2007 AMC 10B Problem 17

The three given lengths are all heights from $P$ to a side. Heights go with areas, so compare areas.

1. Let each side of the big triangle be $s$.
2. Draw $PA$, $PB$ and $PC$. They split triangle $ABC$ into three smaller triangles: $PAB$, $PBC$, $PCA$.
3. Each small triangle has one side of the big triangle as its base (length $s$) and one of the given distances as its height:
- triangle $PAB$: $\frac12 \times s \times 1$,
- triangle $PBC$: $\frac12 \times s \times 2$,
- triangle $PCA$: $\frac12 \times s \times 3$.
4. Together they fill the big triangle, so its area is
$$\tfrac12 \times s \times (1 + 2 + 3) = \tfrac12 \times s \times 6.$$
5. But the big triangle's area is also $\frac12 \times s \times h$, where $h$ is its height. Comparing, $h = 6$.
6. In an equilateral triangle the height is $\frac{\sqrt3}{2}$ times the side (half of it is a $30$-$60$-$90$ triangle). So
$$\frac{\sqrt3}{2}\, s = 6 \quad\text{gives}\quad s = \frac{12}{\sqrt3} = \frac{12\sqrt3}{3} = 4\sqrt3.$$

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

For any point inside an equilateral triangle, the three distances to the sides always add up to the triangle's height (this fact is called Viviani's theorem). The area split above is the whole proof. Whenever you are given distances from an inside point to the sides, think about areas.

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