A three-meter-high large painting hangs on the wall, its lower edge is one meter above the viewer's eye level. The view of the painting is more enjoyable the larger the vertical angle in which it is seen. At what distance from the wall will this angle be the largest?
I. solution. The centers of the viewing circles corresponding to the given angles in the picture are 1 + 3/2 = 2.5 meters above the viewer's eye level. The viewing angle is larger the smaller the radius of the viewing circle (Figure 1). Therefore, we need to find the viewing circle with the smallest radius that has a point at eye level. This is clearly the circle with a radius of 2.5 meters. In the right-angled triangle $OAF$ shown in Figure 2, the leg $OF$ is, according to the Pythagorean theorem, $OF = \sqrt{2.5^2 - 1.5^2} = 2$ meters. This is equal to the distance $ST$, because the quadrilateral $STFO$ is a rectangle.

Therefore, the viewing angle of the picture will be the largest from a distance of 2 meters.

Based on the work of Baur Eszter (Békéscsaba, Rózsa F. Gymn., 11th grade)

II. solution. Let the distance from the point $S$ at eye level to the intersection point $T$ of the line $f$ containing the segment $AB$ representing the picture and the eye level be $x$ (Figure 3). Then $BA = 3$ and $AT = 1$, so if $\angle BSA = \alpha$ and $\angle AST = \beta$, then $\tan(\alpha + \beta) = \frac{4}{x}$ and $\tan \beta = \frac{1}{x}$.

Since $\alpha$ is clearly an acute angle, it will be the largest when its tangent is maximal. According to the tangent addition formula,

$$
\tan \alpha = \tan((\alpha + \beta) - \beta) = \frac{\tan(\alpha + \beta) - \tan \beta}{1 + \tan(\alpha + \beta) \cdot \tan \beta} = \frac{\frac{4}{x} - \frac{1}{x}}{1 + \frac{4}{x^2}} = \frac{3}{x + \frac{4}{x}}
$$

This will be the largest when $x + \frac{4}{x}$ is the smallest. According to the arithmetic and geometric mean inequality,

$$
x + \frac{4}{x} \geq 2 \cdot \sqrt{x \cdot \frac{4}{x}} = 4
$$

and equality holds if and only if $x = \frac{4}{x}$, i.e., if $x = 2$ (we know that $x > 0$).

Therefore, the viewing angle of the picture will be the largest when viewed from a distance of 2 meters.

Fehér Ádám (Győr, Révai M. Gymn., 10th grade)
![](https://cdn.mathpix.com/cropped/2024_05_02_fd63e0440134572be551g-1.jpg?height=1164&width=694&top_left_y=1286&top_left_x=694)

![](https://cdn.mathpix.com/cropped/2024_05_02_fd63e0440134572be551g-2.jpg?height=455&width=293&top_left_y=152&top_left_x=908)