Question

Fifteen distinct points are designated on $\triangle ABC:$ the $3$ vertices $A,$ $B,$ and $C;$ $3$ other points on side $\overline{AB};$ $4$ other points on side $\overline{BC};$ and $5$ other points on side $\overline{CA}.$ Find the number of triangles with positive area whose vertices are among these $15$ points.

Answer

Answer: 390

Concepts:combinations complementary counting

Difficulty rating: 1950

Solution:

There are $\binom{15}{3} = 455$ ways to choose $3$ of the points. A choice fails to give a triangle of positive area exactly when the $3$ points are collinear, which happens only when all three lie on one side of the triangle. Including its endpoints, side $\overline{AB}$ contains $5$ points, $\overline{BC}$ contains $6,$ and $\overline{CA}$ contains $7,$ giving $\binom{5}{3} + \binom{6}{3} + \binom{7}{3}$ $= 10 + 20 + 35 = 65$ collinear triples.

The number of triangles is $455 - 65 = 390.$

2017 AIME I Problem 1

Answer: 390

Difficulty rating: 1950

Solution:

Problem 1 in Other Years