It is easy to see that the minimum number of posts
required is 4. This is because any single light

post will fail to illuminate some great circle. (Imagine that the light post is
at the north pole and

the great circle is the equator.) From that point, it is not possible to
illuminate that great circle

with less than three additional posts.

The simplest solution, and almost certainly the one with the shortest total post
length, comes

from placing the posts at the corners of an inscribed regular tetrahedron as
illustrated in Figure 1.

This gives us a total lamppost length of 8R where R is the radius of the planet.
The trick is to

prove that this is the minimum length. To do this, we make one clever
observation and then use

brute force.

Any minimal lighting solution must correspond to the case of a (not necessarily
regular)

tetrahedron inscribed in the circle (this is not the inscribed tetrahedron
described in the last

paragraph). This is because the light from a single post creates a circular
patch on the sphere that

must contain the “triangular” region given by the space not covered by the other
posts. The

minimum post height comes from minimizing the radius of the circle to the one
that contains the

three corners of the triangles. The light pattern appears as in Figure 2 below.

We are working in the case of four points at random on a sphere given by x^{2} + y
^{2} + z ^{2} = 1. For

any face of the tetrahedron, we can find a circumscribed circle. The lamppost
height

corresponding to that circle is given by similar triangles as illustrated in
Figure 3. We see that the

height of the can be found with the help of the ratio

It will be given by the formula

where h′ = h / R and r′ = r / R . At this point, we will
just assume a sphere of unit radius. We

make use of the past Puzzle Corner where it was shown that the radius of a
circumscribed circle

for a triangle with sides a, b, and c is given by

where K is the area of the triangle. If we use the vectors
as given in Figure 4, then the formula

for the areas of those triangles will be given by
etc. This gives us a final formula of

The brute force part comes from generating random
inscribed tetrahedrons and determining the

post length that they correspond to. We are solving the problem by throwing
darts at it. We pick

the points (without loss of generality, we may pick one point to be the north
pole and one point

to be on the x-axis) and check to make sure that the center of the sphere is
contained in the

tetrahedron. If it is, we measure the post heights. For instance, for 1,000,000
sets of throws, we

get a minimum height of 8.0497001 and the number of values distributed (for each
of 10 equal

intervals) between 8 and 9: 3,9,21,32,47,62,77,88,90,117. This is a clear
indication that the

minimum is at 8 (well, clear enough for me).

Figure 1: The minimal lighting solution. The sphere is inscribed in a regular
tetrahedron. The

bases of the posts are the corners of an inscribed regular tetrahedron (not
shown).

Figure 2: The unlit “triangle” and its circumscribed circle.

Figure 3: Similar triangles giving the relationship
between the radius of a circle (on the surface

of a sphere) and the minimal lamppost height required to illuminate it.

Figure 4: Illustration of the vectors used in Eqn. 4. The
points all lie on the surface of a sphere

with radius R.