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 $\triangle BCP$?

1. By theorem 17, if $x,y,z$ are the distances from $P$ to $BC,CA,AB$ respectively, then $x+y+z=h$ is constant, and $[ABP]:[BCP]:[CAP]=z:x:y$.
2. Normalizing $x+y+z=1$, uniform $P$ corresponds to $(x,y)$ uniform over the triangle $x,y\ge0,x+y\le1$ (area $\frac12$), with $z=1-x-y$.
3. The condition $z>2x, z>2y$ becomes $y<1-3x$ and $y<\frac{1-x}3$; integrating gives a region of area $\frac1{12}$.
4. The probability is $\frac{1/12}{1/2}=\boxed{\frac16}$.