Olympiad problems

2019 Romania National Olympiad, 2

Let $ABCD$ be a square and $E$ a point on the side $(CD)$. Squares $ENMA$ and $EBQP$ are constructed outside the triangle $ABE$. Prove that: a) $ND = PC$ b) $ND\perp PC$.

2001 Junior Balkan Team Selection Tests - Moldova, 3

Tags: geometry , equilateral , equal segments

Let the convex quadrilateral $ABCD$ with $AD = BC$ ¸and $\angle A + \angle B = 120^o$. Take a point $P$ in the plane so that the line $CD$ separates the points $A$ and $P$, and the $DCP$ triangle is equilateral. Show that the triangle $ABP$ is equilateral. It is the true statement for a non-convex quadrilateral?

2011 Saudi Arabia Pre-TST, 1.4

Tags: equal angles , geometry , equal segments

Let $ABC$ be a triangle with $AB=AC$ and $\angle BAC = 40^o$. Points $S$ and $T$ lie on the sides $AB$ and $BC$, such that $\angle BAT = \angle BCS = 10^o$. Lines $AT$ and $CS$ meet at $P$. Prove that $BT = 2PT$.

Ukraine Correspondence MO - geometry, 2011.9

Tags: geometry , hexagon , collinear , equal segments

On the diagonals $AC$ and $CE$ of a regular hexagon $ABCDEF$ with side $1$ we mark points$ M$ and $N$ such that $AM = CN = a$. Find $a$ if the points $B, M, N$ lie on the same line.

2011 Sharygin Geometry Olympiad, 7

Tags: equal segments , circles , angle , common tangent , geometry

Circles $\omega$ and $\Omega$ are inscribed into the same angle. Line $\ell$ meets the sides of angles, $\omega$ and $\Omega$ in points $A$ and $F, B$ and $C, D$ and $E$ respectively (the order of points on the line is $A,B,C,D,E, F$). It is known that$ BC = DE$. Prove that $AB = EF$.

2016 Grand Duchy of Lithuania, 3

Tags: equal segments , isosceles , geometry

Let $ABC$ 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 angle bisector of $\angle BEC$. Prove that $BD = EF$ if and only if $AF = EC$.

2001 Mexico National Olympiad, 5

Tags: geometry , equal segments

$ABC$ is a triangle with $AB < AC$ and $\angle A = 2 \angle C$. $D$ is the point on $AC$ such that $CD = AB$. Let L be the line through $B$ parallel to $AC$. Let $L$ meet the external bisector of $\angle A$ at $M$ and the line through $C$ parallel to $AB$ at $N$. Show that $MD = ND$.

2003 Oral Moscow Geometry Olympiad, 3

Tags: geometry , equal segments , equal angles , concyclic

Inside the segment $AC$, an arbitrary point $B$ is selected and circles with diameters $AB$ and $BC$ are constructed. Points $M$ and $L$ are chosen on the circles (in one half-plane with respect to $AC$), respectively, so that $\angle MBA = \angle LBC$. Points $K$ and $F$ are marked, respectively, on rays $BM$ and $BL$ so that $BK = BC$ and $BF = AB$. Prove that points $M, K, F$ and $L$ lie on the same circle.

2022 Oral Moscow Geometry Olympiad, 3

Tags: equal segments , geometry

In quadrilateral $ABCD$, sides $AB$ and $CD$ are equal (but not parallel), points $M$ and $N$ are the midpoints of $AD$ and $BC$. The perpendicular bisector of $MN$ intersects sides $AB$ and $CD$ at points $P$ and $Q$, respectively. Prove that $AP = CQ$. (M. Kungozhin)

1953 Kurschak Competition, 3

Tags: geometry , equal segments , equal angles , hexagon

$ABCDEF$ is a convex hexagon with all its sides equal. Also $\angle A + \angle C + \angle E = \angle B + \angle D + \angle F$. Show that $\angle A = \angle D$, $\angle B = \angle E$ and $\angle C = \angle F$.

2002 Croatia Team Selection Test, 2

Tags: geometry , incenter , equal segments

A quadrilateral $ABCD$ is circumscribed about a circle. Lines $AC$ and $DC$ meet at point $E$ and lines $DA$ and $BC$ meet at $F$, where $B$ is between $A$ and $E$ and between $C$ and $F$. Let $I_1, I_2$ and $I_3$ be the incenters of triangles $AFB, BEC$ and $ABC$, respectively. The line $I_1I_3$ intersects $EA$ at $K$ and $ED$ at $L$, whereas the line $I_2I_3$ intersects $FC$ at $M$ and $FD$ at $N$. Prove that $EK = EL$ if and only if $FM = FN$

2010 NZMOC Camp Selection Problems, 5

Tags: ratio , geometry , equal segments , equal angles

The diagonals of quadrilateral $ABCD$ intersect in point $E$. Given that $|AB| =|CE|$, $|BE| = |AD|$, and $\angle AED = \angle BAD$, determine the ratio $|BC|:|AD|$.

