2003 AMC 12A Problem 16

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}

The three triangles share equal bases, so their areas depend only on

P

’s distances to the sides

By the threefold symmetry each side yields the largest triangle equally often

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.

Problem 16 in other years