2007 AMC 12B Problems/Problem 14

Problem 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 $AB$ ?

$\mathrm {(A)} 4$ $\mathrm {(B)} 3\sqrt{3}$ $\mathrm {(C)} 6$ $\mathrm {(D)} 4\sqrt{3}$ $\mathrm {(E)} 9$

Solution

Drawing $\overline{PA}$ , $\overline{PB}$ , and $\overline{PC}$ , $\triangle ABC$ is split into three smaller triangles. The altitudes of these triangles are given in the problem as $PQ$ , $PR$ , and $PS$ .

Summing the areas of each of these triangles and equating it to the area of the entire triangle, we get:

$\frac{s(1)}{2} + \frac{s(2)}{2} + \frac{s(3)}{2} = \frac{s^2\sqrt{3}}{4}$ where $s$ is the length of a side

$s = 4\sqrt{3} \Rightarrow \mathrm {(D)}$

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2007 AMC 12B Problems/Problem 14

Problem 14

Solution