2019 AMC 10C Problems/Problem 24

Problem

Let $S$ be the set of points $(x,y,z)$ in the 3-dimensional coordinate plane such that $x$, $y$, and $z$ are integers, $0 \le x \le 2$, $0 \le y \le 2$, and $0 \le z \le 1$. How many tetrahedrons of positive volume can be formed by choosing four points in $S$ as vertices of the tetrahedron?


$\mathrm{(A) \ } 2436\qquad \mathrm{(B) \ } 2472\qquad \mathrm{(C) \ } 2580\qquad \mathrm{(D) \ } 2664\qquad \mathrm{(E) \ } 3060$

Solution

We solve this problem using some casework. Case 1: three points have the same z-coordinate and one point has a different z-coordinate: Trivially $2 \times (\dbinom{9}{3} - 8) \times 9$ ways (since there are $8$ sets of three points on a $3 \times 3$ grid that do not form a triangle with positive area) Case 2: two points that have the same z-coordinate and two points that have the other z-coordinate: There are $\dbinom{9}{2} \times \dbinom{9}{2}$ ways just to select the points, but the number of non-degenerate tetrahedrons is obviously less than that. Subcase 1: the four points are all on a vertical 3x2 rectangle parallel to the x axis or the y axis: $\dbinom{3}{2} \times \dbinom{3}{2} \times 6 = 54$ ways Subcase 2: the four points are all on the vertical 3x2 rectangle that is the diagonal of the figure: $\dbinom{3}{2} \times \dbinom{3}{2} \times 2 = 18$ ways Subcase 3: the four points are all on a vertical rectangle other than the rectangles already listed: 8+4=12 ways Subcase 4: the four points are on a slanted rectangle that forms a 45 degree angle with the "ground": $\dbinom{3}{2} \times \dbinom{3}{2} \times 8 = 72$ ways Subcase 5: the four points are on a slanted rectangle that is formed by opposite sides: $4 \times \dbinom{3}{2} \times \dbinom{3}{2} = 36$ ways Subcase 6: the four points are on the rectangle that has sides parallel to none of the axes (all 45 degrees): 4 ways Subcase 7: the four points are the vertices of a parallelogram with a 45 degree angle to all axes: 16 ways Subcase 8: the four points are the vertices of a trapezoid: 8 ways Subcase 9: the four points are on a "big parallelogram": 8 ways Our total number of ways is just the sum of Case 1 and Case 2 minus all the degenerate cases, which is $2436$.

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