5. The words in the circles in the picture should be replaced with the numbers 1, 2, 3, 4, 5, and 6 so that $a+b+c=c+d+e=a+f+e$ holds. All numbers must be used and must not be repeated. What can the sum $a+b+c$ be? For each possible sum, show one corresponding arrangement of the numbers in the picture.

![](https://cdn.mathpix.com/cropped/2024_06_03_93b46e1519bfbf998b64g-1.jpg?height=414&width=465&top_left_y=2203&top_left_x=841)

## ZUPANIJSKO NATJECANJE IZ MATEMATIKE
5. The words in the circles in the picture should be replaced by the numbers 1, 2, 3, 4, 5, and 6 so that $a+b+c=c+d+e=a+f+e$. All numbers must be used and must not be repeated. What can the sum $a+b+c$ be? For each possible sum, show one corresponding arrangement of the numbers in the picture.

![](https://cdn.mathpix.com/cropped/2024_05_30_5cfeb7f7e539b150a939g-05.jpg?height=425&width=468&top_left_y=387&top_left_x=840)

## Solution.

The sum of all the given numbers is $1+2+3+4+5+6=21$.

From $a+b+c=c+d+e=a+f+e$ we conclude that

$(a+b+c)+(c+d+e)+(a+f+e)=3 \cdot(a+b+c)$.

In the sum on the left side, the numbers $a$, $c$, and $e$ appear twice as addends (these are the numbers at the vertices of the triangle).

Therefore, $3 \cdot(a+b+c)=(a+b+c+d+e+f)+(a+c+e)$,

That is, $3 \cdot(a+b+c)=21+(a+c+e)$,

The sum of the numbers at the vertices $a+c+e$ can be as low as $1+2+3=6$, and as high as $4+5+6=15$.

$1 \text{ POINT}$

From $3 \cdot(a+b+c)=21+6=27$ it follows that the smallest sum $a+b+c=9$,

while from $3 \cdot(a+b+c)=21+15=36$ it follows that the smallest sum $a+b+c=12$.

Thus, the sum $a+b+c$ cannot be less than 9 or greater than 12.

For the sum $a+b+c$ there are 4 different solutions (4 different sums). These are 9, 10, 11, and 12. 2 POINTS
![](https://cdn.mathpix.com/cropped/2024_05_30_5cfeb7f7e539b150a939g-05.jpg?height=734&width=922&top_left_y=1618&top_left_x=540)

Note: If the student does not investigate whether there are other solutions besides these 4, they cannot get full points.

## COUNTY COMPETITION IN MATHEMATICS