We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position. So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve. Sincerely, Adam P. Goucher

Sent: Thursday, June 18, 2015 at 12:56 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle

The minimum radius has no maximum. WFL

On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:

...

Let F be the Fano Plane.

Let FL be the set of 7 lines in F.

Let FP be the set of 7 points in F.

For L in FL, let P(L) be the set of 3 points on L.

Now let SP be a set of 7 points in general position in R^3.

Let m : FP ó SP be a bijection.

For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).

Let SL = C(FL).

Let S = (SL, SP).

S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.

If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?

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Note that this problem can be formulated in a different way, by having 14 points instead of 7 (namely the seven original points and the seven circumcentres). Then the problem is clearly equivalent to: "Is the Heawood graph a unit-distance graph?" Looking at the Wikipedia article for the Heawood graph, it transpires that this is indeed one of its properties. It provided a reference to arxiv: http://arxiv.org/abs/0912.5395 So yes, it is indeed possible for all circles to have unit radius. (I can't decide whether this is the acceptable practice of `reduction to a known problem' or the underhanded technique of `corollary-sniping'...) Sincerely, Adam P. Goucher

Sent: Tuesday, June 23, 2015 at 12:10 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle

<< The minimum radius has no maximum. >> WFL

<< I'm guessing you can't model the Fano plane with seven unit circles. >> DWW

<< This seems decidedly non-trivial to resolve. >> APG

I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.

Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.

Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.

So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.

The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .

Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?

Fred Lunnon

[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];

[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];

On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:

...

We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.

So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.

Sincerely,

Adam P. Goucher

...

Sent: Thursday, June 18, 2015 at 12:56 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle

The minimum radius has no maximum. WFL

On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:

...

Let F be the Fano Plane.

Let FL be the set of 7 lines in F.

Let FP be the set of 7 points in F.

For L in FL, let P(L) be the set of 3 points on L.

Now let SP be a set of 7 points in general position in R^3.

Let m : FP ó SP be a bijection.

For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).

Let SL = C(FL).

Let S = (SL, SP).

S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.

If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?

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Yes, Adam made a beeline to a proof, including the needed lemma. That paper was numerically amazing, in that for each coordinate of the 14 vertices of the Heawood graph that they determined, it was the solution of a polynomial in one variable of degree 79 that involved all 80 possible coefficients! (Or at least the one they displayed as an example did.) In fact, that polynomial was displayed 2 coefficients {c_j} to a line, which were mostly gigantic integers, and the ragged edge on the display's right-hand side came close to following a smooth curve — at least for its 40 distinct "points", which were essentially each determined by length(line_k) = floor( log_10( c_(2k) * c_(2k+1) ) ). —Dan

On Jun 22, 2015, at 6:15 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

<< Does one circle really contain 4 vertices, or does it just look that way? >> DA

I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0 <https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0>

Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!

DWW asks for a properly 3-space Fano configuration, with circles disjoint (except at Fano points). Consider some general set of 7 planes in 3-space; define the 7 Fano points as an appropriate subset of 3-plane intersections, and the 7 Fano lines as the circumcircles in each plane. Specifying each unit radius constitutes 1 constraint; so modulo isometry, solid configuration set (complex) freedom dimension = 7.3 - 6 - 7.1 = 11 ; and its real planar (presumed) limit dimension = 4 (probably). But is a planar configuration in general the limit of properly solid ones? We have to be careful here: for instance modulo isometry, set of (7-fold coincident) unit circles with 7 concyclic points is freedom-6; whence limits of (freedom-4) planar configurations necessarily constitute a proper subset. In mechanical terms, suppose that Fano points can slide along Fano circles. If a planar configuration is picked up and shaken, will it always disorganise properly into 3-space? It's a pity my points & centres relaxation fails to converge, since it could be modified trivially to solid space (albeit sans graphics). Fred Lunnon On 6/23/15, David Wilson <davidwwilson@comcast.net> wrote:

That's pretty neat, I wouldn't have thought to look for planar solutions.

My original hope was a solution with unit circles in R^3 where the circles are Fano lines and circle intersection points the Fano points. You are saying that's not possible?

...

-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred Lunnon Sent: Monday, June 22, 2015 9:15 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle

<< Does one circle really contain 4 vertices, or does it just look that way? >> DA

I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0

Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!

