MathDB
1994 AMC 12 #13

Source:

December 30, 2011
AMC

Problem Statement

In triangle ABCABC, AB=ACAB=AC. If there is a point PP strictly between AA and BB such that AP=PC=CBAP=PC=CB, then A=\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] <spanclass=latexbold>(A)</span> 30<spanclass=latexbold>(B)</span> 36<spanclass=latexbold>(C)</span> 48<spanclass=latexbold>(D)</span> 60<spanclass=latexbold>(E)</span> 72 <span class='latex-bold'>(A)</span>\ 30^{\circ} \qquad<span class='latex-bold'>(B)</span>\ 36^{\circ} \qquad<span class='latex-bold'>(C)</span>\ 48^{\circ} \qquad<span class='latex-bold'>(D)</span>\ 60^{\circ} \qquad<span class='latex-bold'>(E)</span>\ 72^{\circ}