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Ukrainian National Mathematical Olympiad, 4th Round

Geometry Difficulty 5.6 AIME, harder Prove it Ukraine

In a triangle $ABC$ with $AC > BC > AB$ . Points $D$ and $K$ are chosen on the sides $BC$ and $AC$ respectively so that $CD = AB$ , $AK = BC$ . $F$ and $L$ are the midpoints of the segments $BD$ and $KC$ respectively. $R$ and $S$ are the midpoints of the sides $AC$ and $AB$ respectively. The line segments $SL$ and $FR$ intersect at the point $O$ , and it is known that $\angle SOF = 55^\circ$ . Find $\angle BAC$ .

Solution

Fig. 41

$\angle FOS = \angle BJT = 55^\circ \Rightarrow \angle BJC = 125^\circ$ , and because $J$ is the incenter $\angle BJC = 90^\circ + \frac{1}{2}\angle BAC \Rightarrow \angle BAC = 70^\circ$ .

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