1996 USAMO Problems/Problem 5: Difference between revisions

Revision as of 09:49, 29 January 2013

Problem:

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

Solution:

@@ Line 3: / Line 3: @@
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-Prove that the average of the numbers <math> n\sin n^{\circ}\; (n = 2,4,6,\ldots,180) </math> is <math>\cot 1^\circ</math>.
+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=10^\circ</math> , <math>\angle MAC= 40^\circ</math> and <math>\angle MCA=30^\circ</math>. Prove that the triangle is isosceles.
@@ Line 10: / Line 10: @@
 '''Solution:'''
 ----
-First, as <math>180\sin{180^\circ}=0,</math> we omit that term. Now, we multiply by <math>\sin 1^\circ</math> to get, after using product to sum, <math>(\cos 1^\circ-\cos 3^\circ)+2(\cos 3^\circ-\cos5)+\cdots +89(\cos 177^\circ-\cos 179^\circ)</math>.
-This simplifies to <math>\cos 1^\circ+\cos 3^\circ +\cos 5^\circ+\cos 7^\circ+...+\cos 177^\circ-89\cos 179^\circ</math>. Since <math>\cos x=-\cos(180-x),</math> this simplifies to <math>90\cos 1^\circ</math>. We multiplied by <math>\sin 1^\circ</math> in the beginning, so we must divide by it now, and thus the sum is just <math>90\cot 1^\circ</math>, so the average is <math>\cot 1^\circ</math>, as desired.
-<math>\Box</math>

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1996 USAMO Problems/Problem 5: Difference between revisions

Revision as of 09:49, 29 January 2013