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I'd like to divide a unit circle disk into nine parts of equal area, using circle arcs as delimiting lines.

Illustrative figure

The whole setup should be symmetric under the symmetry group of the square, i.e. 4 mirror axes and 4-fold rotational symmetry. The dividing arcs should all be of equal curvature. (Thanks to the comment by i. m. soloveichik for making me aware this latter requirement.) For these reasons, several areas will automatically be of the same size, indicated by a common color in the figure above. There are three different colors corresponding to three different shapes, and the requirement that all three of these should have the same area therefore corresponds to two equations. This agrees nicely with the fact that there are two real parameters one may tune, e.g. the distance $d$ between the center of the figure and the centers of the dividing circles, together with the radius $r$ for these dividing circles. Other combinations are possible.

But how would one obtain the actual numbers for these parameters? Is the solution even unique?

I understand that it might be difficult to give an exact answer to this question. So numeric answers are acceptable as well, as long as they explain how the numbers were obtained, not only what the numbers are.

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There is a unique solution – Henry Oct 24 '12 at 15:56
Are concentric circles ok? – i. m. soloveichik Oct 24 '12 at 16:54
As Hagen writes, the solution will probably have to be numerical, because you want to compare areas involving circular segments, which leads to transcendental equations. One thing you might find useful in keeping the equations manageable, though, is that the radius $r$ of the big circles, their distance $d$ from the centre and the larger coordinate $a$ of the intersections between the circles and the unit circle are related by $d^2+2ad+1=r^2$, which you can derive by subtracting the two circle equations. – joriki Oct 24 '12 at 16:57
@i.m.soloveichik: Concentric circles are not what I was aiming for. So perhaps I should have required the circle arcs to have same curvature. I can see what kind of solution your comment is aiming for, and that that solution would be fairly easy to compute. This is exactly the kind of unexpected other solution that caused me to doubt the uniqueness of this. So thanks for a valuable comment, even as the anser is “no”. – MvG Oct 25 '12 at 5:07

To answer your second question: The solution is unique.

For a fixed distance $d$ of the centers, there is a unique radius $r$ of the dividing circles to makes the area of the green part $\frac19$ of the whole circle. Clearly, with growing $d$, $r$ must grow. In fact, the square determined by the four vertices of the green part must grow because with bigger $r$, the excess of the green shape over that square gets smaller. But that means that the vertices wander towards the outside of the circle and also the arcs become flatter, both in effect making the blue parts smaller. Therefore, there is only one $d$ (with matching $r$) that makes the blue parts $\frac 19$ each, which then automatically solves the complete puzzle.

The solution itself can only be found numerically.

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The first sentence in your main paragraph may not always be true, depending on your considerations so far. For a given $d$ there will still be two different $r$ that lead to a $\frac19$ central area. One will be convex like the one I depicted, while the other will be a concave cushion shape. I'm pretty certain that the latter will always result in the red areas being larger than the blue ones, but that consideration should be made explicit. – MvG Oct 25 '12 at 5:20
Oh, you are right. I took for granted that the figure should look like your sketch, i.e. with $r>d$. I agree that the concave solution looks like not working, but would wait for the numerical result – Hagen von Eitzen Oct 31 '12 at 7:31
up vote 6 down vote accepted

Based on the answer by Hagen, I wrote a bit of code to numerically compute $d$ in an outer loop and $r$ for a given $d$ in an inner loop. The results I obtained look like this:

\begin{align*} r &= 4.740253970598989846488464631691100376659654929999896463057971 \\ d &= 4.441836291757233092492625306779987045065972123154874957376197 \end{align*}

This was computed using arbitrary precision interval arithmetic, so unless I completely garbled up my bisection algorithm, the given digits should be reliable.

The parameters denote the common zero of these two functions, written in Python for Sage and using the comment by joriki:

def area1(d, r):
    """Deficit value of two blue and one red area."""
    x = (r^2 - 1 - d^2)/(2*d)       # intersection as @joriki described it
    a1 = 2*x.arccos()               # angle for the central circle
    a2 = 2*((d + x)/r).arccos()     # angle for the outer circle
    a1 = (a1 - a1.sin())/2          # angle of central circle segment
    a2 = (a2 - a2.sin())/2*r^2      # angle of outer circle segment
    return d.parent().pi()/3 - (a1 - a2) # expected minus actual moon shape

def area2(d, r):
    """Excess value of the green area."""
    x = ((2*r^2 - d^2).sqrt() - d)/2 # intersect outer circle with line x = y
    a2 = 2*((d + x)/r).arccos()      # angle for outer circle
    a2 = (a2 - a2.sin())/2*r^2       # area for one green circle segment
    return ((2*x)^2 + 4*a2) - d.parent().pi()/9 # square + segments - expected

The signs of the results are chosen such that area2 increases with $r$ for fixed $d$, while area1 increases with $d$ for optimal $r$. Approximating the resulting areas using polygons, I could verify the result with resonable precision, so I believe that in this second attempt (see edit history for first mistake), I got the formulas right.

The resulting figure, by the way, looks like this:

Illustration of numeric solution

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