There were errors and misunderstandings in my previous post. The Salamon et al curve is unrelated to Wechsler's problem. The cubic equation for the curve through Hess's viewpoint in sector <r, o, q, p> at the fixed corner origin is correct, but the curve has a crunode (double point) at the origin --- all these curves pass continuously through the origin, when side s = 0 . The congruent curve reflected in centre line x = 1/2 meets my viewpoint in sector <q, p, r, o> . [There is more of this stuff, my friends; much more!] I went searching and turned up the Singmaster article at Gathering for Gardner 11 (2016), as well as an article and slides by Salamon (2014) at https://s3-eu-west-1.amazonaws.com/content.gresham.ac.uk/data/binary/2365/20... https://nms.kcl.ac.uk/simon.salamon/X/dulwich.pdf These sketch a neat-looking treatment which appears to avoid heavy computation, but considers curves generated by viewpoints with two equal angles of a fixed rectangle. In the meantime it dawned on me that the Wechsler curve problem as it stood was incompletely posed: the answer would depend both on choice of fixed origin (at rectangle centre, rather than corner?), and on some relation between rectangle sides ( s + t = 1 , rather than t = 1 ?). At the moment it's unclear what refinement of these choices might best simplify the outcome. That telescope is sure going to have its work cut out! WFL On 7/18/18, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Nice find, HH. http://chesswanks.com/txt/BatterseaPowerStationPuzzle.pdf
It is hard to imagine how Messrs Salomon, Armstrong & Sylvester could have managed to convince themselves about the conic shown in the final figure of Singmaster's document: for instance, their point o = 0 coincides with the origin tower, from where the other angles cannot possibly be equal. [Unless of course BPS has transmogrified into a square floorplan since last I set eyes on it, which would require shifting an awful lot of bricks in addition to the roof.]
But it's comforting to know how many other people this wicked problem has tied into knots. Incidentally in Hess' case where angle v = pi/14 , I find the (nontrivial) locus of viewpoint V = (x,y) for variable distance o to be the cubic curve
4*x^3 - 4*x*y^2 - 3*x^2 + y^2 = 0 ,
which (ummm) also meets the origin!
Wait, wait --- at the origin c_0 c_2 - c_1^2 = 3 > 0 indicating an acnode (isolated point), see https://en.wikipedia.org/wiki/Singular_point_of_a_curve Phew --- so everything is alright? Probably; though ACW's telescope may undergo extensive structural modification at this point ...
I suppose some of this is going to have be included in my screed.
WFL
On 7/18/18, Hans Havermann <gladhobo@bell.net> wrote:
gosper.org/battersea_run.txt gosper.org/battersea.pdf
For reference, I have a slightly edited version of David Singmaster's original posing of the problem (of unknown date) wherein I've added a photo at the bottom of page 3 and an addendum page 4, both taken from an alternate (but presumably subsequent) version of the pdf. I've left out the photo of Elton John.
http://chesswanks.com/txt/BatterseaPowerStationPuzzle.pdf
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