Problem 7.2. Denis divided a triangle into nine smaller triangles, as shown in the figure, and placed numbers in them, with the numbers in the white triangles being equal to the sums of the numbers in the adjacent (by sides) gray triangles. After that, Lesha erased the numbers 1, 2, 3, 4, 5, and 6 and wrote the letters A, B, C, D, E, and F in some order. The resulting arrangement of numbers and letters is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_ff369b3e8ca7495bdf12g-20.jpg?height=303&width=355&top_left_y=874&top_left_x=549)

Where did the original numbers stand?
(a) Instead of the letter A
(1) the number 1 stood.
(b) Instead of the letter B
(2) the number 2 stood.
(c) Instead of the letter C
(3) the number 3 stood.
(d) Instead of the letter D
(4) the number 4 stood.
(e) Instead of the letter E
(5) the number 5 stood.
(f) Instead of the letter F
(6) the number 6 stood.
Answer: a1 b3 c2 d5 e6 f4.

Solution. Note that the number 6 can be uniquely represented as the sum of three numbers from the set of numbers from 1 to 6, which is $6=1+2+3$ (or the same numbers in a different order).

Now let's look at the numbers $B, D$, and $E$. The maximum value of the sum $D+E$ is the sum $5+6=11$, and the maximum value of $B$ is 3. According to the condition, the sum of these numbers is exactly 14. Thus, $B=3$, $D$ and $E$ are 5 and 6 (or vice versa), and $A$ and $C$ are 1 and 2 (or vice versa). By the method of elimination, we get that $F=4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_ff369b3e8ca7495bdf12g-21.jpg?height=319&width=357&top_left_y=83&top_left_x=546)

The maximum value of the sum $C+E$ is the sum $6+2=8$, and according to the condition, $12=C+E+4$. From this, we can uniquely restore the entire arrangement of numbers.

![](https://cdn.mathpix.com/cropped/2024_05_06_ff369b3e8ca7495bdf12g-21.jpg?height=313&width=357&top_left_y=518&top_left_x=548)