According to a non-detailed calculation and interpretation, if we take the viewing angle of the hypotenuse (in radians) from the interior points of an isosceles right triangle, the average of these angles is $\pi-\log 2$. As a rough approximation, we think of the right-angled coordinate system with the grid triangle $(0 ; 0),(6 ; 0),(0 ; 6)$ and ask: what is the average viewing angle in these 10 grid points?

The task is a clever, simple organization and execution of this calculation.
Let $P$ be an internal point of the triangle $OAB$, and let $PX'$ and $PY'$ be the rays through $P$ parallel to the positive $x$ and $y$ axes, respectively (Figure 1a).

![](https://cdn.mathpix.com/cropped/2024_05_02_05927bb2f53ab1b4263ag-1.jpg?height=466&width=415&top_left_y=385&top_left_x=497)
a)

![](https://cdn.mathpix.com/cropped/2024_05_02_05927bb2f53ab1b4263ag-1.jpg?height=694&width=680&top_left_y=263&top_left_x=923)

Then the angle of view as the difference in direction angles of the rays $PB$ and $PA$ is

$$
\angle APB = \angle X'PB - \angle X'PA = \frac{\pi}{2} + \angle Y'PB + \angle APX',
$$

and we need to calculate the average $\varphi$ of this for the grid points $P_1, P_2, \ldots, P_{10}$ of our triangle (Figure 1b):

$$
\varphi = \frac{\pi}{2} + \frac{1}{10} \sum_{i=1}^{10} \left( \angle Y'P_iB + \angle AP_iX' \right) = \frac{\pi}{2} + \frac{2}{10} \sum_{i=1}^{10} \angle AP_iX'
$$

indeed, our shape (the triangle and the set of grid points within it) is symmetric with respect to the line $y=x$, based on which the sums of the two types of terms in the parentheses are equal.

It is convenient to immediately write the angles in radians based on their tangents:

| $i$ | $\operatorname{tg} \angle AP_iX' = t_i$ | $\operatorname{arc} \operatorname{tg} t_i$ | Average |
| ---: | :--- | :--- | :--- |
| 1 | $1 / 2 = 0.5$ | 0.4636 | 0.4636 |
|  |  |  |  |
| 2 | $1 / 3 = 0.3333$ | 0.3217 |  |
| 3 | $2 / 3 = 0.6667$ | 0.5880 | 0.4578 |
|  |  |  |  |
| 4 | $1 / 4 = 0.25$ | 0.2450 |  |
| 5 | $2 / 4 = 0.5$ | 0.4636 |  |
| 6 | $3 / 4 = 0.75$ | 0.6435 | 0.4542 |
|  |  |  |  |
| 7 | $1 / 5 = 0.2$ | 0.1974 |  |
| 8 | $2 / 5 = 0.4$ | 0.3805 |  |
| 9 | $3 / 5 = 0.6$ | 0.5404 |  |
| 10 | $4 / 5 = 0.8$ | 0.6747 | 0.4518 |

The last column of the table shows the average of the radian values obtained up to the boundaries $i=1,3,6,10$. The last of these relates to the original question of the problem, while the first three relate to the corresponding results for $OA=3,4$, and 5. Thus (rounded to three decimal places):

$$
\begin{aligned}
\varphi & = (\varphi_6 =) \frac{\pi}{2} + 2 \cdot 0.4518 = 2.474 \\
\varphi_3 & = 2.498, \quad \varphi_4 = 2.486, \quad \varphi_5 = 2.479
\end{aligned}
$$

while the given value: $\pi - \log 2 = 3.1416 - 0.6931 = 2.4484 \ldots$, is exceeded by $\varphi_6$ by just over $1\%$.

Remark. The averages corresponding to the points $P_1$; $P_2$ and $P_3$; $P_4$ and $P_5$ and $P_6$; and $P_7, P_8, P_9, P_{10}$, respectively, provide individual (usual) approximations to the

$$
\int_{0}^{1} \operatorname{arctg} x \, dx
$$

integral. These are not quite regular upper approximations, since the function value at the last division point $x=1$ does not appear in the average. We can say it is by chance that we still get an upper approximation. In any case, if we refine the partition, we will get the value of the said integral more and more accurately, so the weighted average of these approximate values also tends to the value of the integral. According to the above calculation, the upper approximate values of this integral and the lower approximate values obtained by the trapezoidal rule are summarized in the following table:

| $n$ | Upper approximation | Lower approximation |
| :---: | :---: | :---: |
| 2 | 0.4636 | 0.4282 |
| 3 | 0.4548 | 0.4341 |
| 4 | 0.4507 | 0.4362 |
| 5 | 0.4482 | 0.4371 |

If we know that the

$$
F(x) = x \operatorname{arctg} x - \frac{1}{2} \log \left(1 + x^2\right)
$$

function's derivative is exactly $\operatorname{arctg} x$ (the verification of which requires the known differentiation rules and only the derivatives

$$
(\operatorname{arctg} x)' = \frac{1}{1 + x^2}, \quad (\log x)' = \frac{1}{x}
$$

known from the school table), we can also determine the exact value of the above integral according to the Newton-Leibniz formula:

$$
\int_{0}^{1} \operatorname{arctg} x \, dx = F(1) - F(0) = \operatorname{arctg} 1 - \frac{1}{2} \log 2 = \frac{\pi - \log 4}{4} = 0.4388
$$

The average angle of view in the problem is obtained according to the above solution by adding the angle $\pi / 2$ to twice this, so we indeed get the number $(\pi - \log 2)$.