1968 IMO Problems/Problem 1
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.
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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
[Incomplete, please edit] In a given triangle , let , , and . Then Hence, (with assumptions. This needs clearing up) If is the shortest side, , , No other permutation of , and in terms of size gives integral values to [show]. So there is only one such triangle.
Let be the side lengths of a triangle in which
Extend to such that Then , so and are similar by AA Similarity. Hence, . Then proceed as in Solution 2, as only algebraic manipulations are left.
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