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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...")
 
 
(3 intermediate revisions by 2 users not shown)
Line 1: Line 1:
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==Problem==
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==Solution==
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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>
 
<asy>
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draw((0,0)--(8,0)--(4,12)--cycle);
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import cse5;
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draw((8,0)--(1.6,4.8));
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pathpen=black;
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label("A", (4,12), N);
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pointpen=black;
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label("B", (0,0), W);
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pair A=(4,12),B=(0,0),C=(8,0),P=(1.6,4.8);
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label("C", (8,0), E);
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D(MP("A",A,N)--MP("B",B,W)--MP("C",C,E)--cycle);
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label("P", (1.6,4.8), NW);
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D(C--MP("P",P,NW));
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dot((0,0));
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D(A);D(B);D(C);D(P);
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dot((4,12));
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MA("x",10,B,A,C,1,1);
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dot((8,0));
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MA("x",10,A,C,P,1,1);
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dot((1.6,4.8));
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MA(-20,"180-2x",8,C,P,A,1,2);
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</asy>
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MA("2x",10,B,P,C,1,3,blue);
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<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>
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MA("2x",10,C,B,P,1,3,blue);</asy>
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==Solution==
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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]
 +
 
 +
==See Also==
 +
 
 +
{{AHSME box|year=1994|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 19:54, 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

See Also

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. AMC logo.png

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