# internally and externally tangent circles

See the diagram here: diagram

The diagram shows two circles of radius 1 and 2 tangent to each other and internally tangent to a circle of radius 3. What is the radius of the outlined circle internally tangent to the circle of radius 3 and externally tangent to the circles of radii 1 and 2?

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Let $r$ be the radius of the smallest circle. Let $O_1$ be the center of the circle of radius 1, $O_2$ be the center of the circle of radius 2, $O_3$ be the center of the circle with radius 3, and $O_r$ be the center of the circle of radius $r$.
If you draw $\triangle O_1O_2O_r$, you can easily verify that it has side lengths 3, $1+r$, and $2+r$. In addition, you can also verify that the cevian $O_rO_3$ to the side with length 3 has length $3-r$. We now have enough information to solve for $r$. If we let $\theta=\angle O_1O_3O_r$, then by the Law of Cosines: \begin{align} 4+(3-r)^2+4(3-r)\cos\theta &= (1+r)^2\\ 1+(3-r)^2-2(3-r)\cos\theta &= (2+r)^2. \end{align} Adding the first equation to twice the second yields the equation $6+3(3-r)^2=(1+r)^2+2(2+r)^2$, which has the solution $r=\frac 67$.
Alternatively, you can use Descartes' Theorem, in particular, the following formula: $$\frac1r=\frac 11+\frac12-\frac13\pm2\sqrt{\frac1{1\cdot2}-\frac1{2\cdot3}-\frac1{3\cdot1}}=\frac 76,$$ so $r=\frac67$.