2007 AMC 10B Problem 17

Let the side length be $s$. Draw the segments $PA$, $PB$, $PC$; they cut $\triangle ABC$ into three triangles $PAB$, $PBC$, $PCA$, each having a side of the big triangle as its base and the corresponding perpendicular from $P$ as its height:
$$[PAB] + [PBC] + [PCA] = \tfrac12 s \cdot 1 + \tfrac12 s \cdot 2 + \tfrac12 s \cdot 3 = \tfrac12 s \cdot 6.$$

The whole triangle has area $\tfrac12 s \cdot h$, where $h = \frac{\sqrt3}{2}s$ is its altitude. Comparing, $h = 6$, so
$$\frac{\sqrt3}{2}\, s = 6 \quad\Longrightarrow\quad s = \frac{12}{\sqrt3} = 4\sqrt3.$$

The answer is $\boxed{\textbf{(D)}\ 4\sqrt{3}}$.

This is Viviani's theorem: for any interior point of an equilateral triangle, the distances to the three sides add up to the altitude. The proof is the area decomposition above, and the same trick (connect an interior point to the vertices, add the areas) works for any polygon with equal sides. Whenever perpendiculars from an interior point are given, think "areas."

• Treating the sum of the distances, 6, as the side length rather than the altitude of the equilateral triangle.
• Trying to locate $P$ explicitly with coordinates; the individual distances are irrelevant, only their sum matters.