Assume that 101 distinct points are placed in a square 10×10 such that no three of them lie on a line. Prove that we can choose three of the given points that form a triangle whose area is at most 1.
2026-04-02 23:55:56.1775174156
How to approach this pigeon hole problem?
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HINT: Divide the square into $50$ $2\times 1$ rectangles, and show that if three non-collinear points are in one of these rectangles, they form a triangle whose area is at most $1$.