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Silk Road Mathematics Competition competition problems

Geometry Difficulty 6.6 National Olympiad Prove it Silk Road Mathematics Competition

Let $A$ , $B$ , $C$ be three collinear points such that the point $B$ lies between $A$ and $C$ . Let $AA'$ and $BB'$ be parallel lines such that the points $A'$ and $B'$ lie on the same side of the line $AB$ , and $A'$ , $B'$ , $C$ are not collinear. Let $O_1$ be the center of the circle passing through the points $A$ , $A'$ , $C$ , and $O_2$ be the center of the circle passing through the points $B$ , $B'$ , $C$ . Find all possible values of the angle $CAA'$ , if triangles $A'CB'$ and $O_1CO_2$ have the same area.

Solution

If $\angle CAA'$ is acute, then $\angle CO_1A' = 2\angle CAA' = 2\angle CBB' = \angle CO_2B'$ . And if $\angle CAA'$ is obtuse, then $\angle CO_1A' = 2(180^\circ - \angle CAA') = 2(180^\circ - \angle CBB') = \angle CO_2B'$ . In particular, $\angle CO_1A' = \angle CO_2B'$ . (It can be easily seen that $S_{CA'B'} \neq S_{CO_1O_2}$ is not possible if $\angle CAA' = 90^\circ$ .) Since triangles $CO_1A'$ and $CO_2B$ are both isosceles, it follows that they are similar.

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