AMC 12 A

Without loss of generality let $AB = AC = BC = 1$; then the area of $\triangle ABC$ is $\frac{1}{4}\sqrt{3}$. Let $x = AP$ and $y = AQ$. Then the area of $\triangle APQ$ is
$$\frac{1}{2}xy \cdot \sin 60^\circ = \frac{1}{4}\sqrt{3} \cdot xy.$$
The probability that the area of $\triangle APQ$ is less than half the area of $\triangle ABC$ is therefore the probability that $xy < \frac{1}{2}$. Graph the curve $xy = \frac{1}{2}$ in the unit square whose lower left corner is at the origin, as shown. Note that the curve passes through the points $(\frac{1}{2}, 1)$ and $(1, \frac{1}{2})$ and is concave up on the interval $\frac{1}{2} < x < 1$.

The probability that $xy > \frac{1}{2}$ is the area of the upper right "fat triangular" region with curved longest side, which is less than $\frac{1}{4}$ but greater than $\frac{1}{8}$. Therefore the probability that $xy < \frac{1}{2}$ lies between $1 - \frac{1}{4} = \frac{3}{4}$ and $1 - \frac{1}{8} = \frac{7}{8}$.