You may be making a minor error. 150 years old book should be right !
I think it should be:
r^2 + (2r-t)^2 = (r+t)^2
r^2 = (r+t)^2 - (2r-t)^2
r^2 = (r+t - 2r+t)(r+t+2r-t)
r^2 = (2t - r)(3r)
r = (2t - r)(3)
r = 6t - 3r
r=(3/2) t
Nice question.
-O
We have been assigned this math problem (22) and I'm having some trouble solving it.
Sacred Mathematics: Japanese Temple ... - Google Book Search
It wants me to show that R=2t, and I know so far that the sides of the square are equal to 4r
The answer key says to use the Pythagorean theorem on the little circles to show that r= 3/2 t and I'm having problems with that
This is what I did so far, from diving the square in halves and then connecting the four origins of the small circles inside the square, you then see 4 small triangles. I used the bottom right triangle.
I got that the hypotenuse is r+t and then the side on the horizontal half line is 2r - t and the side on the vertical half line is r for sure. I tried using the pythagorean theorem but all I came to was that 2r^2 - tr = 2t, for the theorem the sides came up to be r^2 + (2r-t)^2 = (r+t)^2
Does anyone know what I'm doing wrong here?