Challenging Geometry Puzzle: Problem 1599. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
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ReplyDeleteDefine points K,L,H,M and angle u per sketch
Let LF=LH= 1
Triangle GLF congruent to EKF ( case SSS)
So EF =GF and GF=EF=FC= sqrt(5)
Using similar triangles we will have MG= 1/sqrt(5) and AM=2/sqrt(5)
And AC= v= 2.sqrt(5)+3/sqrt(5)
Area A1 of triangle ABC= v^2/4
And A2/A1= 16/v^2 where v= 2.sqrt(5)+3/sqrt(5)
Let the 4 squares be GDKL, LKMN, NMEH and NHTF
ReplyDeleteLet AB = BC =a and let each square be of side b
Let < EGH = x and < NGF = y
Now EN^2 = 2b^2 = NL.NG
So NE is tangential to Circle GLE at E and so < LEN = < EGH = x
< ELH = y, it follows that x+y = 45
Also since < FET = y and GH // FT, < EFC = 90 and Triangle GEC is Right Isosceles
Hence GF = FE = FC = V5.b and CE = V10.b
Now since GE // DB, < BDE = x and so
Triangles BDE & GEH are similar and so BE = EH/EG X DE = V10. b/b X DE = 3bV10
So BC = b(3/V10 + V10) and
A1 = BC^2 / 2 = b^2 / 2 (3/V10 + V10)^2 = b^2 / 2 (9 / 10 + 10 + 6) = 169.b^2 / 20
Hence A1/A2 = (169.b^2 /20) / 4.b^2 = 169/80
Sumith Peiris
Moratuwa
Sri Lanka
=80/169
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