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 5 circles in a square part 2 (Posted on 2019-09-08)
Given square ABCD with E on AB, F on BC, G on CD, H on DA, AE=BF=CG=DH. Segments AF, BG, CH, DE dissect the square into 4 triangles, 4 trapezoids and a central square.

If the circle inscribed in the square has the same radius as the incircles of the triangles, find AE/AB.

Note: exact solution is the solution to a cubic equation.

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 Solution Comment 1 of 1
A lot of work from my solution to part 1 also applies here.  In addition to that a few things need to be added.

Continuing from the work already done: IE = LH = KG = JF = AF-AJ = x^2/sqrt(1+x^2).

The The radius of the incircle of AEI can be found using area = base*height/2 = inradius*perimeter/2, which creates the equation (x^2/sqrt(x^2+1)) * (x/sqrt(x^2+1)) / 2 = r * (x^2/sqrt(x^2+1) + (x/sqrt(x^2+1) + x) / 2.

The inradius of the square is still r = (1-x)/(2*sqrt(x^2+1)).  Substituting that into the prior equation and simplifying eventually yields a cubic equation 4x^3+x^2-4x+1 = 0.  This has three real roots x=0.29914, x=0.67996, x=-1.2291.  The only one which satisfies the original problem is x=0.67996.

 Posted by Brian Smith on 2019-09-09 10:59:44

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