26th Junior Balkan Mathematical Olympiad | JBMO 2022 - Problem 2

Problem 2

Let $ABC$ be an acute triangle such that $AH = HD$ , where $H$ is the orthocenter of $ABC$ and $D \in BC$ is the foot of the altitude from the vertex $A$ . Let $\ell$ denote the line through $H$ which is tangent to the circumcircle of the triangle $BHC$ . Let $S$ and $T$ be the intersection points of $\ell$ with $AB$ and $AC$ , respectively. Denote the midpoints of $BH$ and $CH$ by $M$ and $N$ , respectively. Prove that the lines $SM$ and $TN$ are parallel.

Solution

Let L be the midpoint of AC, O be the midpoint of AB, E, F the traces of the heights from A, B respectively.

It is $\angle{ATH} = 90^{\circ} - \angle{EHT} = 90^{\circ} - \angle{SHB} = 90^{\circ} - \angle{HCB} = \angle{B}$ ,

А quadrilateral BCTS is cyclic , so $\angle{ATS} = \angle{B}$ .

Now $LN \parallel BH \leftrightarrow \angle{LNH}=\angle{AHF}=\angle{B}=\angle{ATS}$ .

А quadrilateral THNL is cyclic , so $\angle{HTN}=\angle{HLN}=90^{\circ}$ we get that $HL \parallel BC$ .

We also show that $\angle{HSM}=90^{\circ}$ and we are done.