WFL

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Fred, in your second gif, there is still a point (in the 4th quadrant, near (1,-0.5)) where it looks like four circles intersect. Perhaps it is an ineluctable infelicity? On Mon, Jun 22, 2015 at 9:15 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

<< Does one circle really contain 4 vertices, or does it just look that way? >> DA

I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0

Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!

WFL

On 6/23/15, Adam P. Goucher <apgoucher@gmx.com> wrote:

...

Note that this problem can be formulated in a different way, by having 14 points instead of 7 (namely the seven original points and the seven circumcentres). Then the problem is clearly equivalent to:

"Is the Heawood graph a unit-distance graph?"

Looking at the Wikipedia article for the Heawood graph, it transpires that this is indeed one of its properties. It provided a reference to arxiv:

http://arxiv.org/abs/0912.5395

So yes, it is indeed possible for all circles to have unit radius.

(I can't decide whether this is the acceptable practice of `reduction to a known problem' or the underhanded technique of `corollary-sniping'...)

Sincerely,

Adam P. Goucher

...

Sent: Tuesday, June 23, 2015 at 12:10 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle

<< The minimum radius has no maximum. >> WFL

<< I'm guessing you can't model the Fano plane with seven unit circles.

DWW

<< This seems decidedly non-trivial to resolve. >> APG

I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.

Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.

Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.

So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.

The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .

Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?

Fred Lunnon

[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];

[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];

On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:

...

We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.

So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.

Sincerely,

Adam P. Goucher

...

Sent: Thursday, June 18, 2015 at 12:56 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle

The minimum radius has no maximum. WFL

On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:

...

Let F be the Fano Plane.

Let FL be the set of 7 lines in F.

Let FP be the set of 7 points in F.

For L in FL, let P(L) be the set of 3 points on L.

Now let SP be a set of 7 points in general position in R^3.

Let m : FP ó SP be a bijection.

For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).

Let SL = C(FL).

Let S = (SL, SP).

S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.

If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?

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This time there's no ambiguity; radii now within 1.5e-8 of unity --- https://www.dropbox.com/s/exsev6qcjroo2dx/fano7pt7rg_2%20copy.gif?dl=0 Coordinates posted on request. WFL On 6/23/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:

So there is, and even worse than before. Time for an eyesight check; or maybe more numerical measures from the program.

But it would be remarkable if such circle-point collisions were unavoidable --- I'll run a few more trials, and polish my spectacles.

WFL

On 6/23/15, Allan Wechsler <acwacw@gmail.com> wrote:

...

Fred, in your second gif, there is still a point (in the 4th quadrant, near (1,-0.5)) where it looks like four circles intersect. Perhaps it is an ineluctable infelicity?

On Mon, Jun 22, 2015 at 9:15 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

...

<< Does one circle really contain 4 vertices, or does it just look that way? >> DA

I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0

Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!

WFL

On 6/23/15, Adam P. Goucher <apgoucher@gmx.com> wrote:

...

Note that this problem can be formulated in a different way, by having 14 points instead of 7 (namely the seven original points and the seven circumcentres). Then the problem is clearly equivalent to:

"Is the Heawood graph a unit-distance graph?"

Looking at the Wikipedia article for the Heawood graph, it transpires that this is indeed one of its properties. It provided a reference to arxiv:

http://arxiv.org/abs/0912.5395

So yes, it is indeed possible for all circles to have unit radius.

(I can't decide whether this is the acceptable practice of `reduction to a known problem' or the underhanded technique of `corollary-sniping'...)

Sincerely,

Adam P. Goucher

...

Sent: Tuesday, June 23, 2015 at 12:10 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle

<< The minimum radius has no maximum. >> WFL

<< I'm guessing you can't model the Fano plane with seven unit circles.

DWW

<< This seems decidedly non-trivial to resolve. >> APG

I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.

Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.

Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.

So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.

The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .

Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?

Fred Lunnon

[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];

[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];

On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:

...

We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.

So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.

