1961 AMC 12 Problem 31

Attempt Problem 31 of the 1961 AMC 12 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 1961 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

31.

In triangle ABCABC the ratio AC:CBAC:CB is 3:4.3:4. The bisector of the exterior angle at CC intersects BABA extended at PP (AA is between PP and BB). The ratio PA:ABPA:AB is:
1:31:3
3:43:4
4:34:3
3:13:1
7:17:1
Answer: D
Concepts:angle bisector theoremratio and proportion
Small hint:
Apply the exterior angle bisector theorem to the full lengths PAPA and PBPB
Big hint:
Use PB=PA+ABPB=PA+AB
Solution:
The exterior angle bisector theorem gives PAPB=ACCB=34. \frac{PA}{PB}=\frac{AC}{CB}=\frac34. Since PB=PA+AB,PB=PA+AB, write PA=3t,PA=3t, PB=4t.PB=4t. Then AB=t,AB=t, so PA:AB=3:1.PA:AB=3:1.
Thus, the correct answer is D.
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