2007 AMC 10B Problem 17

Below is the professionally curated solution for Problem 17 of the 2007 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 10B solutions, or check the answer key.

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Concepts:equilateral triangle area decomposition triangle area

Difficulty rating: 1640

17.

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 $AB?$

$4$

$3\sqrt3$

$6$

$4\sqrt3$

$9$

Solution:

Let the side length be $s.$ The perpendiculars from $P$ are the heights of triangles $APB,$ $BPC,$ and $CPA,$ so their areas are $\dfrac{s}{2},$ $s,$ and $\dfrac{3s}{2}.$

Their sum equals the area of $\triangle ABC,$ which is also $\dfrac{\sqrt3}{4}s^2.$ Hence $3s=\dfrac{\sqrt3}{4}s^2.$

The positive solution is $s=4\sqrt3.$

Thus, the correct answer is D.

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