Question

A point $P$ is chosen at random in the interior of equilateral triangle $ABC.$ What is the probability that $\triangle ABP$ has a greater area than each of $\triangle ACP$ and $\triangle BCP?$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

Solution:

The triangles $\triangle ABP,$ $\triangle ACP,$ and $\triangle BCP$ have equal bases (the sides of the equilateral triangle), so their areas are proportional to the distances from $P$ to those sides.

By the threefold symmetry of the equilateral triangle, $\triangle ABP$ is the largest with the same probability as each of the other two, so that probability is $\dfrac13.$

Thus, the correct answer is C.

2003 AMC 12A Problem 16

Solution:

Problem 16 in Other Years