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How to show that the figure has a symmetry center for instance if we have a convex hexagon where opposite sides are of equal lenght and parallel?

Mark
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  • Isn't it visually obvious? Also, you can prove it using the fact that the vertices are given by $(\mbox{Cos}(2\pi k/6),\mbox{Sin}(2\pi k/6))$ for integer $k$, and the fact that negating the entries of the vector yields $(\mbox{Cos}(2\pi (k+3)/6),\mbox{Sin}(2\pi (k+3)/6))$ after applying the usual trigonometric formulae. – DumpsterDoofus Dec 23 '13 at 00:20
  • yes it is, but I wonder how looks the mathematical proof – Mark Dec 23 '13 at 00:21

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From the given information, the vertices of the given hexagon are $$ p,\quad p+u,\quad p+u+v,\quad p+u+v+w,\quad p+v+w,\quad p+w $$ for some $p,u,v,w\in\mathbb{R}^2$. Then the symmetry center is $$ p + \frac12(u+v+w). $$

Jim Belk
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  • I have question what means $\mathbb{R}^2$? And later how you calculate the symmetry center? Have you been using any theorem for this? – Mark Dec 23 '13 at 10:11
  • Here $\mathbb{R}^2$ means the space of all $2$-dimensional vectors whose entries are real numbers. The given point is the symmetry center because the set of vertices of the hexagon is closed under reflection across this point. – Jim Belk Dec 23 '13 at 21:19
  • so for instace $p$ I can interpret as $p=(x_p,y_p)$ ? – Mark Dec 23 '13 at 23:53
  • Yes, $p$ is a vector with two components, as are $u$, $v$, and $w$. – Jim Belk Dec 24 '13 at 02:05