olymmath

AIME 2009a Problem 5

  • Paper
  • Answer

Triangle ABCABC has AC=450AC = 450 and BC=300BC = 300. Points KK and LL are located on AC‾\overline{AC} and AB‾\overline{AB} respectively so that AK=CKAK = CK, and CL‾\overline{CL} is the angle bisector of angle CC. Let PP be the point of intersection of BK‾\overline{BK} and CL‾\overline{CL}, and let MM be the point on line BKBK for which KK is the midpoint of PM‾\overline{PM}. If AM=180AM = 180, find LPLP.

Dataset text AIME 1983 to 2026. Quoted with source.

Show the answer

72

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/2009a/p5 · Tag JSCRNYZ8E6 · Edit this page on GitHub · Keys: j next, k previous