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Estonian Math Competitions

Geometry Difficulty 5.1 AIME, harder Prove it Estonia

Two circles $c$ and $c'$ with centers $O$ and $O'$ lie completely outside each other. Points $A$ , $B$ , and $C$ lie on the circle $c$ and points $A'$ , $B'$ , and $C'$ lie on the circle $c'$ so that segment $AB \parallel A'B'$ , $BC \parallel B'C'$ , and $\angle ABC = \angle A'B'C'$ . The lines $AA'$ , $BB'$ , and $CC'$ are all different and intersect in one point $P$ , which does not coincide with any of the vertices of the triangles $ABC$ or $A'B'C'$ . Prove that $\angle AOB = \angle A'O'B'$ .

Fig. 1

Solution

The triangles $ABP$ and $A'B'P$ are similar, because their corresponding sides are parallel (Fig. 1). Hence $\frac{|AB|}{|A'B'|} = \frac{|BP|}{|B'P|}$ . Likewise the triangles $BCP$ and $B'C'P$ are similar, hence $\frac{|BC|}{|B'C'|} = \frac{|BP|}{|B'P|}$ . Thus $\frac{|AB|}{|A'B'|} = \frac{|BC|}{|B'C'|}$ , and since $\angle ABC = \angle A'B'C'$ , the triangles $ABC$ and $A'B'C'$ are also similar. From the equality of the angles $ACB$ and $A'C'B'$ the equality of the central angles $AOB$ and $A'O'B'$ now follows.

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