Everything Touches, Part 2

Let the radius of the small and big circles be $r$ and $R$ respectively, $M$ and $N$ be the midpoints of $CD$ and $AB$ respectively, and $PQ$ be a diameter of the big circle such that $A$ , $P$ , $O$ , and $Q$ are colinear,

By tangent-secant theorem , we have:

$\begin{aligned} AP \cdot AQ & = AN^2 \\ (16-2R)\cdot 16 & = 8^2 \\ 8-R & = 2 \\ \implies R & = 6 \end{aligned}$

By Pythagorean theorem , we have:

$\begin{aligned} O'A^2 & = AN^2 + O'N^2 \\ (16+r)^2 & = 8^2 + (16-r)^2 \\ (16+r)^2 - (16-r)^2 & = 64 \\ 64 r & = 64 \\ \implies r & = 1 \end{aligned}$

Therefore $r + R = 1 + 6 = \boxed 7$ .

Claim: The larger circle centered at $O$ has radius $6$ . Proof: Note that the points $A, O$ and the point of contact, $G$ are all collinear. Let the radius of the larger circle be $R$ . In the figure above, applying the Pythagorean Theorem in $\triangle AEO$ , $\;\;\;\;\;\;\;\;\;\;AE^2+EO^2=OA^2$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\implies 8^2+R^2=(16-R)^2=16^2-32R+R^2$ $\implies R=\frac{16^2-8^2}{32}=6$

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Claim: The smaller circle centered at $O'$ has radius $1$ . Proof: Note that the points $A, O'$ and the point of contact, $I$ are all collinear. Let the radius of the smaller circle be $r$ . In the figure above, applying the Pythagorean Theorem in $\triangle AHO'$ , $\;\;\;\;\;\;\;\;\;O'H^2+AH^2=O'A^2$ $\;\;\;\;\;\;\;\implies 8^2+(16-r)^2=(16+r)^2$ $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\implies 8^2+16^2-32r+r^2=16^2+32r+r^2$ $\implies r=\frac{8^2}{64}=1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$

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Therefore, the sum of the radii of the two inscribed circles is $R+r=\boxed{7}$