Shortest distance on the surface on an icosahedron

Geometry Level pending

Given a regular icosahedron of unit edge length, find the shortest distance on its surface between two opposite vertices. If the distance is d d then find 1000 d \lfloor 1000 d \rfloor .

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.


The answer is 2645.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

David Vreken
Jun 16, 2020

Unfold the faces and apply the Pythagorean Theorem to the following red triangle, whose one leg is equivalent to 3 3 unit equilateral triangle heights and the other leg is equivalent to 1 2 \frac{1}{2} a unit equilateral side:

Then d = ( 3 2 ) 2 + ( 1 2 ) 2 = 7 d = \sqrt{(\frac{\sqrt{3}}{2})^2 + (\frac{1}{2})^2} = \sqrt{7} , and 1000 d = 2645 \lfloor 1000d \rfloor = \boxed{2645} .

Hosam Hajjir
Jun 12, 2020

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 1 and 3 \sqrt{3} , making an angle of 15 0 150^{\circ} between them, hence,

d = ( 3 ) 2 + 1 2 3 cos ( 15 0 ) = 2.645751311 d = \sqrt{ (\sqrt{3})^2 + 1 - 2 \sqrt{3} \cos(150^{\circ} )} =2.645751311 , which makes the answer 2645 \boxed{2645} .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...