1994 AHSME Problems/Problem 13

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
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
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1994 AHSME Problems/Problem 13

Solution

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