2008 Junior Balkan Team Selection Tests - Moldova, 11

Tags: geometry , congruent triangles , parallelogram , equal segments , isosceles

Let $ABCD$ be a convex quadrilateral with $AD = BC, CD \nparallel AB, AD \nparallel BC$. Points $M$ and $N$ are the midpoints of the sides $CD$ and $AB$, respectively. a) If $E$ and $F$ are points, such that $MCBF$ and $ADME$ are parallelograms, prove that $\vartriangle BF N \equiv \vartriangle AEN$. b) Let $P = MN \cap BC$, $Q = AD \cap MN$, $R = AD \cap BC$. Prove that the triangle $PQR$ is iscosceles.

2017 Oral Moscow Geometry Olympiad, 1

Tags: equal segments , geometry

On side $AB$ of triangle $ABC$ is marked point $K$ such that $AB = CK$. Points $N$ and $M$ are the midpoints of $AK$ and $BC$, respectively. The segments $NM$ and $CK$ intersect in point $P$. Prove that $KN = KP$.

2018 Switzerland - Final Round, 6

Tags: parallel , equal segments , incircle , geometry

Let $k$ be the incircle of the triangle $ABC$ with the center of the incircle $I$. The circle $k$ touches the sides $BC, CA$ and $AB$ in points $D, E$ and $F$. Let $G$ be the intersection of the straight line $AI$ and the circle $k$, which lies between $A$ and $I$. Assume $BE$ and $FG$ are parallel. Show that $BD = EF$.

1995 All-Russian Olympiad Regional Round, 9.6

Tags: geometry , circles , equal segments

Circles $S_1$ and $S_2$ with centers $O_1$ and $O_2$ respectively intersect at $A$ and $B$. The circle passing through $O_1$, $O_2$, and $A$ intersects $S_1$, $S_2$ and line $AB$ again at $D$, $E$, and $C$, respectively. Show that $CD = CB = CE$.

2023 Greece Junior Math Olympiad, 2

Tags: equal segments , geometry

In triangle $ABC$, points $M$, $N$ are the midpoints of sides $AB$, $AC$ respelctively. Let $D$ and $E$ be two points on line segment $BN$ such that $CD \parallel ME$ and $BD <BE$. Prove that $BD=2\cdot EN$.

2018 Estonia Team Selection Test, 7

Tags: geometry , altitude , equal segments , circumcircle , concurrency

Let $AD$ be the altitude $ABC$ of an acute triangle. On the line $AD$ are chosen different points $E$ and $F$ so that $|DE |= |DF|$ and point $E$ is in the interior of triangle $ABC$. The circumcircle of triangle $BEF$ intersects $BC$ and $BA$ for second time at points $K$ and $M$ respectively. The circumcircle of the triangle $CEF$ intersects the $CB$ and $CA$ for the second time at points $L$ and $N$ respectively. Prove that the lines $AD, KM$ and $LN$ intersect at one point.

Durer Math Competition CD Finals - geometry, 2018.C+2

Tags: geometry , incenter , circumcircle , equal segments

Given an $ABC$ triangle. Let $D$ be an extension of section $AB$ beyond $A$ such that that $AD = BC$ and $E$ is the extension of the section $BC$ beyond $B$ such that $BE = AC$. Prove that the circumcircle of triangle $DEB$ passes through the center of the inscribed circle of triangle $ABC$.

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 $AK=AD$.

2016 Hanoi Open Mathematics Competitions, 12

Tags: trapezoid , equal segments , equal angles , geometry

In the trapezoid $ABCD, AB // CD$ and the diagonals intersect at $O$. The points $P, Q$ are on $AD, BC$ respectively such that $\angle AP B = \angle CP D$ and $\angle AQB = \angle CQD$. Show that $OP = OQ$.

2012 Bundeswettbewerb Mathematik, 3

Tags: geometry , incircle , equal segments , diameter

The incircle of the triangle $ABC$ touches the sides $BC, CA$ and $AB$ in points $A_1, B_1$ and $C_1$ respectively. $C_1D$ is a diameter of the incircle. Finally, let $E$ be the intersection of the lines $B_1C_1$ and $A_1D$. Prove that the segments $CE$ and $CB_1$ have equal length.

Kyiv City MO Juniors Round2 2010+ geometry, 2011.8.3

Tags: geometry , right angle , square , equal segments

On the sides $AD , BC$ of the square $ABCD$ the points $M, N$ are selected $N$, respectively, such that $AM = BN$. Point $X$ is the foot of the perpendicular from point $D$ on the line $AN$. Prove that the angle $MXC$ is right. (Mirchev Borislav)

2004 District Olympiad, 2

Tags: geometry , angle bisector , equal segments , parallel

Let $ABC$ be a triangle and $D$ a point on the side $BC$. The angle bisectors of $\angle ADB ,\angle ADC$ intersect $AB ,AC$ at points $M ,N$ respectively. The angle bisectors of $\angle ABD , \angle ACD$ intersects $DM , DN$ at points $K , L$ respectively. Prove that $AM = AN$ if and only if $MN$ and $KL$ are parallel.

Kyiv City MO Juniors 2003+ geometry, 2018.7.4

Tags: geometry , right angle , equal segments , angle

Inside the triangle $ABC $, the point $P $ is selected so that $BC = AP $ and $\angle APC = 180 {} ^ \circ - \angle ABC $. On the side $AB $ there is a point $K $, for which $AK = KB + PC $. Prove that $\angle AKC = 90 {} ^ \circ $. (Danilo Hilko)