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$4$ circles of the same radius $r$ are internally tangent to a bigger circle centered at $O$ with radius $R$. Additionally the circles centered at $M,S$ or $N,T$ are externally tangent.

Given $MN\parallel ST$, the task is to calculate $MN$ for $R=2.5$, $r=0.5$ and $ST=1$.

So far, I have only derived the formula for the distance between $ST$ and $C$: $$d(ST, C) = R - \sqrt{(R-r)^2-\left(\frac{ST}{2}\right)^2}$$ that in my case equals $\frac{5-\sqrt{15}}{2}$. Otherwise, this problem has me stumped .

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  • $\begingroup$ Consider that $MSTN$ is an isosceles trapezoid with three sides with the same unit length. There is a relation between $MN$ and the height $h$ of such trapezoid, so there is a relation between the length of $MN$ and its distance from $O$. Another relation is given by the fact that $OM=ON=R-r$. Can you finish from here? $\endgroup$ Commented Aug 24 at 15:51

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Since $d_1 = 1$ and $r = \frac12$ we conclude that the two small circles with centers $S$ and $T$ are tangent to each other. As a result, angles $\widehat{MOS}$, $\widehat{SOT}$ and $\widehat{TON}$ are equal.

For simplicity let's write $\widehat{SOT} = 2 \alpha$. Then $\widehat{CON} = 3\alpha$.

We are looking for $|MN| = 2 (R - r) \sin{3\alpha}$.

You can do the rest and check your answer here later.

! $$\begin{array}{lr} \sin{\alpha} = \frac14 & \cos{\alpha} = \frac{\sqrt{15}}{4} \\ \sin{2\alpha} = 2\sin{\alpha}\cos{\alpha} = \frac{\sqrt{15}}{8} & \cos{2\alpha} = 1 - 2\sin^2{\alpha} = \frac78 \\ \sin{3\alpha} = \sin{\alpha}\cos{2\alpha} + \cos{\alpha}\sin{2\alpha} = \frac{22}{32} \end{array}$$

Finally we find:

$$|MN| = \frac{11}{4}$$

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