Sincerely,

Adam P. Goucher

> Sent: Thursday, June 18, 2015 at 12:56 AM > From: "Fred Lunnon" <fred.lunnon@gmail.com> > To: math-fun <math-fun@mailman.xmission.com> > Subject: Re: [math-fun] Fano Plane puzzle > > The minimum radius has no maximum. WFL > > > > On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote: > > Let F be the Fano Plane. > > > > Let FL be the set of 7 lines in F. > > > > Let FP be the set of 7 points in F. > > > > For L in FL, let P(L) be the set of 3 points on L. > > > > > > > > Now let SP be a set of 7 points in general position in R^3. > > > > Let m : FP ó SP be a bijection. > > > > For each line of L of F, let C(L) be the circle in R^3 through > > the 3 > > points > > m(P(L)). > > > > Let SL = C(FL). > > > > Let S = (SL, SP). > > > > S is then a model of the Fano Plane in R^3 with circles for Fano > > lines > > and > > points for Fano points. > > > > > > > > If we scale S so that the largest circle has radius 1, how large can > > we > > make > > the radius of the smallest circle of S by judicious choice of > > SP? > > > > _______________________________________________ > > math-fun mailing list > > math-fun@mailman.xmission.com > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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By the way, is this 4th point's closeness really a problem? And, is it really close in R^3 or only close in the displayed projection? —Dan

On Jun 23, 2015, at 12:37 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

So there is, and even worse than before. Time for an eyesight check; or maybe more numerical measures from the program.

But it would be remarkable if such circle-point collisions were unavoidable --- I'll run a few more trials, and polish my spectacles.

WFL

On 6/23/15, Allan Wechsler <acwacw@gmail.com> wrote:

...

Fred, in your second gif, there is still a point (in the 4th quadrant, near (1,-0.5)) where it looks like four circles intersect. Perhaps it is an ineluctable infelicity?

On Mon, Jun 22, 2015 at 9:15 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

...

<< Does one circle really contain 4 vertices, or does it just look that way? >> DA

I hadn't noticed that rather ugly feature --- one point is rather close to a circle to which it does not belong. Another example avoiding that infelicity: https://www.dropbox.com/s/yl819xpiy716qwh/fano7pt7rg_1.gif?dl=0

Adam seems to have nailed the whole problem very neatly --- I haven't checked his reference yet, but it presumably gives an explicit construction. Incidentally, I also tried relaxation based on 14 points and centres, but it was a total failure!

WFL

On 6/23/15, Adam P. Goucher <apgoucher@gmx.com> wrote:

...

Note that this problem can be formulated in a different way, by having 14 points instead of 7 (namely the seven original points and the seven circumcentres). Then the problem is clearly equivalent to:

"Is the Heawood graph a unit-distance graph?"

Looking at the Wikipedia article for the Heawood graph, it transpires that this is indeed one of its properties. It provided a reference to arxiv:

http://arxiv.org/abs/0912.5395

So yes, it is indeed possible for all circles to have unit radius.

(I can't decide whether this is the acceptable practice of `reduction to a known problem' or the underhanded technique of `corollary-sniping'...)

Sincerely,

Adam P. Goucher

...

Sent: Tuesday, June 23, 2015 at 12:10 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle

<< The minimum radius has no maximum. >> WFL

<< I'm guessing you can't model the Fano plane with seven unit circles.

DWW

<< This seems decidedly non-trivial to resolve. >> APG

I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.

Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.

Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.

So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.

The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .

Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?

Fred Lunnon

[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];

[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];

On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:

...

We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.

So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.

Sincerely,

Adam P. Goucher

> Sent: Thursday, June 18, 2015 at 12:56 AM > From: "Fred Lunnon" <fred.lunnon@gmail.com> > To: math-fun <math-fun@mailman.xmission.com> > Subject: Re: [math-fun] Fano Plane puzzle > > The minimum radius has no maximum. WFL > > > > On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote: >> Let F be the Fano Plane. >> >> Let FL be the set of 7 lines in F. >> >> Let FP be the set of 7 points in F. >> >> For L in FL, let P(L) be the set of 3 points on L. >> >> >> >> Now let SP be a set of 7 points in general position in R^3. >> >> Let m : FP ó SP be a bijection. >> >> For each line of L of F, let C(L) be the circle in R^3 through the 3 >> points >> m(P(L)). >> >> Let SL = C(FL). >> >> Let S = (SL, SP). >> >> S is then a model of the Fano Plane in R^3 with circles for Fano >> lines >> and >> points for Fano points. >> >> >> >> If we scale S so that the largest circle has radius 1, how large can >> we >> make >> the radius of the smallest circle of S by judicious choice of SP? >> >> _______________________________________________ >> math-fun mailing list >> math-fun@mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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That's pretty convincing. On Tue, Jun 23, 2015 at 6:13 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

On 6/23/15, Dan Asimov <asimov@msri.org> wrote:

...

