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Geometry Difficulty 6.6 National olympiad Prove it Silk Road Mathematics Competition

Let AA, BB, CC be three collinear points such that the point BB lies between AA and CC. Let AA′AA' and BB′BB' be parallel lines such that the points A′A' and B′B' lie on the same side of the line ABAB, and A′A', B′B', CC are not collinear. Let O1O_1 be the center of the circle passing through the points AA, A′A', CC, and O2O_2 be the center of the circle passing through the points BB, B′B', CC. Find all possible values of the angle CAA′CAA', if triangles A′CB′A'CB' and O1CO2O_1CO_2 have the same area.

Solution

If ∠CAA′\angle CAA' is acute, then ∠CO1A′=2∠CAA′=2∠CBB′=∠CO2B′\angle CO_1A' = 2\angle CAA' = 2\angle CBB' = \angle CO_2B'. And if ∠CAA′\angle CAA' is obtuse, then ∠CO1A′=2(180∘−∠CAA′)=2(180∘−∠CBB′)=∠CO2B′\angle CO_1A' = 2(180^\circ - \angle CAA') = 2(180^\circ - \angle CBB') = \angle CO_2B'. In particular, ∠CO1A′=∠CO2B′\angle CO_1A' = \angle CO_2B'. (It can be easily seen that SCA′B′≠SCO1O2S_{CA'B'} \neq S_{CO_1O_2} is not possible if ∠CAA′=90∘\angle CAA' = 90^\circ.) Since triangles CO1A′CO_1A' and CO2BCO_2B are both isosceles, it follows that they are similar.

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