2017 AIME II Problem 3

The points closer to $B$ than to $A$ lie on one side of the perpendicular bisector of $\overline{AB};$ likewise for $C$

The bisectors are $x = 6$ and $y = \frac{2}{5}x + 1;$ find the area of the quadrilateral they cut off near $B$ and divide by $60$

The points closer to $B$ than to $A$ lie to the right of the perpendicular bisector of $\overline{AB},$ the line $x = 6.$ The points closer to $B$ than to $C$ lie below the perpendicular bisector of $\overline{BC},$ which passes through the midpoint $(10, 5)$ with slope $\frac{2}{5}$ (the negative reciprocal of the slope $-\frac{5}{2}$ of $BC$): the line $y = \frac{2}{5}x + 1.$

Inside the triangle, the favorable region is the quadrilateral with vertices $(6, 0),$ $B(12, 0),$ the midpoint $(10, 5)$ of $\overline{BC},$ and $\left(6, \frac{17}{5}\right),$ where the two bisectors meet. Splitting it along the segment from $(6, 0)$ to $(10, 5),$ its area is $$\begin{aligned} &\frac{1}{2} \cdot \frac{17}{5} \cdot 4 + \frac{1}{2} \cdot 6 \cdot 5 \\ &= \frac{34}{5} + 15 \\ &= \frac{109}{5}. \end{aligned}$$

The triangle has area $\frac{1}{2} \cdot 12 \cdot 10 = 60,$ so the probability is $\frac{\frac{109}{5}}{60} = \frac{109}{300},$ and $p + q = 109 + 300 = 409.$