Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.6 AIME, harder Prove it North Macedonia

One rectangle has side lengths $a$ cm and $b$ cm. If the side with length $a$ cm is enlarged by $b$ cm and the side with length $b$ cm is enlarged by $a$ cm then the resulting square has area of $100\,\mathrm{cm}^2$ . Determine the rectangle that satisfies this condition with smallest area if its side lengths are positive integers.

Solution

The area of the resulting square is $P_{sq} = (a + b)^2 = 100\,\mathrm{cm}^2$ , hence $a + b = 10\,\mathrm{cm}$ . Because $a$ and $b$ are positive integers we have the following cases: $a = 1$ , $b = 9$ ; $a = 2$ , $b = 8$ ; $a = 3$ , $b = 7$ ; $a = 4$ , $b = 6$ ; $a = 5$ , $b = 5$ ; $a = 6$ , $b = 4$ ; $a = 7$ , $b = 3$ ; $a = 8$ , $b = 2$ ; and $a = 9$ , $b = 1$ . The rectangle with smallest area is obtained for $a = 1$ , $b = 9$ or $a = 9$ , $b = 1$ . Its area is $9\,\mathrm{cm}^2$ .

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