## Task 4 - 340714

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0554.jpg?height=275&width=279&top_left_y=2144&top_left_x=363)

The numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 are to be entered into the circles of the drawing. The following condition must be met: In each of the four groups of three, which are connected by straight lines, the same sum \( s \) must be obtained.

a) What is the smallest value of \( s \) for which this condition can be met? Provide an entry where this sum \( s \) is present! Prove that no smaller sum is possible!

b) What is the largest value of \( s \) for which the stated condition can be met? Provide an entry for this as well!

c) Is the condition also met with all natural numbers \( s \) that lie between the smallest and largest values from a) and b)?
For each required entry of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9:

If one adds the four sums mentioned in the problem text (which are equal to each other), then the numbers \(a, b, c\) (see figure) are each used twice, and the other numbers are used once as summands. Therefore, we have

\[
4s = 1 + 2 + \ldots + 9 + a + b + c
\]

The sum \(s\) is thus the smallest or largest for the entries where \(a + b + c\) is the smallest or largest, respectively.

a) The smallest possible sum of three different numbers from 1, 2, ..., 9 is \(1 + 2 + 3 = 6\). If an entry with these numbers as \(a, b, c\) were possible, then from (1) we would get the equation \(4s = 45 + 6\), which is not solvable with an integer \(s\).

The next largest possibility \(1 + 2 + 4 = 7\) leads to \(4s = 45 + 7 = 52\), so \(s = 13\). As shown in Figure a, this value of \(s\) can be achieved by an entry. Thus, 13 is proven to be the smallest value of \(s\), and a corresponding entry is provided.

b) The largest possible sum of three different numbers from 1, 2, ..., 9 is \(7 + 8 + 9 = 24\). Again, an entry with these \(a, b, c\) is not possible, as (1) would lead to \(4s = 45 + 24\). The next smaller possibility \(6 + 8 + 9 = 23\) with \(4s = 45 + 23 = 68\), so \(s = 17\), is achievable by an entry; see, for example, Figure b.

c) Entries with \(s = 14\) and \(s = 16\) are also possible, as shown in Figures c and d. However, there is no entry with \(s = 15\).

For such an entry, according to (1), \(60 = 45 + a + b + c\) would have to hold, so \(a + b + c = 15\) would also have to hold. The number \(c\) would have to be placed in the field between \(a\) and \(b\) because \(s = 15\), which contradicts the requirement that the numbers to be entered are different.

a)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0555.jpg?height=291&width=303&top_left_y=1596&top_left_x=357)
b)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0555.jpg?height=300&width=309&top_left_y=1595&top_left_x=722)
c)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0555.jpg?height=294&width=323&top_left_y=1598&top_left_x=1072)
d)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0555.jpg?height=288&width=292&top_left_y=1598&top_left_x=1453)

Solutions of the 1st Stage 1994 adopted from [5]

# 4.36.2 2nd Round 1994, Class 7