2007 AMC 12B Problem 14

Below is the professionally curated solution for Problem 14 of the 2007 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12B solutions, or check the answer key.

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Concepts:area decomposition equilateral triangle triangle area

Difficulty rating: 1680

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 $AB?$

$4$

$3\sqrt{3}$

$6$

$4\sqrt{3}$

$9$

Solution:

Let $s=AB.$ Joining $P$ to the vertices splits the triangle into $\triangle PAB,$ $\triangle PBC,$ and $\triangle PCA,$ with areas $\tfrac{s}{2},$ $s,$ and $\tfrac{3s}{2}.$

Their total is $3s,$ which must equal the area $\tfrac{\sqrt3}{4}s^2$ of the equilateral triangle. So $3s=\dfrac{\sqrt3}{4}s^2,$ giving $s=\dfrac{12}{\sqrt3}=4\sqrt3.$

Thus, the correct answer is D.

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