<s>[INST] Find the radius of three equal circles, with radius $r$, inscribed in a rectangle. How can we find the radius of the small circles with the same given radius $r$. Three equal circles, with radius $r$, are inscribed in a rectangle in a way only one of them touches the others two, as the figure indicates. The circles centres form an isosceles triangle, is there any theorem can help in finding the value of $r$? Thanks for your help.

 [/INST] If the white curve is a circle the you have three points of the big circle (assume symmetry: Big circle has $(-8,0),(0,6),(8,0)$) so the equation (and radius) can be figured. 
Big Circle is center at $(0, -A)$ with radius $R$ and $8^2 + A^2 = R^2$ and $0^2 + (A+8)^2 = R^2$.
The three little circles have equations $(a_i - x)^2 + (b_i-y)^2 = r^2$.  $a_1=-a_3;b_1=b_3; a_2 = 0$. circle 1: has the three points $(-w,0), (-v,z);$ (which satisfies $v^2 + (z+A)^2 = R^2$) and $(-m,k)$.  
Circle 2: has three point $(-m,k),(m,k)$ and $(0,6)$.  The point $(-m,k)$ shared between circle 1 and circle two is tangent and colinear to the centers.
And circle 3 has three points $(w,0),(v,z),(m,k)$.The point $(m,k)$ shared between circle 1 and circle two is tangent and colinear to the centers.
So solve the following equations.
Big Circle:
$(8-0)^2 + (0+A)^2 = R^2$
$(0-0)^2 + (6 + A)^2 = R^2$
$(m-0)^2 + (k+A)^2 = R^2$
$(v-0)^2 + (z+A)62 = R^2$
Circle 2:
$(m-0)^2 + (k-b_2)^2 = (0-0)^2 + (6-b_2)^2 = r^2$.
Circle 3:
$(a_3-w)^2 + (b_3-0)^2 = (a_3 - v)^2 + (b_3 - z)^2=(a_3-m)^2 + (b_3-k)^2 = r^2$
Colinear tangent of circle 2 and circle 3
$m = \frac {a_3}2$ and $k = \frac {b_2 + b_3}2$.
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