## Task B-2.6.

At a height of $x$ meters above the stage, a spotlight is directed vertically down onto the stage, illuminating a circle of area $P \mathrm{~m}^{2}$. If we raise the spotlight by 1 meter, the illuminated area would increase by $2.25 \mathrm{~m}^{2}$. If we lower the spotlight by 1 meter, the illuminated area would decrease by $1.75 \mathrm{~m}^{2}$ (relative to the original area). At what height above the stage is the spotlight located?
## First Solution.

A beam of light that emerges from a spotlight and illuminates a circle on the stage defines a cone.
![](https://cdn.mathpix.com/cropped/2024_05_30_63d9d2269da4cabd8f72g-11.jpg?height=566&width=1262&top_left_y=1536&top_left_x=293)

The bases of these cones are concentric circles, and the vertices (depending on the position of the spotlight) are located on a line perpendicular to the stage.

The characteristic right triangles of these cones are similar to each other. Therefore, the radii of the concentric circles are in the same ratio as the corresponding heights, and their areas are in the ratio of the squares of these heights.

We write the following system of equations:

$$
\left\{\begin{array}{l}
\left(\frac{x+1}{x}\right)^{2}=\frac{P+2.25}{P} \\
\left(\frac{x-1}{x}\right)^{2}=\frac{P-1.75}{P}
\end{array}\right.
$$

By multiplying by the denominators, we get

$$
\left\{\begin{array}{l}
P x^{2}+2 P x+P=P x^{2}+2.25 x^{2} \\
P x^{2}-2 P x+P=P x^{2}-1.75 x^{2}
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
2 P x+P=2.25 x^{2} \Longrightarrow P=\frac{2.25 x^{2}}{2 x+1} \\
2 P x-P=1.75 x^{2} \Longrightarrow P=\frac{1.75 x^{2}}{2 x-1}
\end{array}\right.
$$

By equating the expressions for \( P \), we get

$$
\frac{2.25 x^{2}}{2 x+1}=\frac{1.75 x^{2}}{2 x-1} /: x^{2} \quad(x \neq 0)
$$

We obtain \( 4.5 x - 3.5 x = 2.25 + 1.75 \).

We conclude that the spotlight is located at a height of \( x = 4 \) meters above the stage.