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# Platonic Planet

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.

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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.

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