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 Solid triangles (Posted on 2005-08-26)
Using only the vertices of a regular icosahedron as the corners, how many equilateral triangles can you make?

What if you could only use the vertices of a regular dodecahedron?

 See The Solution Submitted by Tristan Rating: 3.0000 (1 votes)

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 part 2 -- spoiler | Comment 2 of 9 |

The only way for me to see this was to hold an actual regular dodecahedron in my hand, with one point on top and its diametric opposite on bottom.

There are 5 possible distances of other vertices from the top vertex (and therefore from any given vertex): The three that are connected via an edge and lie at that edge-length distance; the six that are on the two diagonals from that point on each of its three pentagonal faces; the six that lie at this same distance from the opposite vertex; the three that are adjacent to the opposite point (connected by an edge) and the opposite point itself.

There are no triangles (read equilateral, whenever you see triangle here) that can be formed that have the same edge length as the pentagons that make up the dodecahedron.

The triangles whose sides are diagonals of the pentagons can each be associated with one vertex of the dodecahedron.  Since there are 20 vertices, there are 20 of these triangles.

The triangles that are formed at the next distance, cutting across what on the surface of the dodecahedron is a zig-zag set of three edges, can also be associated each with one vertex upon which they are "centered", with the center of the triangle being imbedded within the dodecahedron, but can be projected outward to the nearest vertex of the solid.  So there are 20 of these.

At the next distance, to the points that lie an edge-unit away from the opposite vertex, the travel is across the opposite edge from the original vertex.  But the points that are that distance from the new point are not the same distance from the original point.  They lie a pentagon-diagonal from the original point.

Again, as in the icosahedral case, there are no triangles that can be formed with the diametrically opposite point.

So there are 20+20=40 equilateral triangles, of two sizes, that can be formed.

 Posted by Charlie on 2005-08-26 18:44:37

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