# 2007 AMC 12B Problems/Problem 14

The following problem is from both the 2007 AMC 12B #14 and 2007 AMC 10B #17, so both problems redirect to this page.

## Problem

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 $AB$?

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

## Solution 1

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:

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

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

## Solution 2

The 60 degree angles suggest constructing 30-60-90 triangles. As such, let the foot of the altitude from $Q$ to $\overline{BC}$ be $D$. Also, since $\angle{PQD}=60$, let the foot of the altitude from $P$ to $\overline{QD}$ be $E$. Since $\triangle{QEP}$ is 30-60-90, and $PQ=1$, $QE=\frac{1}{2}$. Also, since $PR=2$, $DE=2$. Thus, $QD=\frac{5}{2}$. Once again, since $\triangle{QBD}$ is 30-60-90, $QB=\frac{5\sqrt{3}}{3}$. Similar reasoning to $\overline{AQ}$ and summing the segments yields $\boxed{(D) 4\sqrt{3}}$