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1961 AHSME Problems/Problem 31

Revision as of 01:38, 7 June 2018 by Rockmanex3 (talk | contribs) (→‎Solution)

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.

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 30 		Followed by
Problem 32
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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