olymmath

AIME 2011a Problem 2

  • Paper
  • Answer

In rectangle ABCDABCD, AB=12AB = 12 and BC=10BC = 10. Points EE and FF lie inside rectangle ABCDABCD so that BE=9BE = 9, DF=8DF = 8, BE‾∥DF‾\overline{BE} \parallel \overline{DF}, EF‾∥AB‾\overline{EF} \parallel \overline{AB}, and line BEBE intersects segment AD‾\overline{AD}. The length EFEF can be expressed in the form mn−pm \sqrt{n} - p, where mm, nn, and pp are positive integers and nn is not divisible by the square of any prime. Find m+n+pm + n + p.

Dataset text AIME 1983 to 2026. Quoted with source.

Show the answer

36

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Locator aime/2011a/p2 · Tag DBYTGA278Y · Edit this page on GitHub · Keys: j next, k previous