By the way, is this 4th point's closeness really a problem?

It's largely an aesthetic matter: such collisions do make the diagram difficult to interpret.

...

And, is it really close in R^3 or only close in the displayed projection?

Note that all configurations so far constructed by myself are _planar_!

[ DWW has raised the question of whether they might be "shaken out" (as it were) so as to properly occupy 3-space --- which appears likely, though I do not currently know for sure. ]

...

Just clicked on this but got a 404 "The file you are looking for has either moved or been deleted: error."

Synching error apparently --- try instead https://www.dropbox.com/s/jeyqj76ng4z42s7/fano7pt7rg_2.gif?dl=0

WFL

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On 6/23/2015 6:13 PM, Fred Lunnon wrote:

[ DWW has raised the question of whether they might be "shaken out" (as it were) so as to properly occupy 3-space --- which appears likely, though I do not currently know for sure. ]

I worked this out yesterday, but I was hoping someone else would save me the trouble of writing it up (and perhaps find something simpler to describe). There is indeed a 3-dimensional arrangement of 7 unit circles with the correct intersections, and it can be chosen to be quite symmetrical (my example has 3-fold rotational symmetry about the z-axis and bilateral symmetry in the xz-plane, so the six-element dihedral group). Start with the unit circle in the xz-plane centered at (1, 0, 0) and add two more by rotating it 120 and 240 degrees about the z-axis (I'll call these circles 1, 2 and 3). They meet at the origin, which is Fano point A. The three lowest points on the circles are points B, C and D, and lie on unit circle 4. To complete the Fano plane, each of the points B, C and D needs to be on a new circle (5, 6 or 7) that meets whichever two of circles 1, 2 and 3 that this point isn't already on; these intersections will be Fano points E, F and G. To locate F and G, for example, start with the circle in the x = 1 plane centered at (1, 0, 0). It passes through B on circle 1, but is far from circles 2 and 3. Fix this by tilting the circle toward the z-axis, keeping the bottom point at B. When it meets circles 2 and 3, those points are F and G. Their 120-degree rotational counterpart is E on circle 1, completing the set; the tilted circle and its rotated versions are circles 5, 6 and 7. (I'm sure that wasn't the easiest thing to follow. If someone who does follow would like to make a picture, that may help.) If I've done the numerical geometry correctly, point E has coordinates (.22893146917093, 0, .63675216588960), approximately. x and z should be algebraic, and in fact they appear to be the real roots of the cubic equations 13x^3 + 3x^2 + 3x - 1 = 0 and 13z^3 + 11z^2 + 5z - 11 = 0. -- Fred W. Helenius fredh@ix.netcom.com

I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions: If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then e_k * e_(k+1) = e_(k+3) (indices mod 7) (e_k)^2 = -1 e_j * e_k = -e_k * e_j These rules plus the fact that the octonions are an algebra over the reals (so additively identical to R^8) are enough to multiply any two elements —Dan

On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!

Impossible, I'm afraid. By well-known theorems of crystallography, Euclidean point symmetry groups with order divisible by 7 exist only in 2-space. Even if you accepted the planar restriction, the only possibility would be concyclic, with points at the corners of a regular heptagon. WFL On 6/24/15, David Wilson <davidwwilson@comcast.net> wrote:

So there's a cyclic incidence relationship between Fano points and lines? Makes me want to hope that perhaps there is a R^3 Fano incarnation with unit circle lines, Cartesian incidence, and 7-way radial symmetry.

...

-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, June 23, 2015 9:53 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle

I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:

If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then

e_k * e_(k+1) = e_(k+3) (indices mod 7)

(e_k)^2 = -1

e_j * e_k = -e_k * e_j

These rules plus the fact that the octonions are an algebra over the reals

(so additively identical to R^8) are enough to multiply any two elements

—Dan

...

On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!

