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1991 AHSME Problems/Problem 29

Equilateral triangle $ABC$ has $P$ on $AB$ and $Q$ on $AC$. The triangle is folded along $PQ$ so that vertex $A$ now rests at $A'$ on side $BC$. If $BA'=1$ and $A'C=2$ then the length of the crease $PQ$ is

(A) $\frac{8}{5}$ (B) $\frac{7}{20}\sqrt{21}$ (C) $\frac{1+\sqrt{5}}{2}$ (D) $\frac{13}{8}$ (E) $\sqrt{3}$ The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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