2019 AMC 10C Problems/Problem 24
Let be the set of points in the 3-dimensional coordinate plane such that , , and are integers, , , and . How many tetrahedrons of positive volume can be formed by choosing four points in as vertices of the tetrahedron?
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 ways (since there are sets of three points on a 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 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: ways Subcase 2: the four points are all on the vertical 3x2 rectangle that is the diagonal of the figure: 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": ways Subcase 5: the four points are on a slanted rectangle that is formed by opposite sides: 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 .