1994 AHSME Problems/Problem 13

Problem

In triangle $ABC$ , $AB=AC$ . If there is a point $P$ strictly between $A$ and $B$ such that $AP=PC=CB$ , then $\angle A =$ $[asy] draw((0,0)--(8,0)--(4,12)--cycle); draw((8,0)--(1.6,4.8)); label("A", (4,12), N); label("B", (0,0), W); label("C", (8,0), E); label("P", (1.6,4.8), NW); dot((0,0)); dot((4,12)); dot((8,0)); dot((1.6,4.8)); [/asy]$ $\textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 36^{\circ} \qquad\textbf{(C)}\ 48^{\circ} \qquad\textbf{(D)}\ 60^{\circ} \qquad\textbf{(E)}\ 72^{\circ}$

Solution

$[asy] import cse5; pathpen=black; pointpen=black; pair A=(4,12),B=(0,0),C=(8,0),P=(1.6,4.8); D(MP("A",A,N)--MP("B",B,W)--MP("C",C,E)--cycle); D(C--MP("P",P,NW)); D(A);D(B);D(C);D(P); MA("x",10,B,A,C,1,1); MA("x",10,A,C,P,1,1); MA(-20,"180-2x",8,C,P,A,1,2); MA("2x",10,B,P,C,1,3,blue); MA("2x",10,C,B,P,1,3,blue);[/asy]$

Let $\angle A=x$ . Since $AP=PC$ , we have $\angle ACP=x$ as well. Then $\angle APC=180-2x\implies\angle BPC=\angle CBP=2x$ . Since $AB=AC$ , we have $\angle CBP=\frac{180-x}{2}$ .

So $2x=\frac{180-x}{2}\implies 5x=180\implies x=\angle A=\boxed{\textbf{(B) }36^\circ.}$

--Solution by TheMaskedMagician

1994 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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1994 AHSME Problems/Problem 13

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