Konyagin S. $\mathbf{\text { K }}$.

In a family album, there are ten photographs. In each of them, three people are depicted: a man stands in the center, to the left of the man is his son, and to the right is his brother. What is the smallest number of different people who can be depicted in these photographs, given that all ten men standing in the center are different?
Let's call the ten men standing in the center of the photographs the main faces. We will divide all men in the photographs into levels. We will assign level 0 to those who do not have fathers in the photographs, and level $k+1$ ($k=0,1$, $2, \ldots$) to men who have fathers at level $k$. Let $r_{k}$ denote the number of main faces, and $t_{k}$ denote the number of all other men at level $k$. The number of fathers of men at level $k+1$ is no more than $1 / 2 r_{k+1} + t_{k+1}$, since each main face has a brother. At the same time, the number of fathers of men at level $k+1$ is no less than $r_{k}$, since each main face has a son. Therefore, $r_{k} \leq 1 / 2 r_{k+1} + t_{k+1}$. We also note that $1 \leq 1 / 2 r_{0} + t_{0}$. Adding all these inequalities, we get $1 / 2 (r_{0} + r_{1} + \ldots) + (t_{0} + t_{1} + \ldots) \geq (r_{0} + r_{1} + \ldots) + 1$, from which $(r_{0} + r_{1} + \ldots) + (t_{0} + t_{1} + \ldots) \geq 3 / 2 (r_{0} + r_{1} + \ldots) + 1 = 3 / 2 \cdot 10 + 1 = 16$.

Thus, there are at least 16 men in the photographs. The diagram below shows 16 men with 10 main faces (numbered from 1 to 10).

![](https://cdn.mathpix.com/cropped/2024_05_06_d1d92653a0a4b11fb209g-09.jpg?height=757&width=1770&top_left_y=2034&top_left_x=127)

Horizontal lines connect brothers, and the other lines lead (from top to bottom) from father to son. Photographs: $(3,1,2),(5,2,1),(7,3,4),(9,4,3)$

## Answer

16 people.