Let $ABC$ be a triangle. Points $P$ and $Q$ are chosen uniformly and independently at random on sides $\overline{AB}$ and $\overline{AC}$, respectively. What is the probability that the area of $\triangle APQ$ is less than one third of the area of $\triangle ABC$?

1. Let $x=\dfrac{AP}{AB}$ and $y=\dfrac{AQ}{AC}$; they are independent and uniform on $[0,1]$, and $\dfrac{[APQ]}{[ABC]}=xy$ (the triangles share the angle at $A$).
2. The region $xy\ge\frac13$ in the unit square lies over $\frac13\le x\le1$, between the curve $y=\frac1{3x}$ and the line $y=1$. Its area is
   $$\int_{1/3}^{1}\left(1-\frac1{3x}\right)dx=\frac23-\frac13\ln3 .$$
3. The required probability is $1-\left(\frac23-\frac13\ln3\right)=\boxed{\dfrac{1+\ln3}{3}}\approx0.70$.