July 2018 - mathfun@mailman.xmission.com

Re: [math-fun] Power Station Problem
by Richard Hess 09 Jul '18

09 Jul '18

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(a)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(a)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(a)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(a)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(a)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(a)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(a)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(a)mailman.xmission.com >>>>>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>>>>> >>>>>>> _______________________________________________ >>>>>>> math-fun mailing list >>>>>>> math-fun(a)mailman.xmission.com >>>>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>>>> >>>>>> >>>>>> _______________________________________________ >>>>>> math-fun mailing list >>>>>> math-fun(a)mailman.xmission.com >>>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>>> >>>>>> _______________________________________________ >>>>> math-fun mailing list >>>>> math-fun(a)mailman.xmission.com >>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>> >>>> >>>> >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

2 4

[math-fun] Redundant trinomials --- are they really necessary?
by Fred Lunnon 08 Jul '18

08 Jul '18

I need an existence proof --- it's the missing link in a theorem establishing full-plane integer zero-free number walls over finite characteristic ( |F_2 excepted). CONJECTURE: Given any r > 1 , there exists over |F_2 a trinomial f(X) = X^s + X^t + 1 = g(X) h(X) , where g(X) is irreducible of degree r . This topic turns out to have been studied previously, since it is of interest in elliptic curve public-key cryptography. Such "redundant trinomials" are used to do arithmetic in binary finite fields |F^(2^r) lacking a trinomial generating polynomial: the idea is to calculate instead in a larger ring using trinomial f(X) with factor some field generator g(X) , afterwards reducing the result to the field. Of course, engineers don't care greatly about proving that such a trinomial exists for every field degree: they're just content that some have been found for every degree up to 10,000, comfortably exceeding anything currently required. I haven't uncovered a great deal about this in the literature: Christophe Doche (2005) <doche(a)ics.mq.edu.au> "Redundant Trinomials for Finite Fields of Characteristic 2 " http://web.science.mq.edu.au/~doche/redundant.pdf is cited by most of the rest; Ryul Kim, Wolfram Koepf <koepf(a)mathematik.uni-kassel.de> "Divisibility of Trinomials by Irreducible Polynomials over |F_2 " https://arxiv.org/pdf/1311.1366.pdf includes results about cofactor h(X) , along with baffling typos. If required I can post lists of sample trinomials and cofactors, currently for prime r < 4096 (in 3 hrs), also all r < 1024 (in 12 mins). Fred Lunnon

2 2

Re: [math-fun] Power Station Problem
by Richard Hess 08 Jul '18

08 Jul '18

I could hardly fail to disagree with you less. Dick Hess Sent from my iPhone > On Jul 7, 2018, at 10:37 PM, James Propp <jamespropp(a)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(a)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(a)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(a)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(a)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(a)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(a)mailman.xmission.com >>>>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>>>> >>>>>> _______________________________________________ >>>>>> math-fun mailing list >>>>>> math-fun(a)mailman.xmission.com >>>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>>> >>>>> >>>>> _______________________________________________ >>>>> math-fun mailing list >>>>> math-fun(a)mailman.xmission.com >>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>>> >>>>> _______________________________________________ >>>> math-fun mailing list >>>> math-fun(a)mailman.xmission.com >>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>> >>> >>> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

1 0

Re: [math-fun] Power Station Problem
by Richard Hess 08 Jul '18

08 Jul '18

There are Z order solutions but they have equal distances to near and far pairs of corners. All such solutions lie along an axis of symmetry of the rectangle. Best Dick Sent from my iPhone > On Jul 7, 2018, at 9:12 PM, Allan Wechsler <acwacw(a)gmail.com> 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(a)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(a)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(a)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(a)mailman.xmission.com >>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

1 0

Re: [math-fun] Power Station Problem
by Richard Hess 08 Jul '18

08 Jul '18

Nice try but the problem stated that the view point is a different distance from each corner. Dick Sent from my iPhone > On Jul 7, 2018, at 9:31 PM, Mike Speciner <ms(a)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(a)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(a)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(a)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(a)mailman.xmission.com >>>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>>> _______________________________________________ >>>> math-fun mailing list >>>> math-fun(a)mailman.xmission.com >>>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> >>> _______________________________________________ >>> math-fun mailing list >>> math-fun(a)mailman.xmission.com >>> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >>> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

1 0

Re: [math-fun] Motly and Morley torus tilings
by Stuart Anderson 08 Jul '18

08 Jul '18

I've sent an email to Geoffrey with your question. Some more references: Grunbaum and Shepard dealt with the cases of two and some of three squares in 'Tilings and Patterns' These papers find all the cases with 3 squares https://www.infona.pl/resource/bwmeta1.element.springer-dd723cfe-ddc9-39d5-… https://arxiv.org/pdf/1101.0223 http://emis.ams.org/journals/BAG/vol.41/no.1/23.html

2 1

Re: [math-fun] Power Station Problem
by Richard Hess 08 Jul '18

08 Jul '18

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(a)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(a)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(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

5 10

[math-fun] Motley torus tilings
by James Propp 08 Jul '18

08 Jul '18

Inspired by Scott Kim’s work on motley dissections ( http://www.gathering4gardner.org/g4g13-videos/) I offer a puzzle that I know the answer to and ask a question I don’t know the answer to: Puzzle: Find a tiling of the square torus by two squares of unequal sizes. Question: Is there a tiling of some (not necessarily square) torus by more than two squares, all of unequal sizes? Jim Propp

4 9

[math-fun] Motly and Morley torus tilings
by Stuart Anderson 08 Jul '18

08 Jul '18

Geoffrey Morley has been investigating these. Here is his MathsJam presentation from 2014. The links in his paper's references to squaring.net won't work as I am updating the website. http://www.mathsjam.com/conference/talks/2014/GeoffreyMorley-Torus-Squaring…

2 1

[math-fun] Power Station Problem
by Richard Hess 07 Jul '18

07 Jul '18

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

2 1