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Characterize all triples $c_1,c_2,c_3$ of circles in the plane such that there are infinitely many unit regular triangles $a_1a_2a_3$ with $a_i\in c_i$ for $i=1,2,3$.

In particular, are there any triples having this property and such that the circles are neither congruent (i.e. have equal radii) nor concentric?

I wonder whether there is an elegant approach and whether the answer is nice as I'm expecting.

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What is a "unit regular triangle"? – Sammy Black Jun 1 '10 at 15:27
equilateral triangle of side equal one, in the most standard (euclidean) metric – filipm Jun 1 '10 at 16:00
There is a degenerate case of three circles neither congruent not concentric satsifying your requirements: Consider an arbitrary first circle $C_1$ and two coinciding circles $C_2=C_3$ of radius $1$ centered at a point of $C_1$. – Roland Bacher Jun 23 '10 at 13:17

There seems to be an obvious geometric approach. Let T be the 3-torus, and take the smooth function F on it to $\mathbb{R}^3$ like this: use three angles on three given circles as the parameters on T, and from the points P, Q, R on the respective circles construct F as the squares of the Euclidean distances from P to Q, Q to R, R to P. So we are interested in the cases where F maps a point of T to the lattice point (1, 1, 1). The inverse image of the lattice point is a closed subset of T. To make it finite, we need by compactness of T only to understand the derivative of F: where it is invertible the inverse function theorem will work for us. So it seems to come down to computing the derivative of F, in explicit terms of the centres and radii of the circles. (The margin here is too small for so much notation.)

Edit: I now understand the problem a bit better, having manipulated the Jacobian of F. It appears to vanish under the following (sufficient) condition. Write x(1), x(2) and x(3) for the centres of the circles, and v(1) etc. for the corresponding "velocity vectors", i.e. the tangent vectors at a given point of a circle of length given by the radius, which are what one finds as the derivative of the position of a rotating point. Key quantities are the scalar products (x(1) - x(2)).(v(1) - v(2)), and so on. Where all three of these scalar products vanish, the derivative of F is not invertible. This can be seen to happen in particular configurations where the centres are at the vertices of an equilateral triangle, and the circles have equal radius. This condition is not clearly necessary, however.

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yes, I think there are several ways to start a computation, but the question is just whether the computation can be actually done in practice... Is there any non-trivial example, apart from three concentric circles with carefully chosen radii or three circles of equal radii centered at vertices of an equilateral triangle of side 1? an interesting observation is that in both cases the center of the triangles are concyclic. maybe it's a good hint. – filipm Jun 2 '10 at 8:27
I assumed this calculation of the Jacobian determinant of F was accessible to computer algebra packages (I don't have access to one). Three of the entries in the determinant are zero, and there is some simplification evident by hand. I had no time last night to work on this. – Charles Matthews Jun 2 '10 at 16:08

I know that the following remark does not connect in an obvious way to your question, but perhaps it is nevertheless helpful: Every simple closed curve has an inscribed unilateral triangle (in fact so many inscribed unilateral triangles that the set of vertices is dense in the curve), cf. e.g. theorem D and E in and the respective proofs.

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Theorem E there does seem to say something interesting about the "equilateral locus" in T, when each circle overlaps another. Then there is a simple closed boundary curve, made up of arcs, and in the "Venn diagram" configuration there will be inscribed equilateral triangles in most places. For small arcs (< 2\pi /3) those have vertices on at least two circles. – Charles Matthews Jun 9 '10 at 10:47

Suppose that we are not in the degenerate case where infinitely many triangles contain a given fixed point $P$, say on the first circle $C_1$ (this would imply that $C_2=C_3$ with center $P$ and radius $1$). Suppose also that two of the circles, say $C_1$ and $C_2$ are not concentric. Expressing the fact that a triangle with given vertices $v_i$ on $C_i$ is regular with edges of length $1$ by algebraic equations, we see that the set of all possible vertices $v_i\in C_i$ associated to such regular triangles is an algebraic subset of $C_i$. The first condition implies that this algebraic subset is infinite and it is thus the whole circle $C_i$. Since this argument holds for every pair of circles, say for $C_1$ and $C_2$, we see that this is only possible if $C_1$ and $C_2$ have the same radius (and are at distance $1$). We get thus easily that $C_1,C_2,C_3$ are three circles with common radius centered at the vertices of a regular triangle with edges of length $1$.

In the remaining case, we get by symmetry that all three circles are concentric with suitable radii.

One nights sleep has revealed me that the above "proof" does not work: The argument that one gets necessarily the whole circle is wrong since $\mathbb R$ is not algebraically closed. One can probably close the gap by a computation but this destroys my hope for a conceptual proof along the above lines.

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