In the plane, there are two circles with different radii, and they have no common interior points. Their radii are $R, r$, and the distance between their centers is $d$. We draw a secant through their external similarity point, then rotate the figure around the line connecting the centers of the circles. Consider the two frustums of cones described by the chords cut by the secant from the circles. What angle does the secant make with the axis of rotation when the difference in the areas of the lateral surfaces is the greatest?
We can assume that $R>r$. Let the external similarity point of the circles be denoted by $H$, the center of the larger circle by $O$, the center of the smaller one by $O^{\prime}$, and the angle between the secant and the line $OH$ by $\alpha$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_140a6087be1380d55651g-1.jpg?height=886&width=1529&top_left_y=262&top_left_x=279)

Figure 1

The two truncated conical surfaces in question are similar (can be mapped to each other by a dilation centered at $H$), the ratio of their corresponding linear dimensions is the same as the ratio of the radii of the circles, and the ratio of their surface areas is the same as the ratio of the squares of the radii. If the surface area of the larger truncated conical surface is $P$, then that of the smaller one is $\frac{r^{2}}{R^{2}} P$, and their difference is $\frac{R^{2}-r^{2}}{R^{2}} P$. It is clear that this is maximized when $P$ itself is maximized. $\left(\frac{R^{2}-r^{2}}{R^{2}}>0\right.$, constant $)$.

Express $P$ as a function of $\alpha$. Let the points of intersection of the secant with the circle of radius $R$ be $A_{1}, A_{2}$, and the feet of the perpendiculars from these points to the line $OH$ be $C_{1}, C_{2}$. According to the known formula for the surface area of a truncated conical surface,

$$
P=\pi\left(A_{1} C_{1}+A_{2} C_{2}\right) A_{1} A_{2}
$$

Let the foot of the perpendicular from $O$ to the secant (which is also the midpoint of the chord $A_{1} A_{2}$) be $F$, and the foot of the perpendicular from $F$ to the line $OH$ be $T$.

$FT$ is the midline of the right trapezoid $A_{1} C_{1} C_{2} A_{2}$, so

$$
A_{1} C_{1}+A_{2} C_{2}=2 F T
$$

$\angle OFT = \alpha$, since they are acute angles of right triangles. Therefore, in the right triangle $OFT$,

$$
F T=O F \cdot \cos \alpha
$$

Furthermore, in the right triangle $OHF$,

$$
O F=O H \cdot \sin \alpha
$$

and by the similarity of the circles,

$$
\frac{O H-d}{O H}=\frac{r}{R}
$$

from which

$$
O H=\frac{d R}{R-r}
$$

Therefore,

$$
F T=\frac{d R}{R-r} \sin \alpha \cos \alpha
$$

In the right triangle $OA_{2}F$,

$$
A_{2} F=\sqrt{R^{2}-O F^{2}}=\sqrt{R^{2}-\frac{d^{2} R^{2}}{(R-r)^{2}} \sin ^{2} \alpha}
$$

The chord $A_{1} A_{2}$ is twice this length.

Based on the obtained relationships, the surface area of the truncated conical surface is:

$$
P=4 \pi \frac{d^{2} R^{2}}{(R-r)^{2}} \sin \alpha \cos \alpha \sqrt{\left(\frac{R-r}{d}\right)^{2}-\sin ^{2} \alpha}
$$

Since $P \geqq 0$, $P$ is maximized at the same time as the positive constant multiple of its square. It is sufficient to examine the expression

$$
\sin ^{2} \alpha \cos ^{2} \alpha\left[\left(\frac{R-r}{d}\right)^{2}-\sin ^{2} \alpha\right]
$$

which, with the substitutions $\sin ^{2} \alpha=x$ and $\left(\frac{R-r}{d}\right)^{2}=a$, takes the form:

$$
x(1-x)(a-x)=x^{3}-(a+1) x^{2}+a x
$$

An important observation from Figure 1 is that $\frac{R-r}{d}=\sin \beta$ where $\beta$ is the angle between the common external tangents of the circles and the line $OH$.

Let us temporarily ignore the meaning of $x$ and $a$, and consider the function

$$
f(x)=x^{3}-(a+1) x^{2}+a x
$$

defined on the real numbers, where $a$ is a positive constant less than 1. The function $f(x)$ has 3 real roots, in increasing order: $x_{1}=0$, $x_{2}=a$, $x_{3}=1$. Since the coefficient of $x^{3}$ is positive, the nature of the graph of the function is as shown in Figure 2.

![](https://cdn.mathpix.com/cropped/2024_05_02_140a6087be1380d55651g-2.jpg?height=441&width=762&top_left_y=1048&top_left_x=665)

Figure 2

However, in our problem, $x=\sin ^{2} \alpha$ and $a=\sin ^{2} \beta$ and $0 \leqq \alpha \leqq \beta$, so only the segment of the graph between 0 and $a$ describes the variation of a real surface area.

We are close to the answer to the posed question. We need to find the zero of the derivative of $f(x)$ between 0 and $a$, from which the desired angle can be determined based on $x=\sin ^{2} \alpha$.

$$
f^{\prime}(x)=3 x^{2}-2(a+1) x+a
$$

The roots of this are

$$
x_{1}=\frac{a+1-\sqrt{a^{2}-a+1}}{3} \text { and } x_{2}=\frac{a+1+\sqrt{a^{2}-a+1}}{3} \text {. }
$$

We will show that $0 < x_{1} < a < x_{2}$. For the first statement, we provide two-sided estimates for the square root expression: on the one hand, $\sqrt{a^{2}-a+1} < 1 - a (>0)$.

For the second statement,

$$
x_{2}>\frac{a+a+\sqrt{(1-a)^{2}+a}}{3}>\frac{2 a+\sqrt{a}}{3}>a
$$

since $1 > a$. Therefore, $x_{1}$ is the point we are looking for.

Based on all of this, the desired angle is:

$$
x_{0}=\operatorname{Arc} \sin \sqrt{\frac{\left(\frac{R-r}{d}\right)^{2}+1-\sqrt{\left(\frac{R-r}{d}\right)^{4}-\left(\frac{R-r}{d}\right)^{2}+1}}{3}}
$$

$\operatorname{Arc} \sin y$ (read: arcsine of $y$) denotes the (radian measure of the) angle $\alpha$ such that $\sin \alpha=y$ and $-\frac{\pi}{2} \leqq \alpha \leqq \frac{\pi}{2}$.

Gyula Károlyi (Budapest, Fazekas M. Gyak. Gimn., IV. o. t.)