Difference between revisions of "1994 AHSME Problems/Problem 13"
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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> | <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); | ||
+ | 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 18:34, 20 July 2014
Problem
In triangle , . If there is a point strictly between and such that , then
Solution
Let . Since , we have as well. Then . Since , we have .
So
--Solution by TheMaskedMagician