A point P is chosen at random in the interior of equilateral triangle $ABC$. What is the probability that $\triangle ABP$ has a greater area than each of $\triangle ACP$ and $\triangle BCP$? $\textbf{(A)}\ \frac{1}{6}\qquad$ $\textbf{(B)}\ \frac{1}{4}\qquad$ $\textbf{(C)}\ \frac{1}{3}\qquad$ $\textbf{(D)}\ \frac{1}{2}\qquad$ $\textbf{(E)}\ \frac{2}{3}$
Solution
By Viviani's theorem, the three sub-triangle areas $[ABP],[BCP],[CAP]$ are proportional to the distances from $P$ to the three sides, and these three distances sum to a constant (the height).
By the three-fold rotational symmetry of the equilateral triangle, each of the three triangles is (strictly) the largest with equal probability, and ties have probability $0$.
So the probability that $[ABP]$ is the largest is (C) $\frac13$.
A point $P$ is chosen at random in the interior of equilateral triangle $ABC$. What is the probability that $\triangle ABP$ has an area more than twice the area of each of $\triangle ACP$ and $\triangle BCP$?
