| Part 1: Size of square:
Say R = 1.
Drop a perpindicular from the center of that circle to the
side of the square that is entirely inside that circle.
Call the length of this perpindicular, x.
By extending this perpindicular to where the circles meet, we have bisected
the square, and seen that the side of the square is 1 + x.
By completing the right triangle started by the perpindicular and half of
the square, we can use pythagoras to express 1/2 of the side of the square
as square_root(1-x**2).
So, since squares will be squares, all sides equal,
1+x = 2*(square_root(1-x**2)), which gives x = .6 (for x>0).
So, side of square is 1.6
Part 2: Size of smaller circle:
Drop a perpindicular from where the circles meet to the mutual
tangent line. (This is half of our square, and so is .8 in length.)
A wonderful fact about tangent circles: The triangle whose vertices
are the two centers and the point on the tangent line we just found is
a right triangle. (To see this, extend the tangent line and the line
between the centers out to where they meet, and start marking equal,
supplementary, and complementary angles.)
The legs of this right triangle are both bases of smaller right triangles,
and their lengths can be calculated as
sqrt(1**2 + .8**2)
and
sqrt(r**2 + .8**2)
This makes the base calculable as both 1 + r and the square root of the sums
of the squares of the two terms above.
Everything cancels nicely:
1+r = sqrt( 1**2 + .8**2 + r**2 + .8**2)
square both sides:
1 + 2r + r**2 = 2.28 + r**2
r = .64.
Which is also the ratio of the two circles.
There is probably a quicker proof, but I wanted to stick to simple geometry
that I thought I was sure of.
|