I got it! The answer is \(({4\over5},{4\over3})\).
Here is how I got it:
Let S be the twice the area of the triangle. Then the side lengths are S, S/2, and S/h. Since the triangle is acute, these side lengths must satisfy these inequalities:
\(S^2 + ({S\over2})^2>({S\over h})^2\)
\(({S\over2})^2+({S\over h})^2 >S^2\)
\(S^2 + ({S\over h})^2>({S\over2})^2\)
We can divide each inequality by \(S^2\) to simplify, and the third inequality will automatically be satisfied, so we are left with:
\(1 + {1\over4} > {1\over h^2}\)
\({1\over4} + {1\over h^2} > 1\)
Simplifying, we get h^2>4/5, and h^2<4/3, so the interval for h^2 is the (4/5, 4/3).
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