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'Platonic Planet' printed from https://nrich.maths.org/
Elijah sent us a clear explanation of how
he tackled this problem, along with some diagrams showing his
solutions.
I made a dodecahedron and made my paths with string and
Blu-Tac.
My first assumption was that Glarsynost was not extremely
freakishly tall so she can't see over the horizon. Each edge is 1
flib long (a real alien word).
From the middle of a face she can see $\frac{1}{12}$ of the planet.
From an edge she can see $\frac{1}{6}$ of the planet. From a vertex
she can see $\frac{1}{4}$ of the planet.
FIRST PATH: Start at a vertex and keep going along a new edge so
you can see a new face every time. The blue path can be seen in the
picture - Path 1. Each face has at least one blue edge. The green
lines show edges that are actually joined up to blue ones. The path
is 12 flibs long.
After that I wanted to cut across the faces so the path would be
shorter and she could see new faces more quickly. So I wondered how
long a diagonal of a pentagon is. My mum told me that it is
$\frac{1 + \sqrt5}{2}$ flibs.
SECOND PATH: Start at a vertex and go across a diagonal so you can
see two new faces. When you have seen them all, get back to the
start. This is Path 2 in the picture. Each face has at least one
blue vertex. The path is 6 diagonals long, $3(1 + \sqrt5)$ flibs
which is about 9.7 flibs. This is the shortest path I could find,
but there are other routes the same length.