Let the area of the larger base of a frustum be $T$, the area of the smaller base be $t$, and its height be $m$. The volume of the frustum is $V=k T m$, where $\frac{1}{3} \leq k \leq 1$. What is the ratio $\lambda(<1)$ of proportionality between the two bases? (The ratio of proportionality, or the proportionality factor: the ratio of corresponding distances in two similar figures.)

Application: a) For a certain frustum, the area of the larger base is $T=532.8 \mathrm{~cm}$, and its height is $m=23.2 \mathrm{~cm}$. The ratio of the corresponding sides of the bases is $3: 5$. What is its volume? - b) For a certain frustum, its volume is half that of a right prism whose base is congruent to the larger base of the frustum and whose height is equal to the height of the frustum. What is the ratio of the areas of the two bases?
The known volume formula for a truncated cone is

$$
k T m=\frac{m}{3}(T+\sqrt{T \cdot t}+t)
$$

that is, $(m \neq 0)$

$$
3 k T=T+\sqrt{T \cdot t}+t
$$

from which $(T \neq 0)$

$$
3 k=1+\sqrt{\frac{t}{T}}+\frac{t}{T}
$$

But it is known that

$$
\frac{t}{T}=\lambda^{2}
$$

and thus we obtain the following quadratic equation for $\lambda$:

$$
\lambda^{2}+\lambda+1-3 k=0
$$

from which (taking only the positive root)

$$
\lambda=\frac{\sqrt{12 k-3}-1}{2}
$$

If $k$ varies from $\frac{1}{3}$ (cone) to 1 (cylinder), then $\lambda$ varies from 0 (for $t=0$) to 1 (for $t=T$).

Application: a) From (1)

$$
k=\frac{\lambda^{2}+\lambda+1}{3}=\frac{\frac{9}{25}+\frac{3}{5}+1}{3}=\frac{9+15+25}{75}=\frac{49}{75}
$$

and thus the sought volume is

$$
v=k \cdot T \cdot m=\frac{49}{75} \cdot 532.8 \cdot 23.2 \approx 8076 \mathrm{~cm}^{3}
$$

b) According to the condition

$$
k T m=\frac{T m}{2}
$$

from which

$$
k=\frac{1}{2}
$$

Substituting this value of $k$ into (2)

$$
\lambda=\frac{\sqrt{3}-1}{2}
$$

and thus

$$
\frac{t}{T}=\lambda^{2}=\left(\frac{\sqrt{3}-1}{2}\right)^{2}=\frac{2-\sqrt{3}}{2}(\approx 0.134)
$$

Endre Szabó (Gyöngyös, Vak Bottyán st. III. fl.)