So the apparent U-shaped solution is a plausible cul-de-sac? A nice final twist to what was already a fine problem! WFL On 7/8/18, Allan Wechsler <acwacw@gmail.com> wrote:
OK, my "easy solution" isn't a solution at all, because the distance to the nearest tower is the length of the shorter dimension of the rectangle, which is 1 x ~4.805.
On Sat, Jul 7, 2018 at 10:38 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Because by the neusis argument I sketched earlier, I'm pretty sure there is another solution that scans the towers in cyclic order, and I thought that more obvious solution must be the one Richard Hess had in mind. But it must not have been, because the "easy" solution I have in mind has angles of pi/9, not pi/14.
On Sat, Jul 7, 2018 at 10:21 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I presume mine is the solution RH intended: why might it not be? WFL
On 7/8/18, Allan Wechsler <acwacw@gmail.com> wrote:
Wait, so you found another solution to Richard Hess's full problem _including_ the distance-to-closest-tower constraint? Whoa.
On Sat, Jul 7, 2018 at 9:26 PM, 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-f un > >>>>> > >>>> _______________________________________________ > >>>> 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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