2011 AIME II Problem 4

Attempt Problem 4 of the 2011 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AIME II solutions, or check the answer key.

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4.

In triangle ABC,ABC, AB=2011AC.AB = \frac{20}{11} AC. The angle bisector of angle AA intersects BC\overline{BC} at point D,D, and point MM is the midpoint of AD.\overline{AD}. Let PP be the point of the intersection of AC\overline{AC} and line BM.BM. The ratio of CPCP to PAPA can be expressed in the form mn,\frac{m}{n}, where mm and nn are relatively prime positive integers. Find m+n.m + n.

Answer: 51
Concepts:mass pointsangle bisector theorem
Difficulty rating: 2270
Solution:

By the angle bisector theorem, BDDC=ABAC=2011.\frac{BD}{DC} = \frac{AB}{AC} = \frac{20}{11}. Use mass points: place mass 1111 at BB and mass 2020 at C,C, so that D,D, which divides BC\overline{BC} with BD:DC=20:11,BD : DC = 20 : 11, is their balance point and carries mass 31.31. Placing mass 3131 at AA makes the balance point of AA and DD exactly the midpoint MM of AD.\overline{AD}.

The center of mass of the whole system therefore lies on line BM,BM, and it also lies on the segment from BB to the balance point of AA and C.C. That balance point is precisely where line BMBM crosses AC,\overline{AC}, namely P,P, and it satisfies 31PA=20CP.31 \cdot PA = 20 \cdot CP.

Hence CPPA=3120,\frac{CP}{PA} = \frac{31}{20}, which is in lowest terms, and m+n=31+20=51.m + n = 31 + 20 = 51.

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