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Question states: "Let $ABCDEF$ be a hexagon in a circle of radius $r$. Show that if $AB = CD = EF = r$, then the midpoints of $BC$, $DE$, and $FA$ are the vertices of an equilateral triangle."

Let $G$ be the midpoint of $BC$, $H$ be midpoint of $DE$, and $I$ be midpoint of $FA$. Also, Let $O$ the center of the circle.

We can use complex number representation.

If we construct a geometric representation of the problem, it becomes evident that triangles $ABO$, $CDO$, and $EFO$ are equilateral triangles. Let $a,b,c,d,e,f,o, g,h,i$ be affixes of $A,B,C,D,E,F,O, G,H$, and $I$, respectively. Then $a$ can be represented as $-ow-bw^2$, $c$ can be represented as $-ow-dw^2$, and $e$ can be represented as $-ow-fw^2$, where $w$ is the cubic root of unity (i.e. $w^3 = 1$).

Also, $g = \frac{b+c}{2}, h = \frac{d+e}{2}$, and $i = \frac{f+a}{2}$.

After this, I can show that the triangle $GHI$ is equilateral by showing that $g = -hw-iw^2$. However, I cannot seem to prove that, however much I try. Is my approach correct? and if so, could you how the calculations?

If anyone has another approach (i.e. without complex number representation), please let me know.

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  • $\begingroup$ In the extreme case with intervening angles $(0,0,\pi) $ when $ (B,C),(D,E) $ coincide, then no equilateral triangle forms ! $\endgroup$ – Narasimham Nov 20 '16 at 23:21
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Choose the origin of the complex plane at $O$ so that its affix is $o=0$. Let $\omega = \text{cis}\,\frac{\pi}{3}$ be the $60^\circ$ rotation, and note that $\omega^3 = -1$ so that $\omega^2 - \omega + 1 = 0$. The problem gives that $b = a \omega$, $d = c \omega$, $f=a \omega$. Then:

$$ \begin{cases} 2(g-h) = b+c-d-e = a \omega + c - c \omega - e = a \omega + c(1-\omega) -e \\ 2(i-g) = f+a-b-c = e \omega + a - a \omega - c = a(1- \omega) - c + e \omega \end{cases} $$

Multiplying the first equation by $\omega$ and adding the two gives:

$$ \require{cancel} \begin{align} 2\big(\omega(g-h) + (i-g)\big) &= a(\omega^2 + 1 - \omega) + c\big(\omega(1-w) - 1\big) + \cancel{e( -\omega + \omega)} \\ & = a(\omega^2 - \omega +1) + c(-\omega^2 + \omega - 1) \\ & = 0 \end{align} $$

Therefore $g-i = \omega(g-h$) so the two sides $GI, GH$ have the same length and the angle between them is $60^\circ$, thus $\triangle GHI$ is equilateral.

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