Challenging Geometry Puzzle: Problem 1594. 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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Let square ABCD of side a and square CGFE be of side b
ReplyDeleteTriangles CGH and CFD are similar, hence
2/b = a/V(a^2+1) from which
4/b^2 = a^2/ (a^2 + 1) ...............(1)
Moreover using Pythagoras on Triangles CEF and CFD,
a^2 + 1 = 2b^2 ...................(2)
(1)/(2) gives {4/b^2 }/ (a^2 + 1) = {a^2 / (a^2 +1) }/ 2b^2
So 4 = a^2 / 2
Hence a = 2V2
Sumith Peiris
Moratuwa
Sri Lanka
An easier way is to realise that CF is tangential to circle AHF at F (< CFH = 45 = < CAF)
DeleteSo CF^2 = CF.CA
Hence a^2 +1 = 2 X V2. a
From which we can get the value of 'a' by solving the straightforward quadratic
Sumith Peiris
Moratuwa
Sri Lanka
a=sqrt2 + 1
ReplyDeletehttps://photos.app.goo.gl/JtBafqPxohVZRD9W6
ReplyDeleteLet AB=CD= x
Note that triangle CHF similar to CFA ( case AA)
So CH/CF=CF/CA => CF^2= CF.CA
So 1+x^2= 2 . x. sqrt(2)
Or x^2-2.sqrt(2).x+1=0
We have 2 solutions x= sqrt(2)+1 and x= sqrt(2)-1
If x=sqrt(2)-1 => points F and H will be outside of the square. It is not acceptable
So AB=x= sqrt(2)+1