Given a regular icosahedron of unit edge length, find the shortest distance on its surface between two opposite vertices. If the distance is d then find ⌊ 1 0 0 0 d ⌋ .
Note: The path you take between the two opposite vertices has to be on the surface of the icosahedron.
Hint: Unfold (unwrap) the faces of the icosahedron, then apply the Law of Cosines.
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The following animation illustrates unfolding of the faces of the icosahedron.
As can be seen from the last frame, the required distance can be obtained by applying the Law of Cosines to the indicated triangle whose two other sides are 1 and 3 , making an angle of 1 5 0 ∘ between them, hence,
d = ( 3 ) 2 + 1 − 2 3 cos ( 1 5 0 ∘ ) = 2 . 6 4 5 7 5 1 3 1 1 , which makes the answer 2 6 4 5 .
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Unfold the faces and apply the Pythagorean Theorem to the following red triangle, whose one leg is equivalent to 3 unit equilateral triangle heights and the other leg is equivalent to 2 1 a unit equilateral side:
Then d = ( 2 3 ) 2 + ( 2 1 ) 2 = 7 , and ⌊ 1 0 0 0 d ⌋ = 2 6 4 5 .