We are approaching a 120-meter tall skyscraper on a horizontal road. After traveling 300 meters, we see the building at an elevation angle that is $45^{\circ}$ greater than at the start of the road. How close have we approached the skyscraper?
Let $x$ denote the sought distance, and $\alpha$ the angle of elevation measured at the beginning of the path.

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\text { 1993-05-215-1.eps }
$$

The following relationships hold:

$$
\begin{gathered}
\operatorname{tg} \alpha=\frac{120}{300+x} \\
\operatorname{tg}\left(\alpha+45^{\circ}\right)=\frac{120}{x}
\end{gathered}
$$

Based on the identity $\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}$, we have $\operatorname{tg}\left(\alpha+45^{\circ}\right)=\frac{\operatorname{tg} \alpha+1}{1-\operatorname{tg} \alpha}$, hence

$$
\frac{\frac{120}{300+x}+1}{1-\frac{120}{300+x}}=\frac{120}{x}
$$

This results in the equation $x^{2}+300 x-21600=0$, whose solutions are $x_{1}=60, x_{2}=-360$. In our case, $x_{1}$ is the appropriate solution. Therefore, we approached the flagpole at a distance of 60 meters.

Décsy Gabriella (Dombóvár, Illyés Gy. Gymn., III. o. t.)