Two Circles Kissing

The radius can be calculated from a Pythagorean Theorem applied to the triangle $ADH$ , once all the sides get expressed in terms of $r$ as shown below:

$AD=KI=2r$

$AF=AD-FD=2r-1$

$AK=AE=AF=2r-1$

$DC=AB=AK+KB=(2r-1)+r=3r-1$

$GI=DC-DG-IC=(3r-1)-1-r=2r-2$

$GH=EH=IH=\frac12\times GI=\frac12(2r-2)=r-1$

$DH=DG+GH=1+(r-1)=r$

$AH=AE+EH=(2r-1)+(r-1)=3r-2$

$AH^2=AD^2+DH^2$

$(3r-2)^2=(2r)^2+r^2$

$3r+2=\sqrt5 r$

$r=\frac2{3-\sqrt5}=\frac{3+\sqrt5}2$

$3+5+2=\boxed{10}$

Choose the sides of the rectangle to be a & b (a<b) and let the radius of the larger circle be x. We can write down the following equations immediately: (a-1-x)²+(b-1-x)²=(1+x)², b=1+y+2√(x) since the common tangent between two circles of radii 1 & x is 2√(x). Moreover, the other common tanget passes through a vertex of the rectangle and thus, a-1=b-x. These three equations can be solved to obtain: a=3+√5, b=a²/4 and x = (3+√5)/2

I choose the point of osculation of the circles as the origin of a coordinate system. Note now that the small circle in the triangle is transformed into the large circle into a larger triangle by point reflection with factor $r$ : $(x,y) \mapsto (-rx,-ry)$ .

The center of the small circle has coordinates $(-c,-s)$ , with $s,c > 0$ and $s^2 + c^2 = 1$ .

The top-left corner of the triangle has coordinates $(-c-1,h)$ . The line from the origin through this corner is tangent to the circles and therefore perpendicular to the radius. This gives the dot-product equation $(-c-1,h)\cdot(-c,-s) = 0$ , with solution $h = (c+c^2)/s$ .

Upon transformation, the bottom of the small circle ( $y = -s-1$ ) is transformed to the top of the large circle ( $y = h$ ). This tells us that $r(s+1) = \frac{c+c^2}s.$ Likewise, the top of the small circle ( $y = 1-s$ ) is transformed to the bottom of the large circle ( $y = -1-s$ ). Thus $r(s-1) = -1-s.$ Adding and subtracting these two equations, and multiplying them by $s$ , we find $\begin{cases} 2rs^2 = c+c^2 -s^2 -s = (c+s+1)(c-s) \\ 2rs = c + c^2 + s^2 + s = c + s + 1\end{cases}$ Divide the two: $s = \frac{2rs^2}{2rs} = \frac{(c+s+1)(c-s)}{c+s+1} = c-s.$ This gives $c = 2s$ so that $c = 2/\sqrt5, s = 1/\sqrt 5$ . Finally, $r = \frac{1+s}{1-s} = \frac{\sqrt 5 + 1}{\sqrt 5 - 1} = \frac{6 + 2\sqrt 5}4 = \frac{3 + \sqrt 5}2.$ The answer, then, is $a + b + c = 3 + 5 + 2 = \boxed{10}$ .