A cylindrical container with a diameter of $36 \mathrm{~cm}$ has a cylindrical insert with a radius of $6 \mathrm{~cm}$ and another with a radius of $12 \mathrm{~cm}$ placed inside it. What is the maximum radius of another cylindrical insert that can be placed inside the container alongside the others?
The two plugged flasks can only be placed in the container in such a way that they touch each other on the diameter due to the condition given by the radii. The third flask will have the largest radius if it is placed in such a way that it touches the side of the container and the previous two plugged flasks. From now on, we will only talk about the base circles of the containers.

Let the center of the circle with a diameter of $36 \mathrm{~cm}$ be $O$, the center of the circle with a radius of $12 \mathrm{~cm}$ be $O_{1}$, the center of the circle with a radius of $6 \mathrm{~cm}$ be $O_{2}$, the center of the circle to be found be $O_{3}$, and its radius be $r$. The circles centered at $O_{1}$ and $O_{2}$ touch each other at point $E$, and the foot of the perpendicular from $O_{3}$ to the line $O_{1} O_{2}$ is $T$. Let $O_{3} T=m$ and $E T=a$.

We write the Pythagorean theorem for the sides of the triangles $\mathrm{TO}_{3} \mathrm{O}_{2}$, $O_{1} \mathrm{O}_{3} T$, and $O O_{3} T$:

\[
\begin{aligned}
m^{2}+(6-a)^{2} & =(6+r)^{2} \\
m^{2}+(12+a)^{2} & =(12+r)^{2} \\
m^{2}+(6+a)^{2} & =(18-r)^{2}
\end{aligned}
\]

From the difference of equations (1) and (2):

\[
\begin{aligned}
m^{2}+36-12 a+a^{2} & =36+12 r+r^{2} \\
m^{2}+144+24 a+a^{2} & =144+24 r+r^{2} \\
-36 a & =-12 r \\
3 a & =r
\end{aligned}
\]

Similarly, from the difference of equations (2) and (3), we get:

\[
a=5 r-24
\]

From (4) and (5):

\[
\frac{r}{3}=5 r-24
\]

Thus, $r=\frac{36}{7} \approx 5.14 \mathrm{~cm}$.

This means that a plugged flask with a diameter of approximately $10 \mathrm{~cm}$ can still be placed in the container.

Turi Adrienn (Budaörs, Illyés Gy. Hatévf. Középisk., 9th grade)