<< Scale diagrams of these solutions might be appreciated here ...? >> Quite so. In a pleasant inversion of the scholastic dictum that a diagram does not constitute a proof, a clutch of accurate diagrams immediately discloses that: The obvious interpretation of a signed sightline distance --- that the correct angle is made only with a line segment directed _away_ from the tower --- was after all perfectly correct. Hence --- assuming an observer lacking eyes in the back of the head --- my second and fourth solutions were duds, and only the other two remain standing (outside the rectangle). Corrected summary & diagrams, and program & results, are posted at https://www.dropbox.com/s/gva6w9ni9y6svp1/battersea.pdf https://www.dropbox.com/s/175plix3vkvxqoj/battersea_run.txt This has turned out to be an unexpectedly involved exercise: in particular, a neat elementary application of Gröbner bases, as well as an excellent topic for an undergraduate project; perhaps in a situation calling for an especially --- ahem --- challenging one ... Fred Lunnon On 7/14/18, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On a clear day at Battersea power station -----------------------------------------
Dick Hess enquired << Imagine a power station with towers of negligible width (= 0) built at the four corners of a rectangle on a flat plane. At a certain viewing point V on the plane, the bases of the four towers are equally spaced in viewing by an angle v . V is at a different distance from each corner and the distance from V to the closest tower equals the length t of the long side of the rectangle. For this case v = 90/7 degrees to 15 places of accuracy, but I’m unable to prove equality.
I wasted an enormous amount of time in a doomed attempt at a relatively simple-minded program generating all possible solutions; but was obliged finally to grasp the nettle, sit down and implement a proper general-purpose multivariant polynomial equation solver, a project which I have been avoiding for several decades. As often happens in such instances, the task proved much simpler and far more satisfactory and effective than my earlier half-cocked efforts.
The equations have a number of solutions which are degenerate, but are conveniently filtered by rejecting those including any zero entry. Three-quarters of the remainder are redundant, since congruent to (visually indistinguishable from) some previous one; those are excluded by requiring rectangle side s and angle cosine u to be positive.
The remaining 13 solutions are all real and reasonably non-trivial: though distances o,p,q,r are equipped with a sign, its significance remains mysterious. Just 4 solutions satisfy the extra constraints that absolute distances be unequal and that side s < t = o = 1, as follows:
x y p q r s u -0.623490 -0.781831 1.801938 2.076521 1.436997 0.512858 0.974928 0.900969 0.433884 0.445042 -0.228243 0.924139 0.639524 0.781831 -0.309017 0.951057 1.618034 1.376382 0.525731 0.525731 0.951057 0.809017 0.587785 -0.618034 0.324920, 0.850651, 0.850651 0.587785
Second & fourth viewpoints are located inside the rectangle!
Various sectors in which viewpoint V = (x,y) may lie are distinguished by 12 orders in which towers might appear: for example <r, o, q, p> denotes the SSW sector where first & third viewpoints lie, with angles v = pi/14, pi/7 resp. Here further redundancy occurs: a corner sector and its opposite or reverse have identical equations, reducing the number of distinct cases to 8.
Case <r, o, q, p> diagram:
(0,s) (t,s) @--------------@ /| t /| / | / | / | s / | s / | / t | / @---/----------@ r / / / / (1,0) / o/ /q / / / / / p /v/ / / //v/v/ /// @ V = (x,y) , u = cos(v) , o = t = 1 ;
View source without proportional spacing! A scale diagram these solutions might be appreciated here ... ?
Magma program and output, including integer polynomials with individual solution entries as roots, is available online at https://www.dropbox.com/s/175plix3vkvxqoj/battersea_run.txt
Fred Lunnon [14/07/18]