Equilateral triangle in a circle

Question

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.

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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? — Narasimham, Mar 16 '15 at 00:28
@Narasimham: That's what I thought at first. Then I realised it's not so simple. — TonyK, Mar 16 '15 at 00:29
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 — TheGeometer, Mar 16 '15 at 00:33
I think it is a very interesting question and I am really eager to find someone who can solve this. I apologize — TheGeometer, Mar 16 '15 at 01:31
No, you cant do that cause then the shape will not satisfy the conditions of the problem — TheGeometer, Mar 16 '15 at 01:41
Your second formulation of the problem is wrong: "Three arcs of unequal length[s] subtend at center of a circle a total of 180 degrees. Show that their mid-points form vertices of an equilateral triangle." That doesn't say that each pair of them is separated by $60^\circ$. It is consistent with moving one of the three chords counterclockwise, thereby moving one of the three vertices while the other two are fixed. The triangle cannot then remain equilateral. ${}\qquad{}$ — Michael Hardy, Mar 16 '15 at 01:44
Yes sorry about that I didnt notice it as I didnt write it thanks for the correction — TheGeometer, Mar 16 '15 at 01:45
Instead of making the three separations between the chord enpoints $60^\circ$ each, suppose you make them three different lengths but keep the total at $180^\circ$. Will the separations still correspond to angles in the triangle? ${}\qquad{}$ — Michael Hardy, Mar 16 '15 at 01:48
I do not think so ... If this is allowed I can construct a counterexample. — TheGeometer, Mar 16 '15 at 01:51
Three of six equal pies are removed from a circle alternatively and disturbed arbitrarily.Show that midpoints of chord of the removed (disturbed) pies always make an equilateral triangle. Can we say so? — Narasimham, Mar 16 '15 at 04:56
The same question had been asked here, which one's to be marked duplicate? — najayaz, Mar 16 '15 at 11:38

Olivier Bégassat · Answer 1 · 2015-03-16T02:01:39.803

4

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.

edited Mar 16 '15 at 02:01

answered Mar 16 '15 at 01:56

Olivier Bégassat

20,248

While your answer is correct, try to solve it by only using geometric arguments. It makes for a far more interesting problem – TheGeometer Mar 16 '15 at 02:15
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@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. – Blue Mar 16 '15 at 03:02
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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. – Olivier Bégassat Mar 16 '15 at 03:20
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. – Narasimham Mar 16 '15 at 17:36

score 3 · Answer 2 · answered Mar 16 '15 at 02:03

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.

Equilateral triangle in a circle

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