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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)

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Some Thoughts The sizes of the triangles | Comment 3 of 9 |

Consider the solids as inscribed within a sphere of unit radius.

Each face of the icosahedron projects onto the sphere an equilateral triangle each of whose angles is 360/5 = 72 degrees.  The sides are therefore, by the law of cosines, 63.43494882292201 degrees. The edges are then twice the sine of this angle, or 1.051462224238267 radii of the circumscribing sphere.

For the larger of the two sizes of equilateral triangle formed from the vertices of the icosahedron, consider that each edge extends from one end point to the other of the creased diamond formed from two adjacent faces. This creased diamond projects onto the sphere as a rhombus, with angles of 72 and 144 degrees.  The line we are concerned with is the long diagonal, which divides the spherical rhombus into two isosceles triangles, with  base angles of 36 degrees and a vertex angle of 144 degrees.  Again using the law of cosines, the base itself is 116.565051177078 degrees. Twice the sine of half this is 1.70130161670408 radii of the circumscribing sphere, or 1.618033988749895 times the edge length of the icosahedron, or side of the smaller triangles (this is phi, the golden ratio).

For the dodecahedron, each pentagonal face projects onto the sphere a pentagon whose angles are each 360/3 = 120 degrees. This can be split into 5 spherical isosceles triangles whose base angles are 60 degrees and vertex angle is 360/5 = 72 degrees.  Solving this triangle gives a base length of 41.81031489577858 degrees.  Twice the sine of half of this is .7136441795461795, so that is the length of one  edge of the dodecahedron in radii of the circumscribing sphere.  As noted previously, there are no equilateral triangles formed with this length.

The smaller of the equilateral triangles formed is one diagonal of a pentagonal face.  This can be solved by plane trigonometry, where the vertex angle of the triangle is the angle of a regular pentagon: 108 degrees. That makes the base angles (180-108)/2 = 36 degrees. The two equal sides of the triangle are each .7136441795461795, so solving for the third side we get 1.154700538379251 radii of the circumscribing circle, which is again, phi, the golden ratio, times the edge length of the dodecahedron.

Each of the 20 larger size equilateral triangles is in fact one face of one of 5 interlocking regular tetrahedra, so we just need to calculate the edge length of such an inscribed tetrahedron.

Each face of the tetrahedron projects onto a spherical equilateral triangle with angles of 360/3 = 120 degrees. This makes the sides equal to 109.4712206344907. Twice the sine of half this angle is 1.632993161855452 radii of the circumscribing circle, making this the length of one side of each of the larger of the formed triangles, equivalent to 2.288245611270738 times the edge length of the dodecahedron, or sqrt(2) times the length of the side of the smaller equilateral triangles formed. (In other words, 2.288245611270738 is phi * sqrt(2).)

  Posted by Charlie on 2005-08-27 13:25:52
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