A certain happy inspiration has been revealed to me, which I will share with you in order to brighten your Monday morning. Perhaps. A rectangle's extended sides and diagonals partition the plane into 16 convex regions (each including shared boundaries). 4 regions interior to the rectangle might (thankfully) be excluded, on the grounds that nobody inside a power station can view its chimneys. [Unless of course some hot-shot developer has bought it up disused, taken its roof off, then promptly gone bankrupt. Like at Battersea?] That leaves 12, each of which might harbour viewpoint P , and require individual analysis. I have dealt with 2 cases at the special vertex; and AW's argument seems to show that the 4 cases along sides must fail on the grounds of symmetry anyway, regardles of whether s < 1 . Which leaves 6 cases remaining, at the other vertices. So fix on a decent notation [this time around!] --- assign each case the permutation in which the chimneys are viewed, starting from the left and numbering them rightwards from the special chimney as #1 --- eg. my "b = 1" case would become "4123" or "dacb". Take a deep breath ... [When am I going to learn that it is invariably fatal to "just take 10 minutes off in order to relax" from a current project with some deceptively innocuous trifle stumbled upon in math-fun?] WFL On 7/8/18, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Aaarghhh! case c = 1 involves transposing 1 with s^2 on the LHS, not t with 2 t^2 - 1 on the RHS. It doesn't affect the results (but that was a dumb notational decision).
More seriously, I neglected to inspect distinct solutions with larger t (smaller angle u ) corresponding to all nontrivial positive roots: in case b = 1 the roots are t ~ 0.43388374, 0.78183148, 0.97492791 , u = pi 3/14, pi 2/14, pi 1/14 ; for the last two we do indeed have s < 1 , and the problem is soluble. In case c = 1 the roots are t ~ 0.58778525, 0.95105652 ; so no improvement there.
Finally then _two_ distinct solutions, with u = pi 1/14, pi 2/14 . Cooked, perhaps?
WFL
On 7/8/18, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Post-afterthoughts:
Wheeling out the heavy artillery in the form of Groebner bases yields a few more surprises. Lines through P are now relabelled a,b,c,d in spatial order irrespective of destination vertices, for case c = 1 involving transposing t with 2 t^2 - 1 in the equations.
Case b = 1 : T_7(t) = 0 , t ~ 0.43388374 , s = 1/(2 t) > 1 ;
Case c = 1 : 16*t^4 - 20*t^2 + 5 = 0 , t ~ 0.58778525 , s = 2 t > 1 .
So the purportedly short side s is always actually longer; and now everybody is in the club, including the setter. Revenge is sweet!
[Magma script and output available on request.]
WFL
On 7/8/18, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Afterthoughts:
AW was correct to check whether s < 1 , rather than assume that a solution exists and is essentially unique. Unfortunately, from my diagram it is clear that a > 2 t ~ 0.8669, so that via equation (A) s^2 > 1 . It therefore appears that my solution is another cul-de-sac, and the diagram should show P further up and left, with lines c, a adjacent. Ho-ho --- join the club!
My solution demonstrated that elementary methods suffice; however for those with access to a comprehensive CAS, there is an easier high-tech method. Compute a Groebner basis for equations (A) -- (F) (or their updated version), with t the final ordered variable: the last basis member then should be polynomial in t alone. This also ensures that the equations are consistent.
WFL
On 7/8/18, Richard Hess <rihess@cox.net> wrote:
So happy to see your approach, Fred. Just what I was looking for. Thanks so much. Dick Hess
Sent from my iPhone
On Jul 8, 2018, at 2:26 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Z-shaped power station problem.
*--------------* /| 1 /| / | / | / | s / | s / | / 1 | / *---/----------* a / / / / / 1/ /c / / / / / d / / / / // / / /// * P : 3 inner angles = u [View without proportional spacing]
Via elementary trigonometry,
cos(u) = t , cos(2 u) = 2 t^2 - 1 , cos(3 u) = 4 t^3 - 3 t ;
now via cosine rule, 6 consistent(!) equations in 5 variables:
(A) s^2 = a^2 + 1 - 2 a t , (B) s^2 = c^2 + d^2 - 2 c d t , (C) 1 = a^2 + c^2 - 2 a c (2 t^2 - 1) , (D) 1 = 1 + d^2 - 2 d (2 t^2 - 1) , (E) 1 + s^2 = 1 + c^2 - 2 c t , (F) 1 + s^2 = a^2 + d^2 - 2 a d (4 t^3 - 3 t) ;
subtracting,
D => 0 = d ( d - 4 t^2 + 2) , E - B => 1 = -2 c t - d^2 + 2 c d t , F - A => 1 = d^2 - 2 a d (4 t^3 - 3 t) - 1 + 2 a t ,
whence d = 4 t^2 - 2 , c = (1 + d^2)/2 t (d - 1) = (16 t^4 - 16 t^2 + 5)/2 t(4 t^2 - 3) , a = (16 t^4 - 16 t^2 + 2)/2 t(16 t^4 - 20 t^2 + 7) ;
substituting into C , 64 t^8 - 176 t^6 + 168 t^4 - 63 t^2 + 7 = 0 ; and the right-hand side = T_7(t) (t^2-1)/t , where T_n denotes Chebyshev polynomial.
