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I'm finding it hard to find the answer to this problem, I suspect it is simple and I'm missing something.

Assume there are three circles with equal radius. The circumference of the circles intersect at one common point - that is, they are Johnson Circles. Their centers are also equidistant from each other; a triangle made from their centers would be equilateral.

In terms of the radius of the circles, what is the distance between any two circle's centers?

And before you get any ideas - none of the circles share a center.

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For anyone coming across who says Maths like this has no practical application, this is for a Venn Diagram I'm coding where the circles can change their radius and it's important to enforce a middle segment or lack thereof in different situations, depending on data.

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  • $\begingroup$ "Their centers are also equidistant from each other" Is this an assumption? $\endgroup$ – mathlove Feb 29 '16 at 6:29
  • $\begingroup$ No, it's part of the problem. If their centers weren't equidistant, I'd have to say which distance I want, right? $\endgroup$ – binderbound Feb 29 '16 at 23:03
  • $\begingroup$ So many equilateral triangles ... $\endgroup$ – David K Feb 29 '16 at 23:20
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The points $B$, $O$, $O_2$ are all equidistant from $O_1$ and $O_3$, so they are collinear. Thus the figure is symmetric under reflection across the line $BO$. Similarly for $AO$ and $CO$. Thus $O$ is the centroid of the equilateral triangle $O_1 O_2 O_3$. Now for some trigonometry...

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Since the circles are equidistant from each other, we can permute them $O_1 \to O_2 \to O_3$ while preserving all features of the figure. This permutation is a rotation of the figure. Three such permutations brings us back to the start, so the angle of rotation (and therefore the angle $\angle O_1OO_2$) is $\frac23\pi$.

Now either divide the rhombus $O_1AO_2O$ into two triangles sharing the edge $OA$, or just simply apply the law of cosines to the triangle $\triangle O_1OO_2$.

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  • $\begingroup$ This is essentially the same answer I've already given in shortened form. $\endgroup$ – binderbound Feb 29 '16 at 23:30
  • $\begingroup$ Slightly different, I think. For one thing, I avoided consideration of angles of measure $\frac\pi6$ (there are two in $\triangle O_1OO_2$, of course, but we don't need to know that). We're all exploiting symmetries of the figure, of course, which are many and highly interrelated. $\endgroup$ – David K Feb 29 '16 at 23:32
  • $\begingroup$ I can see your point, but it is a small difference. Mostly I like using cosines on O1OO2 - that is a good idea $\endgroup$ – binderbound Feb 29 '16 at 23:43
  • $\begingroup$ Really the only reason for posting this at all was the idea of starting with rotational symmetry rather than reflections. Other than that, I agree, not really anything new. $\endgroup$ – David K Feb 29 '16 at 23:52
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I ended up figuring this out. Robert's answer heads down a similar path to what I ended up doing.

As O1, O2 and O3 are equidistant, the triangle from those points is equilateral. Therefore, angle O2O1O3 is pi/3

As O is the centroid of O1O2O3, angle O2O1O is pi/6.

The distance between A and O must logically be exactly bisected by O1O2, so angle AO1O2 is also pi/6.

This means angle AO1O is pi/3. We know that AO1 AND OO1 are also the radius (r), so triangle AO1O must be equilateral.

Therefore it's height (assuming O1 is "up") is colinear with O1O2, and as we could construct an identical triangle between AO2O, its height is half the distance that we are looking for.

To work out height of equilateral triangle AO1O: h = sin(pi/3)*r = sqrt(3/4)*r

Therefore, distance we wanted is sqrt(3)*r.

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Geometric proof without words:

enter image description here

Distance between centers: $r\sqrt3$

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