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Suppose you have a circle and consider three disjoint $60$ degree arcs $A,B,C$ in the circle. (i.e the arcs $A,B,C$ are separated by three arcs $x,y,z$ (with $x+y+z=180$ degrees and $x,y,z>0$)). Now take the chords on $x,y,z$. Call them $X,Y,Z$ respectively. Prove that the triangle that has its vertices at the midpoints of $X,Y,Z$ is an equilateral triangle.

enter image description here

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  • $\begingroup$ Are arcs X,Y and Z arbitrarily placed on the circle? i.e., not as alternate sides in a regular hexagon? ..If so, it is not right.Can you draw a sketch? $\endgroup$ – Narasimham Mar 16 '15 at 0:28
  • $\begingroup$ @Narasimham: That's what I thought at first. Then I realised it's not so simple. $\endgroup$ – TonyK Mar 16 '15 at 0:29
  • $\begingroup$ X,Y,Z are arbitrary as long as they are the chords of three arcs x,y,z separating three 60 degree arcs a,b,c $\endgroup$ – TheGeometer Mar 16 '15 at 0:33
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    $\begingroup$ Picture added in question $\endgroup$ – Joffan Mar 16 '15 at 0:33
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    $\begingroup$ The same question had been asked here, which one's to be marked duplicate? $\endgroup$ – najayaz Mar 16 '15 at 11:38
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You have three points $x,y,z$ on the unit circle, and their images under a $60°$ rotation, which is multiplication by $\omega=e^{\frac{2i\pi}{6}}$. The three midpoints are $$m=\frac{z+\omega y}{2},\quad, m'=\frac{x+\omega z}{2},\quad m''=\frac{y+\omega x}{2}$$ and the claim is that they form an equilateral triangle, i.e., the three sides $$\begin{eqnarray} m'-m&=\frac{x-z}{2}+\omega\,\frac{z-y}{2}\\ m''-m'&=\frac{y-x}{2}+\omega\,\frac{x-z}{2}\\ m-m''&=\frac{z-y}{2}+\omega\,\frac{y-x}{2} \end{eqnarray}$$ have the same modulus. But they are of the form $$\begin{eqnarray} A+\omega B\\ C+\omega A\\ B+\omega C \end{eqnarray}$$ with $A+B+C=0$, and therefore must have the same modulus. Indeed, $$|A+\omega B|=|-(B+C)+\omega B|=|(1-\omega)B+C|=|B+\omega C|$$ since $1-\omega=\overline{\omega}=\omega^{-1}$ and $|\omega|=1$. The other equality is derived similarly. Hence $$|m'-m|=|m''-m'|=|m-m''|$$ and the triangle is equilateral.

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  • $\begingroup$ While your answer is correct, try to solve it by only using geometric arguments. It makes for a far more interesting problem $\endgroup$ – TheGeometer Mar 16 '15 at 2:15
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    $\begingroup$ @TheGeometer: Are you saying that you already have a solution to this problem? If so, then you should indicate that in your question. Also, if you're going to disregard non-geometric solutions, then you should say so explicitly, so that potential answerers don't waste their time. Note that MSE is not really a place to pose mathematical puzzles (even interesting ones like yours); there are plenty of other sites available for that kind of thing. It would be appropriate, however, to post your own solution and ask for alternatives. $\endgroup$ – Blue Mar 16 '15 at 3:02
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    $\begingroup$ You should write down your geometric solution. I for one am perfectly happy with an algebraic argument, it is enough to convince me the proposition is true, but I'd like to read your geometric solution too. $\endgroup$ – Olivier Bégassat Mar 16 '15 at 3:20
  • $\begingroup$ After first demonstration one should be able to express radius, displacement coordinates of the eccentric equilateral triangle in terms of α,β,γ of the deficit sectors which describe angular deviaions from $\pi/3$ of regular hexagon. $\endgroup$ – Narasimham Mar 16 '15 at 17:36
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It can be done (even if it shouldn't be done!) by a mindless calculation.

If $\alpha$ is a cube root of $-1$, other than $-1$, then $1 - \alpha + \alpha^2 = 0.$

Let $u, v, w$ be any three complex numbers. \begin{gather*} 0 = (1 - \alpha + \alpha^2)u = (1 - \alpha + \alpha^2)w \\ \therefore\ (-1 - \alpha^2)w + \alpha^2u + v = \alpha{u} + v - (\alpha{w} + u) \\ \text{i.e. } \alpha^2[(\alpha - 1)w +(u - \alpha{v})] = (\alpha{u} + v) - (\alpha{w} + u) \\ \text{i.e. } \alpha^2[(\alpha{w} + u) - (\alpha{v} + w)] = (\alpha{u} + v) - (\alpha{w} + u) \\ \end{gather*}

Dividing by 2, and taking $\alpha = (1 + \sqrt{3}i)/2$, so that the beginnings and ends of three 60-degree arcs of the unit circle in the complex plane are represented by $u$, $\alpha{u}$, $v$, $\alpha{v}$, $w$, $\alpha{w}$, we see that the complex numbers representing the oriented line segments joining the midpoints $(\alpha{w} + u)/2$, $(\alpha{u} + v)/2$, $(\alpha{v} + w)/2$, are each equal to $\alpha^2$ times the preceding one in cyclic order, i.e. they form an equilateral triangle.

Disjointness doesn't seem to play any part, so long as you take the 6 endpoints in the right order.

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