# Difference between revisions of "1996 USAMO Problems/Problem 5"

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− | + | ==Problem== | |

− | + | Let <math>ABC</math> be a triangle, and <math>M</math> an interior point such that <math>\angle MAB=10^\circ </math>, <math>\angle MBA=20^\circ</math> , <math>\angle MAC= 40^\circ</math> and <math>\angle MCA=30^\circ</math>. Prove that the triangle is isosceles. | |

− | + | ==Solution== | |

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

+ | <cmath> | ||

+ | \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*} | ||

+ | </cmath> | ||

+ | Thus, <math>AB = BC</math>, our desired conclusion. | ||

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{{MAA Notice}} | {{MAA Notice}} |

## Revision as of 09:25, 1 August 2013

## Problem

Let be a triangle, and an interior point such that , , and . Prove that the triangle is isosceles.

## Solution

Clearly, and . Now by the Law of Sines on triangles and , we have and Combining these equations gives us Without loss of generality, let and . Then by the Law of Cosines, we have

Thus, , our desired conclusion.

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