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

(Created page with "==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...")
 
(→‎Solution)
Line 15: Line 15:
 
<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>
 
<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==
 
==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);
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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 <math>\angle A=x</math>. Since <math>AP=PC</math>, we have <math>\angle ACP=x</math> as well. Then <math>\angle APC=180-2x\implies\angle BPC=\angle CBP=2x</math>. Since <math>AB=AC</math>, we have <math>\angle CBP=\frac{180-x}{2}</math>.
 +
 +
So <math>2x=\frac{180-x}{2}\implies 5x=180\implies x=\angle A=\boxed{\textbf{(B) }36^\circ.}</math>
 +
 +
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]

Revision as of 19:34, 20 July 2014

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

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