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AIME 2012 problem

#12 on this year’s AIME, could it possibly be a simple problem with a simple solution?

Let \triangle ABC be a right triangle with right angle at C. Let D and E be points on \overline{AB} with D between A and E such that \overline{CD} and \overline{CE} trisect \angle C. If \frac{DE}{BE} = \frac{8}{15}, then \tan B can be written as \frac{m \sqrt{p}}{n}, where m and n are relatively prime positive integers, and p is a positive integer not divisible by the square of any prime. Find m+n+p.


Comments on: "AIME 2012 problem" (1)

  1. […] was checking out Mess or Math?’s cool blog and her latest post wonders whether problem 12 on the 2012 AIME 1 has a simple solution. Simple is […]

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