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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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Compare $m_af_a$ and $m_bf_b$ while assuming that $c$ is maximal. Use the law of cosines to determine the angle $\gamma$. Do similar calculations for the other angles. Check if the resulting triangle works. – Phira May 13 '11 at 13:24
At least at a quick glance and with the $f$ being the foot of the altitudes and only when the altitude is within the triangle, $m_{a}f_{a}m_{b}f_{b}m_{c}f_{c}$ (in that order) seems to be (nearly?) always self-intersecting. – Isaac May 13 '11 at 13:28
@user9325:If i understand correctly, I've tried what you've suggested, and the resulting triangle is equilateral. But I may have made a mistake. I'll check – kodyv May 13 '11 at 13:28
@Koundinya Don't forget the minus sign in the law of cosines. – Phira May 13 '11 at 13:33
@user9325: Based on what you've suggested, I am getting $C= \frac{2\pi}{3}$. But that doesn't uniquely determine $ABC$. Help? – kodyv May 13 '11 at 13:39
up vote 4 down vote accepted

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. :) – kodyv May 13 '11 at 15:34

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