# Problem #1366

 1366 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 $AB?$ $\textbf{(A) } 4 \qquad\textbf{(B) } 3\sqrt{3} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 4\sqrt{3} \qquad\textbf{(E) } 9$ This problem is copyrighted by the American Mathematics Competitions.
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