As $X O Y$ is a right angle, we measure the distances $O A=a, O B=b(a<b)$ along the ray $O Y$. From point $M$ on line $O X$ $(O M=x)$, at what angle is $A B$ visible? For which value of $x$ will this angle be the largest?
Let $O M A \angle=\alpha, O M B \angle=\beta$ and $A M B \angle=\varphi$, Then

$$
\operatorname{tg} \varphi=\operatorname{tg}(\beta-\alpha)=\frac{\operatorname{tg} \beta-\operatorname{tg} \alpha}{1+\operatorname{tg} \alpha \operatorname{tg} \beta} \quad \text { and } \quad \operatorname{tg} \alpha=\frac{a}{x}, \quad \operatorname{tg} \beta=\frac{b}{x}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_6e59c74ef98ef6418ccbg-1.jpg?height=390&width=412&top_left_y=366&top_left_x=846)

Substituting these into the value of $\operatorname{tg} \varphi$:

$$
\operatorname{tg} \varphi=\frac{(b-a) x}{x^{2}+a b}
$$

Rearranging this equation in terms of $x$:

$$
\operatorname{tg} \varphi \cdot x^{2}-(b-a) x+a b \operatorname{tg} \varphi=0
$$

We get a real solution if

$$
(b-a)^{2}-4 a b \operatorname{tg}^{2} \varphi \geqq 0, \quad \text { i.e., } \quad 0<\operatorname{tg} \varphi \leqq \frac{b-a}{2 \sqrt{a b}}
$$

Accordingly, the maximum value of $\operatorname{tg} \varphi$ is: $\operatorname{tg} \varphi_{m}=\frac{b-a}{2 \sqrt{a b}}$

and the corresponding value of $x$ is: $\frac{b-a}{2 \operatorname{tg} \varphi_{m}}=\sqrt{a b}$.

Steiner Iván (Toldy Ferenc g. VI. o. Bp. II.)

I. Note. $\operatorname{tg} \varphi$ can be written as: $\operatorname{tg} \varphi=\frac{b-a}{x+\frac{a b}{x}}$.

The value of $\operatorname{tg} \varphi$ is the largest when the denominator is the smallest. The product of the two terms in the denominator is $a b$, a constant; the sum of the two terms is the smallest when they are equal, i.e., when $x=\frac{a b}{x}$, or $x^{2}=a b$.

T. i.

$$
\begin{aligned}
\left(x+\frac{a b}{x}\right)^{2} & =\left(x-\frac{a b}{x}\right)^{2}+4 a b \\
x & +\frac{a b}{x} \text { is the smallest when } x-\frac{a b}{x}=0, \text{ i.e., } x^{2}=a b
\end{aligned}
$$

II. Note. Based on the geometric meaning of this solution, we need to construct a circle that passes through points $A$ and $B$ and is tangent to the $O X$ line. For the point of tangency $M_{1}$,

$$
\overline{O M}_{1}^{2}=\overline{O A} \cdot \overline{O B} \quad \text { i.e., } \quad x^{2}=a b
$$

Construct this circle and take a point $M$ on the $O X$ line different from $M_{1}$. Now, the angle $A M B$ is an external excentric angle with respect to the circle, which is smaller than the inscribed angle $A M_{1} B$ (The measure of the latter is half the arc $\widehat{A B}$, the former is $\frac{1}{2}(\widehat{A B}-\widehat{K L})$.