Olympiad problems

1973 All Soviet Union Mathematical Olympiad, 177

Given an angle with the vertex $O$ and a circle touching its sides in the points $A$ and $B$. A ray is drawn from the point $A$ parallel to $[OB)$. It intersects with the circumference in the point $C$. The segment $[OC]$ intersects the circumference in the point $E$. The straight lines $(AE)$ and $(OB)$ intersect in the point $K$. Prove that $|OK| = |KB|$.

2006 Denmark MO - Mohr Contest, 5

Tags: altitude , geometry , similar triangles , equal segments

We consider an acute triangle $ABC$. The altitude from $A$ is $AD$, the altitude from $D$ in triangle $ABD$ is $DE,$ and the altitude from $D$ in triangle $ACD$ is $DF$. a) Prove that the triangles $ABC$ and $AF E$ are similar. b) Prove that the segment $EF$ and the corresponding segments constructed from the vertices $B$ and $C$ all have the same length.

2015 Denmark MO - Mohr Contest, 3

Tags: geometry , equilateral , equal segments

Triangle $ABC$ is equilateral. The point $D$ lies on the extension of $AB$ beyond $B$, the point $E$ lies on the extension of $CB$ beyond $B$, and $|CD| = |DE|$. Prove that $|AD| = |BE|$. [img]https://1.bp.blogspot.com/-QnAXFw3ijn0/XzR0YjqBQ3I/AAAAAAAAMU0/0TvhMQtBNjolYHtgXsQo2OPGJzEYSfCwACLcBGAsYHQ/s0/2015%2BMohr%2Bp3.png[/img]

Kyiv City MO Juniors Round2 2010+ geometry, 2019.7.3

Tags: equal segments , equal angles , geometry

In the quadrilateral $ABCD$ it is known that $\angle ABD= \angle DBC$ and $AD= CD$. Let $DH$ be the altitude of $\vartriangle ABD$. Prove that $| BC - BH | = HA$. (Hilko Danilo)

1957 Moscow Mathematical Olympiad, 354

Tags: midpoint , geometry , parallel , equal segments

In a quadrilateral $ABCD$ points $M$ and $N$ are the midpoints of the diagonals $AC$ and $BD$, respectively. The line through $M$ and $N$ meets $AB$ and $CD$ at $M'$ and $N'$, respectively. Prove that if $MM' = NN'$, then $AD // BC$.

Kyiv City MO Juniors 2003+ geometry, 2021.9.5

Tags: geometry , equal segments , median

Let $BM$ be the median of the triangle $ABC$, in which $AB> BC$. Point $P$ is chosen so that $AB \parallel PC$ and$ PM \perp BM$. The point $Q$ is chosen on the line $BP$ so that $\angle AQC = 90^o$, and the points $B$ and $Q$ lie on opposite sides of the line $AC$. Prove that $AB = BQ$. (Mikhail Standenko)

2019 Saudi Arabia BMO TST, 2

Tags: geometry , trapezoid , right angle , equal segments

Let $ABCD$ is a trapezoid with $\angle A = \angle B = 90^o$ and let $E$ is a point lying on side $CD$. Let the circle $\omega$ is inscribed to triangle $ABE$ and tangents sides $AB, AE$ and $BE$ at points $P, F$ and $K$ respectively. Let $KF$ intersects segments $BC$ and $AD$ at points $M$ and $N$ respectively, as well as $PM$ and $PN$ intersect $\omega$ at points $H$ and $T$ respectively. Prove that $PH = PT$.

2014 Czech-Polish-Slovak Junior Match, 2

Tags: geometry , circumcircle , parallelogram , equal segments

Let $ABCD$ be a parallelogram with $\angle BAD<90^o$ and $AB> BC$ . The angle bisector of $BAD$ intersects line $CD$ at point $P$ and line $BC$ at point $Q$. Prove that the center of the circle circumscirbed around the triangle $CPQ$ is equidistant from points $B$ and $D$.

2010 Contests, 3

Tags: geometry , isosceles , equal segments , angle

Consider triangle $ABC$ with $AB = AC$ and $\angle A = 40 ^o$. The points $S$ and $T$ are on the sides $AB$ and $BC$, respectively, so that $\angle BAT = \angle BCS= 10 ^o$. The lines $AT$ and $CS$ intersect at point $P$. Prove that $BT = 2PT$.

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 $|MA| = |MC|$.

2015 Latvia Baltic Way TST, 16

Tags: equal segments , circles , geometry

Points $X$ , $Y$, $Z$ lie on a line $k$ in this order. Let $\omega_1$, $\omega_2$, $\omega_3$ be three circles of diameters $XZ$, $XY$ , $YZ$ , respectively. Line $\ell$ passing through point $Y$ intersects $\omega_1$ at points $A$ and $D$, $\omega_2$ at $B$ and $\omega_3$ at $C$ in such manner that points $A, B, Y, X, D$ lie on $\ell$ in this order. Prove that $AB =CD$.

