Question

use reasoning to solve a problem. example 4. put the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 in the circles so that each edge adds up to the same number.

Explanation:

Step1: Calculate the sum of 1 - 9

The sum of an arithmetic - series \(S_n=\frac{n(a_1 + a_n)}{2}\), where \(n = 9\), \(a_1=1\), \(a_n = 9\). So \(S=\frac{9\times(1 + 9)}{2}=45\).

Step2: Analyze the structure of the figure

Assume the sum of each edge is \(x\). Since the figure has 3 edges and the numbers at the vertices are counted twice. Let the sum of the numbers at the vertices be \(A\) and the sum of the non - vertex numbers be \(B\), and \(A + B=45\). The sum of the three - edge sums is \(3x\), and \(3x=45 + A\).

Step3: Try combinations

The numbers at the vertices should be chosen such that \(45+A\) is divisible by 3. By trial and error:
If the vertices are 1, 5, 9:
The sum of the vertices \(A=1 + 5+9 = 15\). Then \(3x=45 + 15=60\), so \(x = 20\).
For one edge with vertices 1 and 5, we can find the middle number: \(20-(1 + 5)=14\), and we can use 6 and 8 to get \(1+6 + 8+5=20\).
For the edge with vertices 5 and 9, we can use 2 and 4 to get \(5+2 + 4+9=20\).
For the edge with vertices 1 and 9, we can use 3 and 7 to get \(1+3 + 7+9=20\).
One possible solution is:
Vertices (clock - wise or counter - clockwise) 1, 5, 9. Between 1 and 5: 6, 8; between 5 and 9: 2, 4; between 9 and 1: 3, 7.

Answer:

One possible arrangement: Vertices are 1, 5, 9. On the edge between 1 and 5: 6, 8; on the edge between 5 and 9: 2, 4; on the edge between 9 and 1: 3, 7 (There are other valid arrangements as well).