Problem 55

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Problem 55: (2007 AMC 10b, 17;2007 AMC12b, 14)

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 $AB?$ $\textbf{(A) } 4 \qquad$ $\textbf{(B) } 3\sqrt{3} \qquad$ $\textbf{(C) } 6 \qquad$ $\textbf{(D) } 4\sqrt{3} \qquad$ $\textbf{(E) } 9$

Solution

Exercise 55: Solution

Point $P$ is inside equilateral $\triangle ABC$. Points $Q,R,S$ are the feet of the perpendiculars from $P$ to $\overline{AB},\overline{BC},\overline{CA}$ respectively. Given that $PQ=2, PR=3,$ and $PS=5$, what is $AB$?

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