What is the shortest distance between two opposite points (two vertices separated by 2R) in the surface of a regular icosahedron? The regular icosahedron's edges are 1ft in length.

I don't understand the law of cosines, so here I try to solve it without it.

Take a cross section of the icosohedron, straight through the middle. You get a decagon with sides .5'. Draw a circle around the decagon. Call its radius X. Draw lines from the center of the circle to all the corners and midpoints of the decagon. Each line is 18 degrees away. So, sin 18 equals .25/X.

sin 18=.25/X
X*.3090169944=.25
X=.8090169944

Since X is the radius to the midpoints of the icosahedron's edges, use the Pythagorean Theorum to calculate R, the radius to the vertices.

.8090169944²+.5²=R²
.9045084972=R²
.9510565163=R
1.902113033=2R

So the distance is 1.902113033... feet.

Posted by Tristan on 2003-12-10 19:49:17