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Saudi Arabian Mathematical Competitions

Geometry Difficulty 6.3 National Olympiad Prove it Saudi Arabia

Given triangle $A B C$ inscribed in $(O)$ . Two tangents at $B$ , $C$ of $(O)$ intersect at $P$ . The bisector of angle $A$ intersects $(P, P B)$ (the circle with center $P$ and radius $P B$ ) at point $E$ lying inside triangle $A B C$ . Let $M$ , $N$ be the midpoints of two arcs $B C$ of $(O)$ such that $M$ and $A$ are on different sides of $B C$ . The circle with diameter $B C$ intersects the line segment $E N$ at $F$ . Prove that the orthocenter of triangle $E F M$ belongs to $B C$ .

Solution

Let $I$ be the midpoint of $B C$ . We have $\angle I C M = \angle M A C = \angle M C P$ then $C M$ is the bisector of $\angle I C P$ . This follows that $M$ is the insimilicenter of $(I)$ and $(P)$ . However, $\angle M C N = 90^{\circ}$ hence $N$ is the exsimilicenter of $(I)$ and $(P)$ .

Let $L$ , $T$ be the intersections of $F M$ and $(P)$ ( $L$ is between $F$ and $M$ ). Consider the homothety center at $M$ , ratio $\frac{-P C}{I C}$ , $\mathscr{H}_{M}^{\frac{-P C}{I C}}: F \mapsto T$ , we get $I F \parallel P T$ .

By the same way, consider the homothety $\mathscr{H}_{N}^{\frac{P C}{I C}}: F \mapsto E$ , we get $I F \parallel P E$ .

Therefore $E$ , $P$ , $T$ are collinear or $E T$ is the diameter of $(P)$ . This means $\angle E L P = 90^{\circ}$ .

Denote $J$ the second intersection of $F N$ and $(O)$ .

Since 3 circles $(O)$ , $(P)$ , $(E L M J)$ have $E L$ , $B C$ , $J M$ as their radical axes then $E L$ cuts $J M$ at point $K$ lying on $B C$ .

In conclusion, orthocenter $K$ of triangle $E F M$ lies on $B C$ . $\square$

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