Little explorations with HP calculators (no Prime)

[RMollov]

(03-27-2017 07:39 PM)Gerson W. Barbosa Wrote:  

(03-27-2017 12:14 PM)pier4r Wrote:  

Let's divide one of the right-triangles into four right-triangles (two with sides lengths equal to r and x and two with side lengths r and 1-x) and a square with side r. Then it's area is

S = r*x + r(1-x) + r^2

S = r^2 + r

The area of the larger square, witch we know is equal to 1, is the sum of the areas of these four right-triangles plus the area of the smaller square in the center, with side 2*r:

S = 4(r^2 + r) + (2*r)^2 = 1

Or

8*r^2 + 4*r - 1 = 0

Here I am tempted to just take my wp34s and do 8 ENTER 4 ENTER 1 +/- SLVQ and get a valid numerical answer for the quadratic equation, but I decide to go through a few more steps by hand and get the exact answer:

r = (sqrt(3) - 1)/4

Would you show those steps here? I stopped before them thinking that's enough...

Thanks,

*** edit: I mean only if you derived this solution directly from the geometry and not by playing with the equation WolframAlpha style