Gogeometry Problem 458


From Gogeometry

p458_square_circle

Solution

Using :

458

  • Define O middle of AB, r=ED
  • AD^2+AO^2=DO^2
  • a^2+(a/2)^2=(a/2+r)^2
  • a^2+a^2/4=a^2/4+r^2+ar
  • r^2+ar-a^2=0
  • ∆=a^2+4a^2=5a^2
  • r>0 => r=(a.sqr(5)-a )/2
  • Draw a square symetryc of ABCD by O
  • =>D becomes D’, C becomes C’, E becomes E’
  • => C’D’BA is a square, D, E, O, E’, D’ are colinear
  • => EG=E’G’ and FE=F’E’
  • FG ꓕ DD’ => F’G’ ꓕ DD’
  • ∆ADO is similar to ∆E’DG’ and EDF (aa)
  • =>AD/AO=2=E’D/E’G’=ED/EF
  • E’D=a+r = 2 EG and ED=r=2 EF
  • =>EF+EG=(r+a+r)/2=r+a/2=(a.sqr(5)-a )/2+a/2= a.sqr(5)/2
  • Therefore FG= a.sqr(5)/2

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