On a tower of height $b$, we see a rod of length $a$. At what distance from the tower do we see this rod at the largest angle?
![](https://cdn.mathpix.com/cropped/2024_05_02_8b9c89007d32d76ebd68g-1.jpg?height=504&width=498&top_left_y=122&top_left_x=800)

I. Solution. From the diagram, it is clear that

$$
\operatorname{tg} \delta=\operatorname{tg}(\varphi-\psi)=\frac{\operatorname{tg} \varphi-\operatorname{tg} \psi}{1+\operatorname{tg} \varphi \cdot \operatorname{tg} \psi}=\frac{\frac{x}{b}-\frac{x}{a+b}}{1+\frac{x^{2}}{b(a+b)}}=\frac{a x}{b(a+b)+x^{3}}=m
$$

Rearranging,

$$
m x^{2}-a x+b(a+b) m=0
$$

Thus,

$$
x=\frac{a \pm \sqrt{a^{2}-4 b(a+b) m^{2}}}{2 m}
$$

Since \( x \) represents a distance, it must be that

$$
a^{2} \geqq 4 b(a+b) m^{2}
$$

Therefore,

$$
m_{\max }=\frac{a}{2 \sqrt{b(a+b)}}
$$

Since \( m = \operatorname{tg} \delta \), the maximum value of \( m \) also corresponds to the maximum value of \( \delta \). Thus,

$$
x=\frac{a}{\frac{a}{\sqrt{b(a+b)}}}=\sqrt{b(a+b)}
$$

So \( x \) is the geometric mean of the actual height of the tower and the height increased by the rod.

(István Pivnyik, Nyíregyháza.)

II. Solution. Let \( D \) be the desired point. The angle \( \angle BDA \) is thus a peripheral angle of the circle centered at \( O \) passing through points \( A \), \( B \), and \( D \). This angle is larger when the radius of the circumcircle of triangle \( ABD \) is smaller, given a fixed length of the chord \( AB \).

The circle with the smallest radius passing through points \( A \) and \( B \) and having a common point with the line \( CD \) is the one that is tangent to \( CD \). The angle \( \delta \) is maximized in this circle. The radius of this circle is \( BO = OD = LC = b + \frac{a}{2} \). Therefore, the desired distance is

$$
CD = LO = \sqrt{\overline{BO}^{2} - \overline{BL}^{2}} = \sqrt{b^2 + ab} = \sqrt{b(a+b)}
$$

The point \( O \) is constructed by raising a perpendicular from the midpoint \( L \) of the line \( AB \) and intersecting it with a circle of radius \( b + \frac{a}{2} \) centered at \( B \).

(Imre Bartók, Budapest.)

The problem was also solved by: Baranyó A., Deutsch E., Deutsch I., Enyedi B., Haar A., Jánosy Gy., König D., Liebner A., Ligeti P., Losonczy I., Neidenbach E., Riesz K., Schwemmer I., Weisz P.