# Task 1. (10 points)

Six consecutive natural numbers from 10 to 15 are written in circles on the sides of a triangle in such a way that the sums of the three numbers on each side are equal.

What is the maximum value that this sum can take?

![](https://cdn.mathpix.com/cropped/2024_05_06_3fbb9a43f5593ef9bf58g-1.jpg?height=480&width=548&top_left_y=1002&top_left_x=1325)
# Solution.

Let $a, b, c, d, e, f$ be the given numbers, written in the order of their arrangement in the circles when traversed clockwise, and the numbers $a$, $c$, $e$ are located at the vertices of the triangle. If $S$ is the considered sum, then we have

$$
\left\{\begin{array}{l}
a+b+c=S \\
c+d+e=S \\
e+f+a=S
\end{array}\right.
$$

Adding all the equations of this system, we get,

$$
\begin{gathered}
(a+b+c+d+e+f)+a+c+e=3 S \Rightarrow \\
(10+11+12+13+14+15)+a+c+e=75+a+c+e=3 S \Rightarrow \\
S=25+\frac{a+c+e}{3}
\end{gathered}
$$

Therefore, the number $S$ cannot be greater than the number $25+\frac{15+14+13}{3}=39$.

The figure below shows that the number $S$ can be equal to 39.

![](https://cdn.mathpix.com/cropped/2024_05_06_3fbb9a43f5593ef9bf58g-2.jpg?height=537&width=617&top_left_y=291&top_left_x=774)

Answer: 39.