Gogeometry Problem 276


From Gogeometry

p276_square_circles

Solution

Using :

 

Without using :

Solution

276

  • Define G the center of CircleG with radius x
  • Define the point E as the only point of intersection between CircleG and CircleA => A, E and G are aligned
  • Define K in BC such as KE is tangent to CircleA
  • BC is tangent of circleA on B =>CBE intercepts arc BE of circleA =>2 ∠CBE= ∠BAE
  • EK is tangent of circleA on E =>KEB intercepts arc BE of circleA
  • => ∠KEB= ∠KBE => BKE is isosceles in K => KB=KE and ∠CKE= 2 ∠KBE
  • Define F on BC such as F belongs to CircleG => GF⊥BC
  • => KB=KE and KE⊥GE and KF⊥FG and GF=GE=x
  • KF is tangent to CircleG and KE is tangent to CircleG => KE=KF
  • => ΔKFG is congruent to ΔKEG (SAS)
  • => ∠FKG= ∠EKG => ∠FKE= 2 ∠FKG => ∠FKG= ∠KBE
  • ΔAEK is congruent to ΔABK (SAS) => ∠ABK= ∠FKG
  • =>ΔABK is similar to KFG => AB/BK=KF/FG
  • By symmetry F is in the middle of BC, BK=KF => BK=AB/4=a/4
  • AB/BK=4=KF/FG =>KF=4FG, KF=AB/4=>KF=AB/16
  • Therefore x=a/16

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