Question

In triangle $ABC,$ $AB = \frac{20}{11} AC.$ The angle bisector of angle $A$ intersects $\overline{BC}$ at point $D,$ and point $M$ is the midpoint of $\overline{AD}.$ Let $P$ be the point of the intersection of $\overline{AC}$ and line $BM.$ The ratio of $CP$ to $PA$ can be expressed in the form $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

Solution:

By the angle bisector theorem, $\frac{BD}{DC} = \frac{AB}{AC} = \frac{20}{11}.$ Use mass points: place mass $11$ at $B$ and mass $20$ at $C,$ so that $D,$ which divides $\overline{BC}$ with $BD : DC = 20 : 11,$ is their balance point and carries mass $31.$ Placing mass $31$ at $A$ makes the balance point of $A$ and $D$ exactly the midpoint $M$ of $\overline{AD}.$

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

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

2011 AIME II Problem 4

Solution:

Problem 4 in Other Years