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In a $\Delta ABC$, consider the hexagon $m_{a}f_{a}m_{b}f_{b}m_{c}f_{c}$ where $m$ and $f$ stand for the midpoint and foot of perpendicular of the respective sides. Now it can be shown quite easily, that $m_{a}f_{a}= |\frac{b^2-c^2}{2a}|$ With a little more effort, one can show also that, $m_{c}f_{a}= \frac{R}{2}\frac{\sin(2B-\alpha)}{\cos(\alpha /2)}$ Where $\alpha$ is the angle subtended by the segment $m_{a}f_{a}$ at the center of the hexagon, which is nothing but the nine-point center. Similarly, we can find the side-lengths of the other sides of this hexagon.

Under what conditions is this hexagon regular?

P.S. It can also be shown that $\sin(\alpha /2) = \frac{\sin^2(B) - \sin^2(C)}{\sin^2(A)}$

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    @Koundinya: Now make the same calculations for the smallest side $a$.2011-05-13

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Never. A regular hexagon whose alternate sides lay on the sides of a triangle would force the triangle to be equiangular, and hence the midpoints would coincide with the feet of the perpendiculars.

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    Good. This was the exact answer I was looking for. :)2011-05-13