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5 circles in a square part 1 (Posted on 2019-08-28) Difficulty: 2 of 5
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.

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methodology... so close! | Comment 4 of 6 |

See the figure...

Let's call the center of the square "O" which is at (0,0) and let's say AB=1. Now, we start E near B and keep moving E away from B (as G tracks likewise) until the left-most circle hits the wall. This means that the left circle's circumference and the left wall intersect. Knowing E, we know the center and the diameter of the left circle.  

The left wall (DA) is simply x = -0.5.

The center of the center circle "O" is always (0, 0). The circles' radii are "r"

But what is the angle theta?

Since AE=BF, theta = arctan(1/BF) = arctan(1/AE)

The center of the left circle Z is given by r and theta:

Z_x = 2 r cos (90-theta)

Z_y = 2 r sin (90-theta) 

This circle must intersect the left wall. 

But.., do I really have two unknowns, r and theta? I don't think so......

Edited on August 31, 2019, 10:14 am
  Posted by Steven Lord on 2019-08-30 15:56:00

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