Between two spheres, a point-like light source moves along the central axis. In which position will the sum of the illuminated surfaces on the spheres be the largest?
Let's use the notation from the figure. Suppose the light source is at point $P$, and let $O_{1} O_{2}=d$, and $O_{1} P=x$.

![](https://cdn.mathpix.com/cropped/2024_05_02_238200eb9cac8c8fe0efg-1.jpg?height=466&width=593&top_left_y=220&top_left_x=755)

Due to the physical context of the problem, we assume that

$$
d>r+R
$$

and

$$
r \leq x \leq d-R
$$

conditions are satisfied. Since the surface area of the spherical shell is $2 \pi r \cdot m$, we are looking for the maximum of the function $F(x)=2 \pi(M \cdot R+m \cdot r)$ under conditions (1) and (2). From the similarity of triangles $O_{1} P A$ and $O_{1} A P_{1}$, we have $\frac{r}{x}=\frac{r-m}{r}$, which gives $m=r-\frac{r^{2}}{x}$. Similarly, we find that $M=R-\frac{R^{2}}{d-x}$, and thus

$$
F(x)=2 \pi\left(R^{2}+r^{2}\right)-2 \pi\left(\frac{r^{3}}{x}+\frac{R^{3}}{d-x}\right)
$$

From (3), we see that $F(x)$ is maximal precisely when $G(x)=\frac{r^{3}}{x}+\frac{R^{3}}{d-x}$ is minimal. In this case, either $x=r$, or $x=d-R$, or $G^{\prime}(x)=\frac{R^{3}}{(d-x)^{2}}-\frac{r^{3}}{x^{2}}=0$ holds.

The latter holds precisely when

$$
x=\frac{d}{1+\sqrt{\frac{R^{3}}{r^{3}}}}
$$

If the $x$ from (4) satisfies condition (2), i.e.,

$$
r \leq \frac{d}{1+\sqrt{\frac{R^{3}}{r^{3}}}} \leq d-R
$$

then the second derivative $G^{\prime \prime}(x)=2\left(\frac{r^{3}}{x^{3}}+\frac{R^{3}}{(d-x)^{3}}\right)$ is positive at $x$, so $G(x)$ has a local minimum at this point, and $F(x)$ has a local maximum. The left inequality in (5) can be written as

$$
d \geq r+R \sqrt{\frac{R}{r}}
$$

while the right inequality is

$$
d \geq R+r \sqrt{\frac{r}{R}}
$$

Without loss of generality, we can assume that $r \leq R$. Then we see that (6) is a stricter condition than (1), and (7) is weaker, so it can be omitted. If condition (6) is not satisfied, then $d<R+R \sqrt{\frac{R}{r}}$, and thus

$$
x=\frac{d}{1+\sqrt{\frac{R^{3}}{r^{3}}}}<r
$$

Here, $G^{\prime \prime}(x)$ is also positive, so $G(x)$ has a local minimum, and $F(x)$ has a local maximum, but $x$ does not satisfy condition (2). Therefore, on the interval satisfying (2), $F(x)$ is monotonically decreasing, and it attains its maximum at the left endpoint of the interval, $x=r$.

Thus, we have found that if, in addition to condition (1), condition (6) is also satisfied, then the total illuminated surface area is maximized when $x=\frac{d}{1+\sqrt{\frac{R^{3}}{r^{3}}}}$. If, however, $r+R<d<r+R \sqrt{\frac{R}{r}}$, then the maximum is attained at $x=r$. In the latter case (and also if equality holds in (6)), the point $P$ will be on the smaller sphere, and the illuminated surface area on this sphere will be zero.

Specifically, if $r=R$, the total illuminated surface area is maximized when $x=\frac{d}{2}$.

Based on the work of Kovács Vera (Szolnok, Verseghy F, Gymnasium, 4th grade)