1961 AHSME Problems/Problem 31

Problem

In $\triangle ABC$ the ratio $AC:CB$ is $3:4$ . The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$ ( $A$ is between $P$ and $B$ ). The ratio $PA:AB$ is:

$\textbf{(A)}\ 1:3 \qquad \textbf{(B)}\ 3:4 \qquad \textbf{(C)}\ 4:3 \qquad \textbf{(D)}\ 3:1 \qquad \textbf{(E)}\ 7:1$

Solution

$[asy] draw((0,0)--(40,0)--(16,18)--(0,0)); draw((40,0)--(64,72)--(16,18)); draw((40,0)--(160,0)--(64,72),dotted); dot((0,0)); label("B",(0,0),SW); dot((16,18)); label("A",(16,18),NW); dot((40,0)); label("C",(40,0),S); dot((64,72)); label("P",(64,72),N); dot((160,0)); label("X",(160,0),SE); label("$4n$",(20,0),S); label("$3n$",(33,17)); label("$4an-4n$",(100,0),S); label("$3an$",(112,36),NE); [/asy]$ Let $AC = 3n$ and $BC = 4n$ . Draw $X$ , where $X$ is on $BC$ and $AC \parallel PX$ . By AA Similarity, $\triangle ABC \sim \triangle PBX$ , so $PX = 3an$ , $BX = 4an$ , and $CX = 4an - 4n$ .

Also, let $\angle ABC = a$ and $\angle BAC = b$ . Since the angles of a triangle add up to $180^{\circ}$ , $\angle BCA = 180-a-b$ . By Exterior Angle Theorem, $\angle ACX = a+b$ , and since $CP$ bisects $\angle ACX$ , $\angle PCX = \frac{a+b}{2}$ . Because $AC \parallel PX$ , $\angle BXP = 180 - a - b$ . Thus, $\angle CPX = \frac{a+b}{2}$ , making $\triangle CPX$ an isosceles triangle.

Because $\triangle CPX$ is isosceles, $PX = CX$ , so $4an - 4n = 3an$ . That means $a = 4$ , so $PB = 4 \cdot AB$ . Thus, $PA = PB - AB = 3 \cdot AB$ , so $PA : AB = 3:1$ . The answer is $\boxed{\textbf{(D)}}$ , and it can be verified (or obtained) by making $ABC$ a 3-4-5 right triangle.

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 30	Followed by Problem 32
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1961 AHSME Problems/Problem 31

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