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Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO, 2005

Geometry Difficulty 4.9 AIME Prove it Italy

$ABC$ is a triangle with $AC = BC$ and $\angle ACB < 60^{\circ}$ . Let $A'$ and $B'$ be two points on the sides $BC$ and $AC$ respectively such that $AA' = BB' = AB$ . Let $C'$ be the intersection of $AA'$ with $BB'$ . Knowing that $AC' = AB'$ and $BC' = BA'$ , what is the measure in degrees of the angle $\angle ACB$ ?

Solution

Solution:

The answer is $36$ . Indeed, the triangles $CAB$ and $BAB'$ are isosceles sharing an angle adjacent to the base, so they are similar. Similarly $BAB'$ and $AB'C'$ are similar. Reapplying the same reasoning to the triangles $ABA'$ and $BA'C'$ , we find that all five such triangles are similar. Then $\angle ACB = \angle B'AC' = \angle A'AB = \angle ABB' = \angle C'BA'$ . Moreover, the sum of these five angles equals the sum of the interior angles of $ABC$ , and hence $=180^{\circ}$ . Therefore $\angle ACB = \frac{180^{\circ}}{5} = 36^{\circ}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.