be a triangle with altitude $AD$ , angle bisectors $AE$ and $BZ$ that intersecting at point $I$. From point $I$ let $IT$ be a perpendicular on $AC$. Also let line $(e)$ be perpendicular on $AC$ at point $A$. Extension of $ET$ intersects line $(e)$ at point $K$. Prove that $AK=AD$.