Solutions: triangle area ratio problem
March 19, 2008 am31 3:54 am
Consider equilateral triangle ABC. Circles with centers A, B, and C and radii 
Now consider the largest equilateral triangle, DEF, that can be inscribed in that gap. What is the ratio of the areas ABC:DEF?
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JD2718 
I found this to be a pretty wicked problem. The solution I get is 7+4*sqrt(3). I had to resort to analytical geometry to get this, but there should be a solution based on plane geometry, and a much simpler one than I could generate.
As with most problems, this problem turns out to be easy once you know how to do it. I found a method using similar triangles and trigonometry, it gives me the same answer, which means either that it is correct, or that I have made very similar mistakes in the two approaches.
I agree with your solutions, though I’m struggling to justify the required triangle being the one tangent to all three circles with sides parallel to the original other than by saying “obviously, by symmetry, …”. It seems to me that this is the hard part of the problem to do really convincingly. I think I can see an analytic method, but it’s unshiny.
Once you get that, the two triangles have a common centroid, which is 2/3 of the way down each median (also of course an altitude), and the solution drops out almost immediately since the ratio of the areas is the square of the ratio of the medians.
OK. How’s this, for that first part: Clearly the target triangle touches all three circles, as it can otherwise be made bigger. Consider an equilateral triangle V whose vertices all touch the circles. Note that it has the same centroid H as the original triangle (this is an easy exercise, for example using vectors). Now consider the triangle T whose sides are the tangents to circles through the vertices of V. It contains V entirely and fits in the desired space, so is “better”. V is an inscribed equilateral triangle of T, and so they have the same centroid (another easy exercise). Thus our desired triangle has sides tangent to the circles and the same centroid H as the original triangle. Now, the altitude (also passing through the centroid since equilateral) of T must be made as big as possible; so the foot K of the altitude must be as far from H as possible. Let G be the middle of the arc of the circle; then HK is maximised when K=G (since HKG is right with hypotenuse HG). Thus the triangle DEF should be, indeed, the one with sides parallel to the original triangle ABC.
Nothing hard here, but I think I prefer “it’s obvious by symmetry” ;).