5. The length of $\overline{P R}$ is the diameter of circle $k_{1}$, while the lengths of $\overline{P O}$ and $\overline{O R}$ are the diameters of circles $k_{2}$ and $k_{3}$ as shown in the figure. Circle $k_{4}$ is tangent to circles $k_{2}$ and $k_{3}$ externally and to circle $k_{1}$ internally. If the radius of circle $k_{4}$ is $18 \mathrm{~cm}$, calculate the length of the diameter $\overline{P R}$.

![](https://cdn.mathpix.com/cropped/2024_06_03_b1f0e0547660f03869deg-1.jpg?height=591&width=605&top_left_y=1869&top_left_x=791)

Each problem is scored out of 10 points.

The use of a pocket calculator or any reference materials is not allowed.
5. Let $x=\left|\mathrm{PO}_{2}\right|=\left|\mathrm{O}_{2} \mathrm{O}\right|=\left|\mathrm{OO}_{3}\right|=\left|O_{3} R\right|$

Then $\left|O_{2} \mathrm{O}_{4}\right|=x+18$.

1 POINT

Consider the right triangle $\mathrm{O}_{4} \mathrm{OO}_{2}$. By applying the Pythagorean theorem, we get

![](https://cdn.mathpix.com/cropped/2024_05_30_704a435344239ccd44e4g-3.jpg?height=551&width=562&top_left_y=547&top_left_x=747)

$$
\begin{array}{ll}
\left|\mathrm{O}_{2} \mathrm{O}_{4}\right|^{2}=\left|\mathrm{O}_{2} \mathrm{O}\right|^{2}+\left|\mathrm{OO}_{4}\right|^{2} & \\
(x+18)^{2}=x^{2}+(2 x-18)^{2} & \\
& \\
4 x^{2}-108 x=0 & 2 \text { POINTS } \\
4 x(x-27)=0 & 2 \text { POINTS }
\end{array}
$$

Thus, $x_{1}=0$ which is not possible and

$$
x_{2}=27
$$

1 POINT

Finally, $|P R|=4 x=108 \mathrm{~cm}$.