## An Intersecting Square and Triangle

Draw a square and triangle such that each side of either figure intersects at least one side of the other figure. There are two different ways to do this.

Now place unique positive integers at the intersection points and the vertices such that the sum of the numbers on any of the seven straight lines is the same. Try to minimize that sum.

Source: Original.

One solution to one configuration was received from Philippe Fondanaiche:
Let's call ABCD the square and EFG the triangle. The maximum number of intersection points is obtained, for example, with the following drawing:
EF cuts AD at H and CD at I
FG cuts CD at J and BC at K
EG cuts AB at L and AD at M
Globally there are 13 points on 7 straight lines. With a sum per line of 26 which is the minimum, we obtain the following valuation of each point included between 1 and 13. For example, among many other solutions:
A=7 B=13 C=9 D=5 E=1 F=12 G=8 H=3 I=10 J=2 K=4 L=6 M=11
The second configuration was solved by Richard Winterstein:
```Let's call ABCD the square and EFG the triangle. The minimum number of
intersection points is obtained, for example, with the following drawing:
EF cuts AB at H and AD at I
FG cuts CD at J
EG cuts BC at K
(Note that for the min case EFG has one point inside the square, for the max
case no points lie inside the square.)
Globally there are 11 points on 7 straight lines. With a sum per line of 20
which is the minimum, we obtain the following valuation of each point. For
example, among many other solutions:
A=12 B=7 C=5 D=2 E=9 F=4 G=3 H=1 I=6 J=13 K=8
The numbers 10 and 11 are not used.  There doesn't appear to be any way to
satisfy the second condition of the problem using only the numbers 1-11.  13
is the smallest maximum number of all possible solution sets.
```

KD: Actually, the second figure can be numbered with the sequential numbers 2-12. This gives a common sum of 22, so it isn't as small as Richard's above, but it does reduce the smallest maximum number to 12. With the same labeling as the last solution, there are only eight solutions. In all of them, C=11. Here are two of them:
A=12 B=4 C=11 D=2 E=5 F=3 G=10 H=6 I=8 J=9 K=7
A=10 B=5 C=11 D=3 E=4 F=2 G=12 H=7 I=9 J=8 K=6
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