Given square ABCD with E on AB, F on BC, G on CD, H on DA, AE=BF=CG=DH. Segments AF, BG, CH, CE dissect the square into 4 triangles, 4 trapezoids and a central square.
If circles can be inscribed in the trapezoids which have the same radius as the inscribed circle of the central square, find AE/AB.
I will let the main square have unit length sides and AE=BF=CG=DH=x, this makes x equal to the ratio AE/AB. Then AF=BG=CH=DE=sqrt(1+x^2).
Let DE and AF intersect at I, AF and BG intersect at J, BG and CH intersect at K, and CH and DE intersect at L. Then the square is IJKL, and the four trapezoids are BEIJ, FJKC, GKLD, and HLIA.
Each of the four trapezoids are part of triangles ABJ, BCK, CDL, and DAI respectively. Each trapezoid has the same incircle as its corresponding triangle.
Triangles AJB and ABF are similar triangles, which implies BJ/BF = AJ/AB = AB/AF. Then AJ=BK=CL=DI=1/sqrt(x^2+1) and AI=BJ=CK=DL=x/sqrt(1+x^2).
The sides of the square IJ =JK = KL = LI = AJ-AI = (1-x)/sqrt(x^2+1).
The radius of the incircle of IJKL is half its side length implying r = (1-x)/(2*sqrt(x^2+1)).
The radius of the incircle of ABF can be found using area = base*height/2 = inradius*perimeter/2, which creates the equation (1/sqrt(x^2+1)) * (x/sqrt(x^2+1)) / 2 = r * (1/sqrt(x^2+1) + (x/sqrt(x^2+1) + 1) / 2.
Substute the inradius from the square into this equation and simplify to get 6x^3 - 2x = 0. This has one positive root x=1/sqrt(3).
Solution
