Difference between revisions of "1994 AHSME Problems/Problem 13"

Revision as of 18:54, 7 May 2023

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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-==Problem==
-In triangle <math>ABC</math>, <math>AB=AC</math>. If there is a point <math>P</math> strictly between <math>A</math> and <math>B</math> such that <math>AP=PC=CB</math>, then <math>\angle A =</math>
-<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>
-<math> \textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 36^{\circ} \qquad\textbf{(C)}\ 48^{\circ} \qquad\textbf{(D)}\ 60^{\circ} \qquad\textbf{(E)}\ 72^{\circ} </math>
 ==Solution==
 <asy>

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Difference between revisions of "1994 AHSME Problems/Problem 13"

Revision as of 18:54, 7 May 2023

Solution

See Also