1991 USAMO Problems/Problem 1
(diagram by integralarefun)
After drawing the triangle, also draw the angle bisector of , and let it intersect at . Notice that , and let . Now from similarity, However, from the angle bisector theorem, we have but is isosceles, so so all sets of side lengths which satisfy the conditions also meet the boxed condition.
Notice that or else we can form a triangle by dividing by their greatest common divisor to get smaller integer side lengths, contradicting the perimeter minimality. Since is squared, must also be a square because if it isn't, then must share a common factor with , meaning it also shares a common factor with , which means share a common factor—a contradiction. Thus we let , so , and we want the minimal pair .
By the Law of Cosines,
Substituting yields . Since , . For there are no integer solutions. For , we have that works, so the side lengths are and the minimal perimeter is .
In let . From the law of sines, we have Thus the ratio We can simplify Likewise, Letting , rewrite
We find that to satisfy the conditions for an obtuse triangle, and therefore .
The rational number with minimum denominator (in order to minimize scaling to obtain integer solutions) above is , which also has a denominator divisible by 2 (to take advantage of the coefficients of 2 and 4 in the ratio and further minimize scaling).
Inserting into the ratio, we find . When scaled minimally to obtain integer side lengths, we find and that the perimeter is .
(note by integralarefun: The part of the solution about finding is not rigorous and would likely require further proof in an actual test.)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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