What is the size of the largest square which can fit inside a regular pentagon with a side-length of 1?
I should have tried putting a corner of the square into a corner of the pentagon. The problem this time is I have the square standing up straight. This is an improvement but not the best yet.
The height of the pentagon is sin(72)+sin(36)
if you divide this by the √2 diagonal of the square you get
.25(√(5+√5)+√(5-√5))≈1.0881
You can improve this by rotating the square by 9º
(divide by cos(9º))
≈ 1.1017
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Posted by Jer
on 2013-12-02 16:15:39 |
More likely to be correct solution.
