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 √(`x`^{2} + (1+`y`^{2}))
if `y` <= `x`

Left-Top so √((1+`x`)^{2} + `y`^{2})
if `y` >= `x`.

By symmetry the shortest route to points on the Back face is

Bottom-Back so √(`x`^{2}
+ (1+`z`^{2})) if `z` <= `x`

Left-Back so √((1+`x`)^{2}
+ `z`^{2}) if `z` >= `x`

and the shortest route to points on the Right face is

Front-Right so √((2+`y`)^{2}
+ `z`^{2})
if `y` <= 3-3`z` and `y` <= `z`

Front-Top-Right so √((1+`y`)^{2}
+ (3-`z`)^{2})
if `y` >= 3-3`z` and `y` <= `z`

Bottom-Right so √(`y`^{2}
+ (2+`z`)^{2})
if `z` <= 3-3`y` and `y` >= `z`

Bottom-Back-Right so √((3-`y`)^{2}
+ (1+`z`)^{2})
if `z` >= 3-3`y` 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} + `z`^{2} = 8

y^{2} + (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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