Edited by Richard Walker, Saturday 6 June 2026 at 00:49
This follows on from my last post.
It is a well-known result that in any triangle the centroid divides each median in the ratio 2:1. The length of CG is 1 because it a radius of the circumcircle and so we have GE = and CE = .
Because U and V are midpoints F must be the midpoint of CE, so CF = and FG = .
Note also that from the fact that V is the midpoint of BC and the symmetry of the equilateral triangle we have GV = GE = .
Now we can apply Pythagoras' Theorem, first in triangle GFV and then in GFW, to find lengths FV and FW.
In GFV, FV2 = GV2 - GF2 = . So FV = .
In GFW, FW2 = GW2 - GF2 = 1 - = . So FW = = .
Finally, UV = 2 x FV = ; UW = UV + FW = + ; and so = ( + ) ÷ () = = , as claimed.
Proving Odom's Golden Ratio Construction
This follows on from my last post.
It is a well-known result that in any triangle the centroid divides each median in the ratio 2:1. The length of CG is 1 because it a radius of the circumcircle and so we have GE = and CE = .
Because U and V are midpoints F must be the midpoint of CE, so CF = and FG = .
Note also that from the fact that V is the midpoint of BC and the symmetry of the equilateral triangle we have GV = GE = .
Now we can apply Pythagoras' Theorem, first in triangle GFV and then in GFW, to find lengths FV and FW.
In GFV, FV2 = GV2 - GF2 = . So FV = .
In GFW, FW2 = GW2 - GF2 = 1 - = . So FW = = .
Finally, UV = 2 x FV = ; UW = UV + FW = + ; and so = ( + ) ÷ () = = , as claimed.