3. The bride and groom ordered a thirteen-layer cake consisting of three equally high cakes placed on interconnected and equally spaced stands. The distance between adjacent stands was $11 \mathrm{~cm}$, and the thirteen-layer cake was $30 \mathrm{~cm}$ high. The bottom cake was the largest, and the diameter of each subsequent cake was $6 \mathrm{~cm}$ shorter than the diameter of the previous one.

With the thirteen-layer cake, they fed themselves and 28 guests. Consider that each person on average ate $24 \mathrm{dag}$ of cake and that $77 \pi \mathrm{cm}^{3}$ of cake weighs $10 \mathrm{dag}$. What was the radius of the bottom (largest) cake?
3. The torte was eaten by 30 people. Since each person on average ate 24 dag, the thirteen-layer torte weighed $30 \cdot 24=720$ dag. Its volume was $\frac{77 \pi \cdot 720}{10}=5544 \pi \mathrm{cm}^{3}$.

If the height of each layer of the torte is $v$, and the distance between each layer and the stand above it is $x$, then $v+x=11$ and $3 v+2 x=30$. From these two equations, we calculate $v=8 \mathrm{~cm}$.

Let the radii of the individual layers be $r, r-3$, and $r-6$. Then $\pi r^{2} \cdot v+\pi(r-3)^{2} \cdot v+\pi(r-6)^{2} \cdot v=5544 \pi$. Simplifying the equation (considering $v=8 \mathrm{~cm}$) gives $3 r^{2}-18 r+45=693$ or $r^{2}-6 r-216=0$, which factors to $(r-18)(r+12)=0$. The only meaningful solution to the equation is $r=18$. The radius of the bottom (largest) layer was $18 \mathrm{~cm}$.

Mass of the torte: $T=n \cdot 24$ dag $=30 \cdot 24=720 \mathrm{dag}$... 1 point Volume of the torte:

$$
\begin{gathered}
10 \mathrm{dag} \ldots \ldots \ldots . . . .77 \pi \mathrm{cm}^{3} \\
720 \mathrm{dag} \ldots \ldots \ldots . \ldots \mathrm{cm}^{3}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_06_07_06886e410b39c0216094g-13.jpg?height=76&width=1642&top_left_y=656&top_left_x=204)

![](https://cdn.mathpix.com/cropped/2024_06_07_06886e410b39c0216094g-13.jpg?height=60&width=1639&top_left_y=707&top_left_x=206)

Volume of the bottom layer: $V_{1}=\pi r^{2} v$

Volume of the second layer: $V_{2}=\pi(r-3)^{2} v$

Volume of the third layer: $V_{3}=\pi(r-6)^{2} v$. . . 1 point

Together:

$V=V_{1}+V_{2}+V_{3}$

$V=\pi v\left(r^{2}+(r-3)^{2}+(r-6)^{2}\right)$

$V=\pi \cdot 8 \cdot\left(3 r^{2}-18 r+45\right)$

$V=\pi \cdot 8 \cdot 3 \cdot\left(r^{2}-6 r+15\right)$

$5544 \pi=\pi \cdot 24\left(r^{2}-6 r+15\right)$

1 point

$231=r^{2}-6 r+15$

$r^{2}-6 r-216=0$

Solution $(r-18)(r+12)=0 \Rightarrow r=18 \mathrm{~cm}$

.1 point