We have an old problem
Equilateral Triangle which asked can an equilateral triangle have all three of its vertices on lattice points on a 2-D plane or in 3-D space. Those answers were "no" and "yes", respectively.
I ask can a regular pentagon be placed in 2-D with all five of its vertices on lattice points? If that is not possible what about 3-D space?
Consider connecting 3 consecutive vertices of the pentagon to create an isosceles triangle with vertex angle 108.
The vector angle formula is cos(theta) = a.b/(|a||b|)
cos(108) = (1-sqrt(5))/4
a.b is a dot product which will be an integer call it d.
|a| and is the square root of an integer and |b|=|a| so call this 2sqrt(c).
(1-sqrt(5))/4 = d/(2sqrt(c))
square both sides
(3-sqrt(5))/8 = d^2/(4c)
The left is irrational but the right is rational so such a solution is not possible.
(The reason this works with equilateral triangles in 3-D is cos(60) is rational.)
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Posted by Jer
on 2024-04-05 13:30:54 |
3D solution?