There is essentially only one solution with acute angles: u = pi/14 . ***
Fred Lunnon
On 7/7/18, James Propp <jamespropp@gmail.com> wrote: Note the rarely-spotted-in-the-wild triple-negative in “I'm still not convinced there aren't solutions that use the non-cyclic order.”
Jim Propp
> On Saturday, July 7, 2018, Allan Wechsler <acwacw@gmail.com> wrote: > > Mike Speciner's trivial case does not satisfy the extra constraint > about > the distance to the nearest vertex. If you start with that case, > though, > and budge the viewpoint off center along one of the orthogonal axes > of > the > square, you can then fudge the aspect ratio so that the viewing > angles > are > still equal. Keep doing that, moving the viewpoint further and > further > off > center and adjusting the aspect ratio as required. Eventually you > will > hit > a point where Hess's extra constraint is satisfied: this is, I > think, > the > solution that Hess intended. All of these cases involve scanning the > vertices in cyclic order. I'm still not convinced there aren't > solutions > that use the non-cyclic order. > >> On Sat, Jul 7, 2018 at 4:31 PM, Mike Speciner <ms@alum.mit.edu> >> wrote: >> >> There is the trivial case of a square with the viewpoint at the >> center. >> >> >> >>> On 07-Jul-18 16:12, Allan Wechsler wrote: >>> >>> Standing at the viewpoint, as you scan from left to right, you >>> must >>> be >>> enumerating the vertices of the rectangle in one of two kinds of >>> order: >>> "U" >>> order, going around the perimeter of the rectangle, or "Z" order, >>> traversing one of the rectangle's diagonals. >>> >>> I am guessing that Richard Hess's problem involves "U" order, and >>> that > the >>> viewpoint is on one of the axes of symmetry of the rectangle. >>> >>> But it is not obvious to me that there isn't another solution, >>> utilizing >>> "Z" order. Probably the extra condition, that the distance to the > closest >>> vertex from the viewpoint is equal to the long dimension of the > rectangle, >>> eliminates this class of solutions, but I haven't been able to >>> prove >>> it. >>> >>> On Sat, Jul 7, 2018 at 3:40 PM, Richard Hess <rihess@cox.net> >>> wrote: >>> >>> The conditions of the problem uniquely determine the ratio of side > lengths >>>> of the rectangle as .512858431636277... >>>> >>>> If t=tan(theta)^2 then define >>>> r = (3-t)/8/(1-t) >>>> s = 1-r >>>> then rectangle ratio = .25/sqrt(rs) >>>> >>>> >>>> Sent from my iPhone >>>> >>>>> On Jul 7, 2018, at 6:39 PM, James Propp <jamespropp@gmail.com> >>>>> wrote: >>>>> >>>>> Dick, >>>>> >>>>> Can you tell us the conjectural aspect ratio of the rectangle? >>>>> (I’m >>>>> assuming that it’s unique, or that it takes on only finitely >>>>> many >>>>> values, >>>>> related to the cosines and sines of multiples of 90/7 degrees.) >>>>> >>>>> Jim Propp >>>>> >>>>>> On Saturday, July 7, 2018, Richard Hess <rihess@cox.net> wrote: >>>>>> >>>>>> Imagine a power station with towers of negligible (=0)width >>>>>> built >>>>>> at >>>>>> the >>>>>> four corners of a rectangle on a flat plane. At a certain >>>>>> viewing >>>>>> >>>>> point, P, >>>> >>>>> on the plane, the bases of the four towers are equally spaced in > viewing >>>>>> angle by an angle, theta. P is at a different distance from >>>>>> each > corner >>>>>> >>>>> and >>>> >>>>> the distance from P to the closest tower is equals the length of >>>>> the >>>>>> >>>>> long >>>> >>>>> side of the rectangle. For this case theta equals 90/7 degrees >>>>> to >>>>> 15 >>>>>> >>>>> places >>>> >>>>> of accuracy but I’m unable to prove equality. >>>>>> >>>>>> Any takers in finding a proof? >>>>>> >>>>>> Dick Hess >>>>>> >>>>>> Sent from my iPhone >>>>>> >>>>>> _______________________________________________ >>>>>> 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 >>>>> >>>> >>>> _______________________________________________ >>>> 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 >>> >> >> > _______________________________________________ > 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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