2017 AMC 10A Problem 23

There are $25$ lattice points, so $\binom{25}{3} = 2300$ ways to choose three of them. A choice fails to be a triangle exactly when the three points are collinear, so count collinear triples and subtract.

A line through the grid containing $k$ of the points contributes $\binom{k}{3}$ collinear triples. Go through the possible directions.

Horizontal: $5$ rows of $5$ points, $5 \cdot \binom{5}{3} = 50$. Vertical: likewise $50$.

Slope $\pm1$: for each sign there is one diagonal with $5$ points, two with $4$, and two with $3$ (shorter diagonals hold fewer than $3$ points). Per sign: $\binom{5}{3} + 2\binom{4}{3} + 2\binom{3}{3} = 10 + 8 + 2 = 20$. Both signs: $40$.

Slope $\pm2$ and $\pm\tfrac12$: a line of slope $\tfrac12$ through lattice points steps $2$ right and $1$ up, so it can hold at most $3$ grid points, e.g. $(1,1), (3,2), (5,3)$. For slope $\tfrac12$ the three points have $x = 1, 3, 5$ and $y = a, a+1, a+2$ with $a = 1, 2, 3$: $3$ lines. By symmetry (reflect and swap axes) each of the four slopes $\pm2, \pm\tfrac12$ gives $3$ lines, for $12$ collinear triples.

Any other slope (such as $\tfrac13$ or $\tfrac23$) would need a horizontal or vertical span of at least $6$ to fit three lattice points, impossible here.

Collinear triples: $50 + 50 + 40 + 12 = 152$. Triangles: $2300 - 152 = 2148$.

The answer is $\boxed{\textbf{(B)}\ 2148}$.

"Triangles with positive area" is "three points minus degenerate triples," so complementary counting is the natural frame. The hard part is exhaustively listing the lines with at least three lattice points; organizing by slope, and checking that a slope $p/q$ in lowest terms needs a span of $2q$ (and $2p$) to hold three points, guarantees nothing is missed.

• Forgetting the collinear triples on lines of slope 2 or 1/2 (12 of them), which gives 2160, choice (C).
• Counting only the two main diagonals and missing the shorter diagonals with $4$ or $3$ points.