Let's calculate the following fraction without using a calculator or function tables:

$$
\frac{\pi-\operatorname{arctg} \frac{8}{15}}{2 \cdot \operatorname{arctg} 4}
$$

(arc tg $x$ represents the angle in the interval $\left(-\frac{\pi}{2} ; \frac{\pi}{2}\right)$ whose tangent is $x$.)
I. solution. Let's draw a right-angled triangle with legs of 8 and 15 units, the acute angle opposite the 8-unit leg is $\alpha$, the other acute angle is $\beta$, and the hypotenuse $AB = \sqrt{8^2 + 15^2} = 17$. Rotate the hypotenuse $AB$ onto the ray $CA$. In the isosceles triangle $ABD$, $AD = 17$, $\angle BDA = \angle ABD = \frac{\alpha}{2}$.

From the triangle $ABC$, $\tan \alpha = \frac{8}{15}$, since $\arctan \frac{8}{15} = \alpha$.

From the triangle $BCD$,

$$
\tan\left(\frac{\alpha}{2} + \beta\right) = \frac{32}{8} = 4, \quad \text{hence} \quad \arctan 4 = \frac{\alpha}{2} + \beta
$$

Since

$$
\begin{aligned}
\beta & = \frac{\pi}{2} - \alpha \\
\arctan 4 = \frac{\alpha}{2} + \beta & = \frac{\alpha}{2} + \frac{\pi}{2} - \alpha = \frac{\pi}{2} - \frac{\alpha}{2}
\end{aligned}
$$

Thus,

$$
2 \arctan 4 = \pi - \alpha = \pi - \arctan \frac{8}{15}
$$

Substituting this into the value of the ratio:

$$
\frac{\pi - \arctan \frac{8}{15}}{\pi - \arctan \frac{8}{15}} = 1
$$

Gábor Egyed, (Szombathely, Orlay F. Károly Secondary School, 3rd grade)

II. solution. Let $\arctan \frac{8}{15} = \alpha$, $\arctan 4 = \beta$. Then $\tan \alpha = \frac{8}{15}$ and $\tan \beta = 4$.

Write the known relationship for $\tan 2\beta$:

$$
\tan 2\beta = \frac{2 \tan \beta}{1 - \tan^2 \beta} = \frac{8}{1 - 4^2} = -\frac{8}{15} = -\tan \alpha
$$

Since $-\tan \alpha = \tan(\pi - \alpha)$, $2\beta = \pi - \alpha$ (where $0 < \alpha, \beta < \frac{\pi}{2}$). Substituting the values of $\alpha$ and $\beta$,

$$
2 \arctan 4 = 2\beta = \pi - \alpha = \pi - \arctan \frac{8}{15}
$$

Thus, the value of the ratio is 1.

Remark. In the solution, we used the fact that $\pi - \alpha$ and $2\beta$ fall into the interval $(0, \pi)$, otherwise it would not be true that if $\tan(\pi - \alpha) = \tan 2\beta$, then $\pi - \alpha = 2\beta$.

![](https://cdn.mathpix.com/cropped/2024_05_02_dcf4a228cf4589b3ee7ag-1.jpg?height=249&width=699&top_left_y=1694&top_left_x=710)