Re: [math-fun] Power Station Problem
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 >
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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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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 >
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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
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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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participants (2)
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Fred Lunnon -
Richard Hess