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I am working on the problem of finding "rational" dodecahedrons, and I have run across an interesting subproblem: How do you tell if three circles have a common intersection point?

Specifically, given the points: $P_{ij}=(x_{ij},y_{ij}), 1\leq i,j \leq 3$, such that the circles $C_i$ are determined by $P_{ij}$, is there a formula based on the points $P_{ij}$ to determine whether the circles $C_i$ all intersect at one common point? If it helps, for the problem in question one may take $P_{ij}=P_{ji}$.

What would be particularly pleasing is if the formula could be represented as a determinant of a matrix (or potentially a determinant like object for a tensor). This would be reminiscent of the determinant condition for four points to be concyclic.

Edit: Given that the general problem has in some sense been solved, I would like to focus on a particular set of coordinates. The specific matrix I want to work with is:

\begin{pmatrix}(0,-y)&(0,y)&(x,0) \\ (0,y)&(-c,d)&(c,d) \\ (x,0)&(c,d)&(\frac{c^2+d^2}x,0)\end{pmatrix}

I am hoping to arrive at a diophantine equation in c, d, x, and y of small degree that would allow a full characterization (or perhaps an infinite family) of rational dodecahedron.

Edit 2: So I have used Gro-Tsen's degree six equation, and plugged in the points from the dodecahedron matrix. The resulting equation is of degree 8, and after removing the trivial factors $cy(c^2+d^2-x^2)$, the final equation is: $cxy^2-d^2y^2-c^2y^2+dx^2y+d^3y+c^2dy-d^2x^2-c^2x^2+cd^2x+c^3x=0$. You can rewrite this more simply as $xy(cy+dx)+(c^2+d^2)(cx+dy-x^2-y^2)=0$. This degree 4 Diophantine equation is the condition for a rational dodecahedron.

For anyone interested, I have programmed in desmos what the stereographic projection of the rational dodecahedron actually looks like. The points (c,d), (x,0), and (0,y) (which have been changed to (f,0) and (0,g)) are all adjustable. The final equation is the condition for co-intersection.

One final thing I would like to know... Is Gro-Tsen's formula for the symmetric 3x3 set of coordinates actually a determinant of a 6x6 matrix? It seems likely, given that it is degree six and has 720 terms (of which exactly half are negative). If so, what is that matrix?

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  • $\begingroup$ To put it differently, you are asking for a condition on complex numbers $(z_{i,j})_{1≤i≤3,1≤j≤3}$ (sufficiently general) such that all three of $h_1,h_2,h_3$ where $h_i = ((t-z_{i,2})(z_{i,3}-z_{i,1})) / ((t-z_{i,1})(z_{i,3}-z_{i,2}))$ are simultaneously real for some $t\in\mathbb{C}$. $\endgroup$
    – Gro-Tsen
    Jun 15 at 14:41

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If the circles are given by their centers $O_1,O_2,O_3$ and radii $r_1,r_2,r_3$ respectively, then the condition is that the tetrahedron $A_1A_2A_3A_4$ with edge lengths $A_iA_j=O_iO_j$ and $A_4A_i=r_i$ when $i, j<4$ is planar. In other words, its volume is zero. The squared volume is given by the Caley - Menger determinant. Your problem reduces to this data by finding the centers and radii of circumcircles which is rather standard.

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    $\begingroup$ If you write this out fully, this gives an impossibly complicated expression in the $3\times 3\times 2$ point coordinates, however. (Or at least, a variation I tried, writing the circle equations in the form $x^2 + y^2 + ux + vy + w = 0$, solving for $u,v,w$ and using elimination to find a condition for three circles to meet, gave something impossibly complicated, see here.) We are left to wonder if this expression can be factored and/or simplified. $\endgroup$
    – Gro-Tsen
    Jun 16 at 8:24
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    $\begingroup$ Is there a way to represent this with a larger determinant in terms of the coordinates? @Gro-Tsen, if we were to use the constraint that the 3x3 matrix of coordinates is symmetric, does the formula simplify/factorize? $\endgroup$ Jun 17 at 21:16
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    $\begingroup$ @ThomasBlok: Yes, with the additional symmetry constraint, the polynomial factorizes into $4$ factors (I imagine only the largest is of any interest). The Sage code and results are here (you're probably only interested in the last four lines, or even the very last). $\endgroup$
    – Gro-Tsen
    Jun 17 at 22:03
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    $\begingroup$ @Gro-Tsen: The symmetry constraint forces the off diagonal terms to be the other intersection points of the circles. I would guess the degree 4 factors correspond to when the triple intersection is one of those points. The final factor of degree 6 and number of terms 720 makes me suspect there is some 6x6 matrix determinant in the background. If you would be willing to do one further simplification... Could you plug in the matrix of coordinates that I have given in my question edit? Perhaps it factorizes further. I'm hoping to arrive at a solvable diophantine equation in a, b, x, y. $\endgroup$ Jun 18 at 1:00
  • $\begingroup$ A surprising and elegant solution! But, as in the other comments, probably human-computation-intractable! Interesting... :) $\endgroup$ Jun 18 at 3:04

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