## Task B-4.7.

Leda and Una are playing with clay in the shape of a cylinder whose height is 6 times greater than the diameter of its base. Leda took a part of that clay and made a larger ball, while Una made a smaller ball from the remainder. How many times greater is the volume of Leda's ball compared to the volume of Una's ball, if the sum of their radii is 3 times greater than the radius of the cylinder's base?
## Solution.

Let $r$ be the radius of the base of the cylinder. The height of the cylinder is 6 diameters, or 12 radii of the cylinder, so the volume of the cylinder is $V_{V}=r^{2} \pi v=r^{2} \pi 12 r=12 r^{3} \pi$.

Let $R_{L}$ be the radius of the sphere made by Leda, and $R_{U}$ be the radius of the sphere made by Una. The sum of the volumes of these spheres is equal to the volume of the cylinder.

$$
\frac{4}{3} R_{L}^{3} \pi+\frac{4}{3} R_{U}^{3} \pi=12 r^{3} \pi
$$

It follows that $R_{L}^{3}+R_{U}^{3}=9 r^{3}$, and from the condition in the problem, we also have $R_{L}+R_{U}=3 r$.

Let's solve this system of equations.

$$
\begin{aligned}
& \left(3 r-R_{U}\right)^{3}+R_{U}^{3}=9 r^{3} \\
& 27 r^{3}-27 r^{2} R_{U}+9 r R_{U}^{2}=9 r^{3} \\
& 3 r^{2}-3 r R_{U}+R_{U}^{2}=r^{2} \\
& R_{U}^{2}-3 r R_{U}+2 r^{2}=0
\end{aligned}
$$

From this, we get $R_{U}=2 r$ or $R_{U}=r$. 1 point

Since Leda's sphere has the larger radius, it follows that $R_{U}=r, R_{L}=3 r-r=2 r$.

Now we see that the ratio of the volumes is $\frac{V_{L}}{V_{U}}=\left(\frac{R_{L}}{R_{U}}\right)^{3}=\left(\frac{2 r}{r}\right)^{3}=8$.

The volume of Leda's sphere is 8 times greater than the volume of Una's sphere.

Note: In the equation $R_{L}^{3}+R_{U}^{3}=9 r^{3}$, we can recognize the sum of cubes and use $R_{L}+R_{U}=3 r$. After that, we also get a quadratic equation, either in terms of $R_{U}$ or $R_{L}$ or the ratio $\frac{R_{L}}{R_{U}}$.