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But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles. —Dan

On Jun 24, 2015, at 7:11 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

Impossible, I'm afraid. By well-known theorems of crystallography, Euclidean point symmetry groups with order divisible by 7 exist only in 2-space. Even if you accepted the planar restriction, the only possibility would be concyclic, with points at the corners of a regular heptagon. WFL

On 6/24/15, David Wilson <davidwwilson@comcast.net> wrote:

...

So there's a cyclic incidence relationship between Fano points and lines? Makes me want to hope that perhaps there is a R^3 Fano incarnation with unit circle lines, Cartesian incidence, and 7-way radial symmetry.

...

-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, June 23, 2015 9:53 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle

I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:

If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then

e_k * e_(k+1) = e_(k+3) (indices mod 7)

(e_k)^2 = -1

e_j * e_k = -e_k * e_j

These rules plus the fact that the octonions are an algebra over the reals

(so additively identical to R^8) are enough to multiply any two elements

—Dan

...

On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!

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Richard's comment says that if the 7 vertices ("points") of the regular 6-simplex s_6 (in R^6 if you like) are numbered by the elements of Z/7Z, then all triples of vertices {k,k+1,k+3} will form 7 "lines" such that the incidence relations of the points and lines make a Fano plane. All 168 automorphisms of the Fano plane are realizable by rigid motions of the simplex. Are they all rotations, i.e, are the permutations all even? I think so. —Dan P.S. By substituting the centroids of the 7 "lines" for the triangles, plus possibly a little tweaking, there ought to be 14 points in R^6, representing the 7 "points" and 7 "lines" connected with 21 edges (not "lines") making this into the bipartite Heawood graph, in such a way as there are rotations of R^6 that interchanges the "points" points and the "lines" points. Let's see, how many rotations of this kind are there ...

On Jun 24, 2015, at 12:50 PM, rkg <rkg@ucalgary.ca> wrote:

Dear all, I've been watching this with interest. I'm hesitant to rush (well, creep) in where angels fear to tread, but I wonder if there's just a chance that not everyone is familiar with the fact that the Fano configuration (due to Kirkman & Woolhouse, by the way) has two quite different descriptions.

By rotating the difference set {0,1,3} mod 7.

By nim-addition of the set {1,2,4}.

You don't get from one to the other by ``adding 1''.

You can ``see'' that the groups SL(2,7) and GL(2,3) are isomorphic, but it's a different matter to prove this. R.

On Wed, 24 Jun 2015, Dan Asimov wrote:

...

But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles.

I wasn't happy with the known unit distance embeddings of the Heawood graph, so I made one of my own. --Ed Pegg Jr On Wed, Jun 24, 2015 at 5:22 PM, David Wilson <davidwwilson@comcast.net> wrote:

Sadly, it would difficult to make an office toy in R^6.

...

-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Wednesday, June 24, 2015 3:30 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle

But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles.

—Dan

...

On Jun 24, 2015, at 7:11 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

Impossible, I'm afraid. By well-known theorems of crystallography, Euclidean point symmetry groups with order divisible by 7 exist only in 2-space. Even if you accepted the planar restriction, the only possibility would be concyclic, with points at the corners of a regular heptagon. WFL

On 6/24/15, David Wilson <davidwwilson@comcast.net> wrote:

...

So there's a cyclic incidence relationship between Fano points and lines? Makes me want to hope that perhaps there is a R^3 Fano incarnation with unit circle lines, Cartesian incidence, and 7-way radial symmetry.

...

-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Dan Asimov Sent: Tuesday, June 23, 2015 9:53 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle

I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:

If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then

e_k * e_(k+1) = e_(k+3) (indices mod 7)

(e_k)^2 = -1

e_j * e_k = -e_k * e_j

These rules plus the fact that the octonions are an algebra over the reals

(so additively identical to R^8) are enough to multiply any two elements

—Dan

...

On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!

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On 2015-06-24 17:07, Dan Asimov wrote:

...

On Jun 24, 2015, at 3:22 PM, David Wilson <davidwwilson@comcast.net> wrote:

Sadly, it would difficult to make an office toy in R^6.

Not at all. In R^6 they have even better machine shops than we do.

