A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:
$\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle}\qquad\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle}\qquad\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity}$
Begin by drawing the triangle, the point, the altitudes from the point to the sides, and the segments connecting the point to the vertices.  Let the triangle be $ABC$ with $AB=BC=AC=s$.  We will call the aforementioned point $P$.  Call altitude from $P$ to $BC$ $PA'$.  Similarly, we will name the other two altitudes $PB'$ and $PC'$.  We can see that 
\[\frac{1}{2}sPA'+\frac{1}{2}sPB'+\frac{1}{2}sPC'=\frac{1}{2}sh\]
Where h is the altitude. Multiplying both sides by $2$ and dividing both sides by $s$ gives us
\[PA'+PB'+PC'=h\]
The answer is $\textbf{(C)}$