Kyiv City MO Seniors 2003+ geometry, 2006.11.3

Tags: midpoint , circumcircle , equal segments , geometry

Let $O$ be the center of the circle $\omega$ circumscribed around the acute-angled triangle $\vartriangle ABC$, and $W$ be the midpoint of the arc $BC$ of the circle $\omega$, which does not contain the point $A$, and $H$ be the point of intersection of the heights of the triangle $\vartriangle ABC$. Find the angle $\angle BAC$, if $WO = WH$. (O. Clurman)

2012 Dutch BxMO/EGMO TST, 4

Tags: geometry , equal segments , area

Let $ABCD$ a convex quadrilateral (this means that all interior angles are smaller than $180^o$), such that there exist a point $M$ on line segment $AB$ and a point $N$ on line segment $BC$ having the property that $AN$ cuts the quadrilateral in two parts of equal area, and such that the same property holds for $CM$. Prove that $MN$ cuts the diagonal $BD$ in two segments of equal length.

Swiss NMO - geometry, 2018.6

Tags: geometry , parallel , equal segments , incircle

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 $BD = EF$.

2015 Indonesia MO, 6

Tags: geometry , equal segments , circumcircle , perpendicular

Let $ABC$ be an acute angled triangle with circumcircle $O$. Line $AO$ intersects the circumcircle of triangle $ABC$ again at point $D$. Let $P$ be a point on the side $BC$. Line passing through $P$ perpendicular to $AP$ intersects lines $DB$ and $DC$ at $E$ and $F$ respectively . Line passing through $D$ perpendicular to $BC$ intersects $EF$ at point $Q$. Prove that $EQ = FQ$ if and only if $BP = CP$.

Estonia Open Senior - geometry, 2002.1.4

Tags: geometry , equal segments , angle

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 $\angle A$.

Kyiv City MO Juniors Round2 2010+ geometry, 2019.8.41

Tags: geometry , parallelogram , equal segments

Through the vertices $A, B$ of the parallelogram $ABCD$ passes a circle that intersects for the second time diagonals $BD$ and $AC$ at points $X$ and $Y$, respectively. The circumsccribed circle of $\vartriangle ADX$ intersects diagonal $AC$ for the second time at the point $Z$. Prove that $AY = CZ$.

2018 Dutch IMO TST, 3

Tags: geometry , right angle , equal segments

Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude through $A$. On $AD$, there are distinct points $E$ and $F$ such that $|AE| = |BE|$ and $|AF| =|CF|$. A point$ T \ne D$ satises $\angle BTE = \angle CTF = 90^o$. Show that $|TA|^2 =|TB| \cdot |TC|$.

2016 Romania National Olympiad, 2

Tags: geometry , angle bisector , equal segments , angle

Consider the triangle $ABC$, where $\angle B= 30^o, \angle C = 15^o$, and $M$ is the midpoint of the side $[BC]$. Let point $N \in (BC)$ be such that $[NC] = [AB]$. Show that $[AN$ is the angle bisector of $MAC$

Kyiv City MO Juniors 2003+ geometry, 2021.8.41

Tags: geometry , equal segments

On the sides $AB$ and $BC$ of the triangle $ABC$, the points $K$ and $M$ are chosen so that $KM \parallel AC$. The segments $AM$ and $KC$ intersect at the point $O$. It is known that $AK =AO$ and $KM =MC$. Prove that $AM=KB$.

2019 District Olympiad, 2

Tags: geometry , equal angles , equal segments , angle

Consider $D$ the midpoint of the base $[BC]$ of the isosceles triangle ABC in which $\angle BAC < 90^o$. On the perpendicular from $B$ on the line $BC$ consider the point $E$ such that $\angle EAB= \angle BAC$, and on the line passing though $C$ parallel to the line $AB$ we consider the point $F$ such that $F$ and $D$ are on different side of the line $AC$ and $\angle FAC = \angle CAD$. Prove that $AE = CF$ and $BF = EF$

2020 Chile National Olympiad, 3

Tags: ratio , geometry , equal segments , isosceles

Given the isosceles triangle $ABC$ with $| AB | = | AC | = 10$ and $| BC | = 15$. Let points $P$ in $BC$ and $Q$ in $AC$ chosen such that $| AQ | = | QP | = | P C |$. Calculate the ratio of areas of the triangles $(PQA): (ABC)$.

2009 Thailand Mathematical Olympiad, 3

Tags: geometry , equal segments , parallel

Let $ABCD$ be a convex quadrilateral with the property that $MA \cdot MC + MA \cdot CD = MB \cdot MD$, where $M$ is the intersection of the diagonals $AC$ and $BD$. The angle bisector of $\angle ACD$ is drawn intersecting ray $\overrightarrow{BA}$ at $K$. Prove that $BC = DK$ if and only if $AB \parallel CD$.

Denmark (Mohr) - geometry, 2023.4

Tags: geometry , equal segments , regular

In the $9$-gon $ABCDEFGHI$, all sides have equal lengths and all angles are equal. Prove that $|AB| + |AC| = |AE|$. [img]https://cdn.artofproblemsolving.com/attachments/6/2/8c82e8a87bf8a557baaf6ac72b3d18d2ba3965.png[/img]

2014 Oral Moscow Geometry Olympiad, 2

Tags: parallelogram , equal segments , geometry

Let $ABCD$ be a parallelogram. On side $AB$, point $M$ is taken so that $AD = DM$. On side $AD$ point $N$ is taken so that $AB = BN$. Prove that $CM = CN$.