Suppose a 5 × 9 rectangle is partitioned into a set of 10 rectangles with integer dimensions. How can we prove that some two of these smaller rectangles are congruent?

| SelectClick for Answer | | --- | | List all the possible rectangles in the set in order of increasing area:{1 × 1, 1 × 2, 1 × 3, 1 × 4, 2 × 2, 1 × 5, 1 × 6, 2 × 3, 1 × 7, 1 × 8, 2 × 4, …}These have areas, respectively, of {1, 2, 3, 4, 4, 5, 6, 6, 7, 8, 8, …}. If all the 10 small rectangles had different areas, then their minimum total area would be 1 + 2 + 3 + 4 + 4 + 5 + 6 + 6 + 7 + 8 = 46, which is greater than that of the larger rectangle in which they must fit (5 × 9 = 45). So some duplication is unavoidable.From Quantum, via Ross Honsberger’s Mathematical Delights, 2019. |

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