# Problem #1324

 1324 Point $P$ is inside equilateral $\triangle ABC$. Points $Q$, $R$, and $S$ are the feet of the perpendiculars from $P$ to $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$, respectively. Given that $PQ=1$, $PR=2$, and $PS=3$, what is $AB$? $\mathrm {(A)} 4\qquad \mathrm {(B)} 3\sqrt{3}\qquad \mathrm {(C)} 6\qquad \mathrm {(D)} 4\sqrt{3}\qquad \mathrm {(E)} 9$ This problem is copyrighted by the American Mathematics Competitions.
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