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Ukrainian National Mathematical Olympiad - Third Round

Geometry Difficulty 5.5 AIME, harder Prove it Ukraine

Circles $w_1$ and $w_2$ centered at points $O_1$ and $O_2$ respectively intersect at points $A$ and $B$ . Let $w$ be the circumscribed circle of $O_1O_2B$ centered at $O$ , which intersects $w_1$ and $w_2$ again at points $K$ and $L$ respectively. The straight line $OA$ intersects $w_1$ and $w_2$ at points $M$ and $N$ respectively. Denote by $P$ the intersection point of lines $MK$ and $NL$ . Prove that $P$ lies on $w$ and $PM = PN$ .
(Vadym Mytrofanov)

Solution

We use the following lemma by Archimedes:

Lemma (Archimedes). Circles $w_1$ and $w_2$ intersect at points $A$ and $B$ , with the center of $w_2$ lying on $w_1$ . A chord $AC$ of $w_2$ intersects $w_1$ again at a point $M$ . Then $CM = CB$ .

Put $\alpha = \angle KBA = \angle KMA$ , since they intercept the same arc in $w_1$ . Similarly, $\beta = \angle ABL = \angle ANL$ , because they intercept the same arc in $w_2$ . Hence $\angle MPN = 180^\circ - \alpha - \beta$ , which implies that $P \in w$ (Fig. 8).

By Archimedes' lemma, for circles $w_1$ and $w_2$ we can write $KJ = JA = LJ$ , but then $\alpha = \beta$ , and from the circle $w$ we have that $PM = PN$ .

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