Magic Triangle

Suppose this triangle could be completed successfully, and let $x$ be the sum of the numbers on each line. As there are a total of 7 'lines', (one of which is a circle), the sum of all 7 'line sums' is $7x$ . As each of the 7 boxes is included in 3 line sums, and as within each box is a distinct digit from $1$ through $7$ , we then require that

$7x = 3 \times (1 + 2 + 3 + 4 + 5 + 6 + 7) \Longrightarrow 7x = 3 \times 28 \Longrightarrow x = 12$ .

But using the digits from $1$ through $7$ taken three at a time, we can only achieve a sum of $12$ in 5 ways, namely

$(1,5,6), (1,4,7), (2,3,7), (2,4,6)$ and $(3,4,5)$ .

Since we require 7 different ways to achieve a sum of $12$ , we must conclude that such a triangle cannot be completed successfully.

[When I first created this problem, I thought it was possible to do so. I was surprised when I couldn't arrange the digits.]

The sum of all 7 numbers is $1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$ .
Each number appears in 3 colored lines, hence if each of the 7 colored lines have the same sum, then it would be equal to $\frac{ 28 \times 3 } { 7 } = 12$ .
Now, consider where the 7 goes. The 3 lines that go through it will cover all the other 6 numbers. Pick the line which also has 6. Then, this line would have a sum of at least $7 + 6 + 1 = 14 > 12$ .

So it is not possible for all of the colored lines to have the same sum.

Note that this figure has the interesting property that every point is intersected by exactly three paths, and every path intersects exactly three points. (This figure has many interesting properties; it is called the Fano Plane .)

Realize that if we take the three paths that a single point $P$ shares, the sum of the two numbers that are not on $P$ in each path have to be equal to one another. Thus, the sum of these $6$ other numbers has to be a multiple of three. Thus, if $x$ is the number on $P$ , then $1+2+3+4+5+6+7-x$ has to be a multiple of three, which clearly is not satisfied for $x=2, 3, 5, 6$ . Thus, there is no way to place the numbers.

Let's index the order of box from top to bottom and left to right as from 1 to 7. From segment 1-7 there are 4 odd and 3 even numbers, you can easily figure that the box 1,4,5,7 holding odd number to make sure that every paths in the graph make up the same sum. Now all the boxes in graph are in symmetric position, generally we put in 2 to box 2, 4 to box 3 and 6 to box 6. As all the paths make the same sum we have:

$x + y + 2 = z + y + 6 = x + z + 4\\ \begin{cases} x = z + 4 \\ x = y + 2 \end{cases}$

Possible $(x,y,z) = {(7,5,3), (5,3,1)}$ but the light blue and orange path isnt in compliance with the problem.