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Ukrainian National Mathematical Olympiad

Geometry Difficulty 5.0 AIME Prove it Ukraine

Let $M$ be the middle point of the side $AC$ of triangle $ABC$ . Inside $\triangle BMC$ , there is such point $P$ that $\angle BMP = 90^\circ$ , $\angle ABC + \angle APC = 180^\circ$ . Prove that $\angle PBM + \angle CBM = \angle PCA$ .

(Anton Tryhub)

Solution

We construct point $B'$ such that $ABCB'$ is a parallelogram. Then (Fig. 26)
$\angle ABC + \angle APC = 180^\circ = \angle AB'C + \angle APC = 180^\circ,$
hence, quadrilateral $APCB'$ is inscribed. Obviously, $\triangle BPB'$ is isosceles, which yields
$\angle PCA = \angle PB'A = \angle PB'M + \angle MB'A = \\ = \angle PBM + \angle MBC, \text{ Q.E.D.}$

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