Cut the square into four smaller squares, each of which has a side length of 1/2. Now one of these smaller squares must contain at least three of the points. And the largest triangle that these points can form has an area half that of the small square — or 1/8.

05/16/2024 This is an old Martin Gardner puzzle, intended to illustrate the pigeonhole principle. Reader Jon Jerome points out that the three points in the small square might be collinear, but if we accept a degenerate triangle with zero area, then the proof holds.

06/02/2024 Reader Drake Thomas writes:

“An alternate approach to nine points in the square: sort the points by *y* coordinate. There are four disjoint intervals [*y*_{1}, *y*_{3}], [*y*_{3}, *y*_{5}], [*y*_{5}, *y*_{7}], [*y*_{7}, *y*_{9}], so at least one of those intervals [*y*_{i}, *y*_{i+2}] is of length at most 1/4. Then the triangle *y*_{i}, *y*_{i+1}, *y*_{i+2} occupies at most half of the rectangle of dimensions 1 × 1/4, so is area at most 1/8.”

(Thanks, Drake.)