Re: [math-fun] Power Station Problem
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
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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
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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
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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
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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
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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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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-fun >> > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >
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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-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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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-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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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
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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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participants (5)
-
Allan Wechsler -
Fred Lunnon -
James Propp -
Mike Speciner -
Richard Hess