—Dan

And the wealth to operate them. Like I was just telling the kids, On Wed, Jun 24, 2015 at 7:19 AM, Bill Gosper <billgosper@gmail.com> wrote: Subj: Scrooge McDuck was said to have three cubic acres of money. How much is that in square gallons? In[169]:= UnitConvert[Quantity[3, "Acres"]^3, Quantity["Gallons"]^2] Out[169]= Quantity[95610810544128000000000000000000000000000000000000000/ 765615812545937377500183749853000049, ("Gallons")^2] Oops, that's 27 cubic acres! Should have been In[170]:= UnitConvert[Quantity[3, "Acres"^3], Quantity["Gallons"]^2] Out[170]= Quantity[10623423393792000000000000000000000000000000000000000/ 765615812545937377500183749853000049, ("Gallons")^2] I hate it when ducks exaggerate. --Bill

...

...

-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com <mailto:math-fun-bounces@mailman.xmission.com>] Dan Asimov Sent: Wednesday, June 24, 2015 3:30 PM

But there could be beautiful configurations for the Fano plane in higher dimensions. Maybe in 6-space, where the vertices of the unit simplex s_6 form 140 unit equilateral triangles.

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I prefer shifting the indices of either the points or the lines so that {0,1,3} becomes {1,2,4}. Then we have: P_i meets L_j if and only if (i-j) is a quadratic residue mod 7; L_i meets P_j if and only if (i-j) is a quadratic nonresidue mod 7. Sincerely, Adam P. Goucher

Sent: Wednesday, June 24, 2015 at 2:53 AM From: "Dan Asimov" <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle

I've often used Fred's mnemonic's "0,1,3" sequence in John Conways's mnemonic for multiplying octonions:

If 7 orthonormal unit vectors perpendicular to the reals are denoted e_0, e_1, ..., e_6, then

e_k * e_(k+1) = e_(k+3) (indices mod 7)

(e_k)^2 = -1

e_j * e_k = -e_k * e_j

These rules plus the fact that the octonions are an algebra over the reals

(so additively identical to R^8) are enough to multiply any two elements

—Dan

...

On Jun 23, 2015, at 5:45 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

It might be appropriate to remind everyone of a canonical notation for points P_i and lines L_i of the Fano plane --- index i = 0,...,6 (mod 7) ; L_i meets { P_i, P_(i+1), P_(i+3) } ; P_i meets { L_i, L_(i-1), L_(i-3) } . Anyone cursed with memory and clerical accuracy as unreliable as my own can save substantial futile blaspheming by observing it!

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-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred W. Helenius Sent: Tuesday, June 23, 2015 7:14 PM To: math-fun Subject: Re: [math-fun] Fano Plane puzzle

On 6/23/2015 6:13 PM, Fred Lunnon wrote:

...

[ DWW has raised the question of whether they might be "shaken out" (as it were) so as to properly occupy 3-space --- which appears likely, though I do not currently know for sure. ]

I worked this out yesterday, but I was hoping someone else would save me the trouble of writing it up (and perhaps find something simpler to describe). There is indeed a 3-dimensional arrangement of 7 unit circles with the correct intersections, and it can be chosen to be quite symmetrical (my example has 3-fold rotational symmetry about the z-axis and bilateral symmetry in the xz- plane, so the six-element dihedral group).

Start with the unit circle in the xz-plane centered at (1, 0, 0) and add two more by rotating it 120 and 240 degrees about the z-axis (I'll call these circles 1, 2 and 3). They meet at the origin, which is Fano point A. The three lowest points on the circles are points B, C and D, and lie on unit circle 4.

To complete the Fano plane, each of the points B, C and D needs to be on a new circle (5, 6 or 7) that meets whichever two of circles 1, 2 and 3 that this point isn't already on; these intersections will be Fano points E, F and G. To locate F and G, for example, start with the circle in the x = 1 plane centered at (1, 0, 0). It passes through B on circle 1, but is far from circles 2 and

Thanks, Dr. Helenius. I was always dissatisfied with the standard 2d picture because the lines have spurious R^2 intersection points that aren't Fano points. Your construction provides a neat model of with congruent lines and Fano topology. 3. Fix this

by tilting the circle toward the z-axis, keeping the bottom point at B. When it meets circles 2 and 3, those points are F and G. Their 120-degree rotational counterpart is E on circle 1, completing the set; the tilted circle and its rotated versions are circles 5, 6 and 7.

(I'm sure that wasn't the easiest thing to follow. If someone who does follow would like to make a picture, that may help.)

