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Let P be a convex polygon with area A(P), and to each side of P, attach the largest area triangle possible that lies entirely within P. Must the sum S(P) of the areas of these triangles always satisfy $S(P) \geq 2A(P)$? If P is a rectangle, then $S(P) = 2A(P)$.

Supposing this is true, is there a lower bound with some parameter that would detect, for instance, that for P a triangle, $S(P) = 3A(P)$?

A friend mentioned this question to me several months ago, and it has been bothering me (and several other people) since. I've heard claims of a Fourier-analytic proof, so I believe it isn't open. On the other hand, it sounds like a problem from an IMO or Putnam competition, and it feels like it ought to have a clean solution.

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Have other variants of this type of problem been studied where instead of attaching a triangle to each side of the polygon one attaches some other shape? –  Joseph Malkevitch Aug 8 '10 at 15:28
    
I don't know, but I seriously doubt it. I think that maybe an easier question would involve quantifying the IMO solution; for instance, I don't think it's possible to achieve the lower bound S(P) = 2A(P) except in the case of a quadrilateral, but pentagons that look like quadrilaterals get arbitrarily close, so nothing naive based on side count can possible work. –  Eric Tressler Aug 8 '10 at 18:13

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up vote 4 down vote accepted

(This should be a comment, but I can't comment yet)

Indeed, the problem is from an IMO. You can look at the solutions posted here

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Thanks. I was able to find the official solution at georgmohr.dk/imo/imo06pbsol.pdf, which (for anyone interested) has a better write-up. –  Eric Tressler Aug 7 '10 at 23:35

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