Question

A triangle has vertices $A(0, 0),$ $B(12, 0),$ and $C(8, 10).$ The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$

Answer

Solution:

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 $\frac{1}{2} \cdot \frac{17}{5} \cdot 4 + \frac{1}{2} \cdot 6 \cdot 5 = \frac{34}{5} + 15 = \frac{109}{5}.$

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

2017 AIME II Problem 3

Solution:

Problem 3 in Other Years