Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Prove it North Macedonia

Let be given the $\triangle ABC$ . On the arc $\text{BC}$ of the circumcircle of $\triangle ABC$ , which does not contain the point $A$ , points $X$ and $Y$ are chosen, such that $\angle BAX = \angle CAY$ . Let $M$ be the middle point of the chord $AX$ . Prove that $\overline{BM} + \overline{CM} > \overline{AY}$ .

Solution

Let $O$ be the circumcenter of the circumcircle of $\triangle ABC$ . Then $OM \perp AX$ . We draw a normal line from the point $B$ at $OM$ and let it intersect the circumcircle in the point $Z$ . Since $BZ \perp OM$ we have that $OM$ is a line of symmetry of $BZ$ . According to this, $\overline{MZ} = \overline{MB}$ . Now, from the triangle inequality we have that
$\text{BM + MC = ZM + MC > CZ .}$
But, $BZ \parallel AX$ , so
$\text{AZ = BX = CY}$
where from we get
$\text{ZA C = ZI A + AC = YC + CA = YC A}$

i.e. $\text{CZ = AY}$ . That is why $\text{BM + CM > AY}$ .

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