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Alhazen's problem expanded


BMAD    62
Consider a terrestrial spheroid to be a sphere, and let our given points be A of which the latitude =x1, and longitude y1, and B of which the latitude =x2 and longitude y2. The shortest distance between these points on the surface of the sphere is along the arc of the great circle joining them. Suppose this great circle track passes, north of a given parallel of latitude x; Find the minimum path between A and B on the spherical surface which does not pass to the north of latitude x. The shortest path consists of two arcs of great circles drawn from A and B to a point P in latitude x in such a manner as to make equal angles with the parallel x at P. Determine this point P.
note: I couldn't figure how to write the correct symbol so, x = post-53485-0-05194300-1365874717.png

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bonanova    76

In the plane the basic Alhazen's problem can be stated three ways:

  1. In a given circle:
    find an isosceles triangle whose legs pass through two given points.
  2. In optics,
    find the point on a (concave) spherical mirror where light incident from a source will reach an observer.
  3. Most intuitively, perhaps, is the billiards analogy:
    find the point on a circular billiards table to stroke a ball so that on one carom it will hit an object ball.

It is by no means a simple problem, and extending it to the sphere seems more like a research project than a brain teaser. I looked at it for a few minutes to see whether the great circles involved in the extended problem could be brought in a single projection to the plane in a way that they all became lines. I convinced myself that they can't.

Not to leave the question without some reward for the minor research, however, I refer the interested puzzler to this simulation of the Billiards problem. Happy shooting. Hint: press the space bar.

Edit: A circular table might screw up your billiards game, but it does have the advantage of taking less space from your man cave.


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