Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National olympiad Prove it

0.11 to ** Prove Morley's Theorem: Let $\triangle A B C$ have three points $D, E, F$ inside it such that $\angle D B C = \angle F B A = \frac{1}{3} \angle A B C, \angle F A B = \angle E A C = \frac{1}{3} \angle B A C, \angle E C A = \angle D C B = \frac{1}{2} \angle A C B$ , then $\triangle D E F$ is an equilateral triangle.

Solution

Let's assume $\triangle A B C$ has corresponding angles $\angle A, \angle B, \angle C$ , and $R$ is the circumradius of $\triangle A B C$ . By the Law of Sines, it is easy to know that $A F = 8 R \sin \left(60^{\circ}+\frac{C}{3}\right) \sin \frac{B}{3} \sin \frac{C}{3}, A E = 8 R \sin \left(60^{\circ}+\right.$ $\left.\frac{B}{3}\right) \sin \frac{B}{3} \sin \frac{C}{3}$ . Therefore, $E F^{2} = A E^{2} + A F^{2} - 2 A E \cdot A F \cos \frac{A}{3} = 64 R^{2} \sin ^{2} \frac{A}{3} \sin ^{2} \frac{B}{3} \sin ^{2} \frac{C}{3}$ , so $E F = 8 R \sin \frac{A}{3} \sin \frac{B}{3} \sin \frac{C}{3}$ . Similarly, $F D$ and $D E$ are also this value.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.