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)
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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
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