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
[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 sqrt(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 sqrt(x2 + (1+y2)) if y <= x
Left-Top so sqrt((1+x)2 + y2) if y >= x.
By symmetry the shortest route to points on the Back face is
Bottom-Back so sqrt(x2 + (1+z2)) if z <= x
Left-Back so sqrt((1+x)2 + z2) if z >= x
and the shortest route to points on the Right face is
Front-Right so sqrt((2+y)2 + z2) if y <= 3-3z and y <= z
Front-Top-Right so sqrt((1+y)2 + (3-z)2) if y >= 3-3z and y <= z
Bottom-Right so sqrt(y2 + (2+z)2) if z <= 3-3y and y >= z
Bottom-Back-Right so sqrt((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 sqrt(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?
Back to puzzles
This page is maintained by Thomas Bending,
and was last modified on 23 January 2021.
Comments, criticisms and suggestions are welcome. Copyright © Thomas Bending 2021