Solution to Problem 273

Correct solutions came from Kai Grinde Myrann, Norway; Brian Bradie, USA; Philippe Fondanaiche, France; Bill Webb, USA; John Snyder, USA; Ahron Teitelman, Israel; Ron Welch, USA; Farid Lian, Colombia; Claudio Baiocchi, Italy; Lou Cairoli, USA; Paul Lee; David Stigant, USA; Stefan Gatachiu, Romania; Juan Balonga, Spain; Jerome Cherry, USA; Phillip Adkins, USA; Dustin Smith, Bradley University; Jean Moreau de Saint-Martin, France; Allen Druze, USA; Francesc Suņol, Spain.

The maximum value is 3sqrt(3)/2.

Here is a straightforward approach: Consider sin(a)+sin(b)+sin(pi-a-b). At a maximum, the derivatives of this with respect to a and b must be zero. But these are cos(a)-cos(pi-a-b) and cos(b)-cos(pi-a-b). Since cosine is 1-1 for angles between 0 and pi, we have a=b=pi-a-b=pi/3. (One can check in various ways that this is, in fact, the maximum.)

Another interesting approach is suggested by Lou Cairoli and Stefan Gatachiu: Recall Jensen's Inequality, which states that, for a concave-down function (as sin(x) is for 0<=x<=pi), an average of function values is no more than the function of the average. Thus sin(a)+sin(b)+sin(g) <= 3sin((a+b+g)/3) = 3sin(pi/3).

Here is a third, suggested by Claudio Baiocchi: In a circle of center O and radius 1, draw diameters A-A', B-B', C-C' such that angles AOB, BOC and COA' are a, b and g, respectively. Observe that the problem is now to maximize the area of hexagon ABCA'B'C'. By symmetry, the hexagon must be regular.

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ã2007 Alberto L. Delgado