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