If I've done the numerical geometry correctly, point E has coordinates (.22893146917093, 0, .63675216588960), approximately. x and z should be algebraic, and in fact they appear to be the real roots of the cubic equations 13x^3 + 3x^2 + 3x - 1 = 0 and 13z^3 + 11z^2 + 5z - 11 = 0.

-- Fred W. Helenius fredh@ix.netcom.com

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Nice!!! Does one circle really contain 4 vertices, or does it just look that way? —Dan

On Jun 22, 2015, at 4:10 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:

<< The minimum radius has no maximum. >> WFL

<< I'm guessing you can't model the Fano plane with seven unit circles. >> DWW

<< This seems decidedly non-trivial to resolve. >> APG

I initially overlooked the possibility that there might exist nondegenerate Fano configurations of unit circles in which all 7 points and all 7 circles are distinct, notwithstanding the trivial degenerate limit where the circles coincide.

Formulating the problem in terms of a semi-algebraic set, we have 14 variables --- x- and y-components of Fano plane points --- and 7 sextic equations of form (uvw)^2 - (2A)^2 = 0 , where A,u,v,w denote area and sides of some triangle formed by 3 collinear/concylic points; along with 42 quadratic inequalities asserting no points and or centres coincide.

Modulo isometry, the freedom remaining in this system has dimension 14 - 7 - 3 = 4 . In principle it is decidable whether there are real solutions via Tarski's algorithm, say using SolveTools[SemiAlgebraic]() in Maple; while such an attack would be easy to launch, its prospect of success is remote.

So after a couple of failed attempts to construct a symmetric solution manually, I drafted an iterative search program which starts from a random point set, and simply attempts to relax it progressively towards unit radii.

The first half-dozen configurations generated were indeed degenerate; then out popped https://www.dropbox.com/s/ymf2f9k4mcgmhac/fano7pt7rg.gif?dl=0 with evidently well-separated parts, and radii within 2.0E-6 of unity --- at which error bound my program terminates, though that could easily be improved. The data is attached below in the format [1, x, y, r] .

Clearly an approximation of this nature does not constitute a proof of existence, however intuitively convincing. So where to from here?

Fred Lunnon

[[1, 0, 0, 0], [1, -0.5668481233, 0.6075796485, 0], [1, -0.3575661931, -0.7701296409, 0], [1, 1.094644573, 0.2231048189, 0], [1, -0.2475264260, 0.1937463870, 0], [1, 0.2495133582, -0.8697660570, 0], [1, -0.2053263145, -1.296737218, 0]];

[[1, 0.3816724709, 0.9242989585, 1.000001220], [1, -1.171379414, -0.1890023228, -1.000000460], [1, 0.1000783952, 0.1190048358, 0.9999993435], [1, 0.4397688784, -0.5326334070, -1.000001220], [1, -0.7324710115, -0.6807994545, -1.000000840], [1, -0.6281769140, -0.3905378926, 0.9999999230], [1, 0.6424250025, -0.7663464105, -0.9999983530]];

On 6/18/15, Adam P. Goucher <apgoucher@gmx.com> wrote:

...

We can get arbitrarily close to 1 by beginning with SP being the vertices of a regular 7-gon of unit circumradius, and perturbing the points by epsilon so that they're in general position.

So the question is: can we get equality to hold? That is to say, is it possible for all circles to have unit radius? This seems decidedly non-trivial to resolve.

Sincerely,

Adam P. Goucher

...

Sent: Thursday, June 18, 2015 at 12:56 AM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Fano Plane puzzle

The minimum radius has no maximum. WFL

On 6/18/15, David Wilson <davidwwilson@comcast.net> wrote:

...

Let F be the Fano Plane.

Let FL be the set of 7 lines in F.

Let FP be the set of 7 points in F.

For L in FL, let P(L) be the set of 3 points on L.

Now let SP be a set of 7 points in general position in R^3.

Let m : FP ó SP be a bijection.

For each line of L of F, let C(L) be the circle in R^3 through the 3 points m(P(L)).

Let SL = C(FL).

Let S = (SL, SP).

S is then a model of the Fano Plane in R^3 with circles for Fano lines and points for Fano points.

If we scale S so that the largest circle has radius 1, how large can we make the radius of the smallest circle of S by judicious choice of SP?

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