55rd Ukrainian National Mathematical Olympiad - Third Round

Question

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)

Accepted Answer

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

A C

of

w_{2}

intersects

w_{1}

again at a point

M

. Then

C M = C B

.

Put

α = ∠ K B A = ∠ K M A

, since they intercept the same arc in

w_{1}

. Similarly,

β = ∠ A B L = ∠ A N L

, because they intercept the same arc in

w_{2}

. Hence

∠ M P N = 18 0^{\circ} - α - β

, which implies that

P \in w

(Fig. 8).

By Archimedes' lemma, for circles

w_{1}

and

w_{2}

we can write

K J = J A = L J

, but then

α = β

, and from the circle

w

we have that

P M = P N

.