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.