1996 USAMO Problems/Problem 5

Problem

Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$ , $\angle MBA=20^\circ$ , $\angle MAC= 40^\circ$ and $\angle MCA=30^\circ$ . Prove that the triangle is isosceles.

Solution

Clearly, $\angle AMB = 150^\circ$ and $\angle AMC = 110^\circ$ . Now by the Law of Sines on triangles $ABM$ and $ACM$ , we have $\frac{AB}{\sin 150^\circ} = \frac{AM}{\sin 20^\circ}$ and $\frac{AC}{\sin 110^\circ} = \frac{AM}{\sin 30^\circ}.$ Combining these equations gives us $\frac{AB}{AC} = \frac{\sin 150^\circ \sin 30^\circ}{\sin 20^\circ \sin 110^\circ}.$ Without loss of generality, let $AB = \sin 150^\circ \sin 30^\circ = \frac{1}{4}$ and $AC = \sin 20^\circ \sin 110^\circ$ . Then by the Law of Cosines, we have

$\begin{align*} BC^2 &= AB^2 + AC^2 - 2(AB)(BC)\cos\angle BAC\\ &= \frac{1}{16} + \sin^2 20^\circ\sin^2 110^\circ - 2\left(\frac{1}{4}\right)\sin 20^\circ\sin 110^\circ\cos 50^\circ \\ &= \frac{1}{16} + \sin^2 20^\circ \sin^2 110^\circ - \frac{1}{2}\sin 20^\circ\sin 110^\circ\sin 40^\circ \\ &= \frac{1}{16} + \sin^2 20^\circ \sin^2 110^\circ - \sin 20^\circ\sin 110^\circ\sin 20^\circ\cos 20^\circ \\ &= \frac{1}{16} \end{align*}$

Thus, $AB = BC$ , our desired conclusion.

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1996 USAMO Problems/Problem 5

Problem

Solution