No, JeffM, the problem is perfectly clear:

The goal is to prove that the ONLY

triangle that satisfies the conditions in the picture (namely, that the three angles marked as 30° are so) is the equilateral triangle.

It appears to be a very interesting problem, not a standard geometry problem. In playing with the figure in GeoGebra, I find that it appears to be an extremum problem in disguise, because if two of the angles marked as 30° are so, the other is always less than 30° unless the triangle is equilateral.

When I follow Jonathan's suggestion, I obtain the following

equation: [tex]sin\alpha \cdot sin\beta \cdot sin\gamma = \frac{1}{8}[/tex]. I have not dug into this, but I think 1/8 is the maximum value, and is attained only when the three angles are equal (to 30°).

So, Navneet, can you tell us the source of the problem, and whether a multivariable calculus solution seems appropriate? There may be a geometric inequality that can be applied, but I haven't seen it yet.

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