# Platonic solids

A 12 inch long wire is to be cut into a number of pieces. Together, these pieces will trace all the edges of a cube having 1 inch per edge. What is the minimum number of pieces necessary? While using the minimum number of pieces necessary, what's the longest piece you can make?

Solution:

Three edges meet at each corner of the cube, so there must be at least one wire end at each corner. There are eight corners, so there are at least eight ends, and thus, at least four pieces of wire. But four pieces suffice, as shown in the figure below. Each piece is at least one inch long, so the longest piece is at most 12 - 3 * 1 = 9 inches. This can be achieved, as seen in the figure below.

For each of the remaining Platonic solids: What is the minimum number of pieces necessary? While using the minimum number of pieces necessary, what's the longest piece you can make? For trying different solutions, you will likely want to trace either the wireframe of each Platonic solid, or its graph. Here you can see the wireframe of each Platonic solid on the left, and its corresponding graph on the right. Personally, I prefer tracing the graphs. The graphs feel less tangled than the wireframes, as edges never cross.

Any other polyhedral graphs that are interesting? Can you find an algorithm that works for any polyhedral graph?