# A triangle problem

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Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another

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In triangle , let    , and . Using the Law of Sines gives that Therefore . Using the Law of Cosines gives that This can be simplified to . Since  , and are positive integers, . Note that if is between and , then is relatively prime to and , and cannot possibly divide . Therefore is either the least of the three consecutive integers or the greatest.

Assume that is the least of the three consecutive integers. Then either or , depending on if or . If , then is 1 or 2. couldn’t be 1, for if it was then the triangle would be degenerate. If is 2, then , but and must be 3 and 4 in some order, which means that this triangle doesn’t exist. therefore cannot divide , and so must divide . If then , so is 1, 2, or 4. Clearly cannot be 1 or 2, so must be 4. Therefore . This shows that and , and the triangle has sides that measure 4, 5, and 6.

Now assume that is the greatest of the three consecutive integers. Then either or , depending on if or  is absurd, so , and . Therefore is 1, 2, or 4. However, all of these cases are either degenerate or have been previously ruled out, so cannot be the greatest of the three consecutive integers. This shows that there is exactly one triangle with this property – and it has side lengths of 4, 5, and 6. 