Olympiad problems

Denmark (Mohr) - geometry, 2013.5

The angle bisector of $A$ in triangle $ABC$ intersects $BC$ in the point $D$. The point $E$ lies on the side $AC$, and the lines $AD$ and $BE$ intersect in the point $F$. Furthermore, $\frac{|AF|}{|F D|}= 3$ and $\frac{|BF|}{|F E|}=\frac{5}{3}$. Prove that $|AB| = |AC|$. [img]https://1.bp.blogspot.com/-evofDCeJWPY/XzT9dmxXzVI/AAAAAAAAMVY/ZN87X3Cg8iMiULwvMhgFrXbdd_f1f-JWwCLcBGAsYHQ/s0/2013%2BMohr%2Bp5.png[/img]

2014 Thailand Mathematical Olympiad, 1

Tags: geometry , equal segments

Let $\vartriangle ABC$ be an isosceles triangle with $\angle BAC = 100^o$. Let $D, E$ be points on ray $\overrightarrow{AB}$ so that $BC = AD = BE$. Show that $BC \cdot DE = BD \cdot CE$

2013 Tournament of Towns, 4

Tags: equal segments , isosceles , angle , geometry

Let $ABC$ be an isosceles triangle. Suppose that points $K$ and $L$ are chosen on lateral sides $AB$ and $AC$ respectively so that $AK = CL$ and $\angle ALK + \angle LKB = 60^o$. Prove that $KL = BC$.

2019 Romania National Olympiad, 2

Tags: equal segments , geometry , square , perpendicular

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) $ND\perp PC$.

2010 Contests, 3

Tags: equal segments , isosceles , geometry , 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 Indonesia MO Shortlist, G4

Tags: geometry , circumcircle , equal segments , isosceles

Given an isosceles triangle $ABC$ with $AB = AC$, suppose $D$ is the midpoint of the $AC$. The circumcircle of the $DBC$ triangle intersects the altitude from $A$ at point $E$ inside the triangle $ABC$, and the circumcircle of the triangle $AEB$ cuts the side $BD$ at point $F$. If $CF$ cuts $AE$ at point $G$, prove that $AE = EG$.

2014 Czech-Polish-Slovak Junior Match, 1

Tags: parallelogram , circles , geometry , equal segments

On the plane circles $k$ and $\ell$ are intersected at points $C$ and $D$, where circle $k$ passes through the center $L$ of circle $\ell$. The straight line passing through point $D$ intersects circles $k$ and $\ell$ for the second time at points $A$ and $B$ respectively in such a way that $D$ is the interior point of segment $AB$. Show that $AB = AC$.

2006 Denmark MO - Mohr Contest, 5

Tags: geometry , altitude , 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.

2019 Balkan MO Shortlist, G5

Tags: circumcircle , geometry , equal segments

Let $ABC$ ($BC > AC$) be an acute triangle with circumcircle $k$ centered at $O$. The tangent to $k$ at $C$ intersects the line $AB$ at the point $D$. The circumcircles of triangles $BCD, OCD$ and $AOB$ intersect the ray $CA$ (beyond $A$) at the points $Q, P$ and $K$, respectively, such that $P \in (AK)$ and $K \in (PQ)$. The line $PD$ intersects the circumcircle of triangle $BKQ$ at the point $T$, so that $P$ and $T$ are in different halfplanes with respect to $BQ$. Prove that $TB = TQ$.