Library / National Math Olympiad - Final Round /6 of 9

National Math Olympiad - Final Round, 2013

Geometry Difficulty 5.8 AIME, harder Prove it Slovenia

Let $ABC$ be a triangle with $D$ a point on the side $AB$ and $E$ a point on the side $AC$ such that $|AE| = |ED| = |DB|$ and $|AD| = |DC| = |CB|$ . Determine the sizes of the angles of the triangle $ABC$ .

Solution

Write $\angle EBD = \alpha$ . Then $\angle DEB = \alpha$ , so $\angle EDA = 2\alpha$ and $\angle DAE = 2\alpha$ . This implies $\angle DEC = 4\alpha$ , or $\angle BEC = 3\alpha$ . At the same time we have $\angle ACD = \angle DAC = 2\alpha$ , so $\angle BDC = 4\alpha$ and $\angle CBD = 4\alpha$ , or $\angle CBE = 3\alpha$ . It follows that the triangle $EBC$ is isosceles with the apex at $C$ , so $|CE| = |CD|$ and therefore $\angle CDE = \angle DEC = 4\alpha$ . So, $180^\circ = \angle BDC + \angle CDE + \angle EDA = 4\alpha + 4\alpha + 2\alpha = 10\alpha$ ,
or $\alpha = 18^\circ$ . From here we conclude that $\angle BAC = 2\alpha = 36^\circ$ , $\angle CBA = 4\alpha = 72^\circ$ and $\angle ACB = 180^\circ - \angle BAC - \angle CBA = 72^\circ$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.