All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes
Solid triangles (Posted on 2005-08-26) Difficulty: 3 of 5
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts part 1 -- spoiler | Comment 1 of 9

In the regular icosahedron there are only 3 distances that points can be separated from one another: the edge length of the icosahedron; the distance from a vertex, past an edge to the next vertex encountered; and the distance diametrically across the icosahedron.

There are of course 20 triangles of edge length equal to the edge length of the icosahedron: the faces of the icosahedron itself.

There can't be any triangles with the side length equal to the diameter, as each point is involved in only one such distance.

From any one point there are 5 points that are the intermediate distance, and from each of those 5 points there are 2 points that lie the intermediate distance from each of these two points.  That's 10 triangles associated with each of the 12 vertices, but each triangle is associated with 3 vertices, so there are 40 triangles altogether of this size.

So there are 20 + 40 equilateral triangles that can be formed from the vertices of the regular icosahedron.

  Posted by Charlie on 2005-08-26 18:13:54
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information