Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.4 AIME, harder Prove it Bulgaria

Let $M$ and $N$ be the midpoints of the sides $AC$ and $BC$ of $\triangle ABC$ ( $AC > BC$ ) and let the bisector of $\angle B$ intersect the segment $MN$ at a point $P$ . The incircle of $\triangle ABC$ has center $I$ and is tangent to $BC$ at a point $Q$ . Denote by $R$ the intersection point of the perpendiculars from $P$ and $Q$ to $MN$ and $BC$ , respectively, and by $S$ the intersection point of the lines $AB$ and $RN$ .

a) Prove that the quadrilateral *PCQI* is cyclic.
b) Express the length of the segment *BS* by the lengths *a*, *b*, *c* of the sides of $\triangle ABC$ .

Solution

a) Obviously
$\vDash ABP = \vDash BPN = \vDash PBN = \beta/2.$
Therefore $BN = CN = PN$ , whence $\vDash BPC = 90^\circ$ . Since $\vDash CQI = 90^\circ$ , the quadrilateral *PCQI* is cyclic.

b) Answer. $BS = \frac{b+c}{2}$ .

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