, 2019

Solution:

We can see that no two blue points can have the same $x$ or $y$ coordinate. The blue points then must make a permutation of $1,2,3,4,5$ that avoid the pattern of $3$-$4$-$5$ triangles. It is not hard to use complementary counting to get the answer from here.

There are $8$ possible pairs of points that are a distance of $5$ apart while not being in the same row or column (i.e. a pair that is in the $3$-$4$-$5$ position). If such a pair of points is included in the choice of five points, then there are $3! = 6$ possibilities for the remaining three points, yielding $8 \times 6 = 48$ configurations that have violations. However, we now need to consider overcounting.

The only way to have more than one violation in one configuration is to have a corner point and then two points adjacent to the opposite corner, e.g. $(1,1), (4,5), (5,4)$. In each such case, there are exactly $2! = 2$ possibilities for the other two points, and there are exactly two violations so there are a total of $2 \times 4 = 8$ configurations that are double-counted.

Therefore, there are $48 - 8 = 40$ permutations that violate the no-integer-condition, leaving $120 - 40 = 80$ good configurations.