How many distinct triangles with at least one side of length $\sqrt{2}$ units can be drawn using three lattice points for the vertices?

Question

This is supposed to be a middle school level math problem, I thought it should be ${}_9 C_3 - 8$ where 8 is the number of straight lines, but not sure. Thanks for any help!

Answer 1

The quantity $\binom93-8$ should give the total number of triangles, but ignores the restriction that one side must have length $\sqrt2$.

If we include that restriction, then:

• There are $8$ different segments of length $\sqrt2$ that can be formed using these lattice points, and $7$ ways in each case to choose a third vertex of the triangle, for $8\cdot7 = 56$ triangles.
• However, there are $8$ triangles with sides $\sqrt2, \sqrt2, 2$ that are counted twice, because they have two sides of length $\sqrt2$; we should subtract $8$ to account for these.
• Additionally, there are $2$ diagonals, which are degenerate triangles with side lengths $\sqrt2, \sqrt2, 2\sqrt2$. These are each counted twice and should not be counted at all, so we should subtract $4$ to account for these.

Altogether, we get $56 - 8 - 4 = 44$ triangles.