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Fly on 2 x 1 x 1 brick

Fly on 2 x 1 x 1 brick

Q: A (pointlike) fly is sitting at a corner of a 2 x 1 x 1 brick. What's the furthest point from it, measuring along the surface?
[rec.puzzles, 15 Apr 1996]

A: Suppose it's at the origin, with a long edge of the brick along the positive x-axis and other edges along the positive axes. Then the furthest point is (2, 3/4, 3/4) which is at distance √(65/8) = 2.850+.

Some details: consider the x-y plane to be horizontal, and suppose we're observing the brick from y = -infinity, so that for example the Right face is x = 2, and the Top face is z = 1. Then the shortest route to points on the Top face is
Front-Top so √(x2 + (1+y2)) if y <= x
Left-Top so √((1+x)2 + y2) if y >= x.

By symmetry the shortest route to points on the Back face is
Bottom-Back so √(x2 + (1+z2)) if z <= x
Left-Back so √((1+x)2 + z2) if z >= x
and the shortest route to points on the Right face is
Front-Right so √((2+y)2 + z2) if y <= 3-3z and y <= z
Front-Top-Right so √((1+y)2 + (3-z)2) if y >= 3-3z and y <= z
Bottom-Right so √(y2 + (2+z)2) if z <= 3-3y and y >= z
Bottom-Back-Right so √((3-y)2 + (1+z)2) if z >= 3-3y and y >= z.

Although intuitively one expects the furthest point to be the far corner (2, 1, 1), this is at distance only √8 = 2.828+. The set of points further away than this form a kite-shaped region on the Right face. The kite's long axis lies along the line y = z, with the head at (2, 0.73205, 0.73205) and the tail at (2, 1, 1), and the cross-piece ends at (2, 0.72486, 0.75838) and (2, 0.75838, 0.72486). The kite is bounded by the concave circular arcs
(1 + y)2 + (3 - z)2 = 8
(3 - y)2 + (1 + z)2 = 8
(2 + y)2 + z2 = 8
y2 + (2 + z)2 = 8.

Q: ??? On a centrally-symmetric body, must it be the case that every, or indeed any, longest geodesic's endpoints are antipodes?

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