Problem:
For positive real numbers , let denote the set of all obtuse triangles that have area and two sides with lengths and . The set of all for which is nonempty, but all triangles in are congruent, is an interval . Find .
Solution:
We are given that two side lengths of a triangle are fixed: and , and we are to consider the set of all obtuse triangles with area and those two sides. We are to find all values of for which is nonempty, but all triangles in are congruent. This means that for such values of , there exists exactly one obtuse triangle (up to congruence) with area and two sides of length and .
Let the triangle have sides , , and variable third side . By the triangle inequality, the third side must satisfy , so .
Now we determine for which values of the triangle is obtuse. A triangle is obtuse if the square of one side is greater than the sum of the squares of the other two. That occurs in two cases:
Case 1:
If , then the angle opposite side is obtuse. This happens when .
Case 2:
If , then the angle opposite side is obtuse. This happens when , so .
Therefore, the triangle is obtuse if and only if .
Now, for a fixed area , there may be multiple triangles (with the same side lengths and ) that realize that area, corresponding to different angles between those two sides. However, for values of that correspond to in the two intervals above, the triangle is always obtuse, and only one such triangle exists for each such value (since the triangle is uniquely determined by , , and ).
Let us now find the range of area values corresponding to these two intervals.
We use Heron's formula: let , , and be variable. The semi-perimeter is . The area is given by .
We are not required to simplify the exact function, but we note that the maximum area occurs when (transition point between obtuse and acute), and the minimum occurs as or (degenerate triangle, area goes to 0).
At , we can compute the area using the formula . The angle is a right angle, so , and the area is . Thus, the upper bound for is .
At , we compute using the Law of Cosines: . Then , so .
The area is then . Squaring gives .
Therefore, the range of for which is nonempty and all triangles in are congruent is , so the final answer is .
The problems on this page are the property of the MAA's American Mathematics Competitions.