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
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