Slicing a cube

The following picture shows a plane (in red) passing through the vertices $(10, 0, 0)$ , $(0, 10, 0)$ , and $(0, 0, 10)$ of a $10 \times 10 \times 10$ cube:

If the $10 \times 10 \times 10$ cube is broken up into 100 unit cubes as indicated in the problem, then the plane would intersect top front vertex of the following unit cubes:

These include the unit cubes $(7, 0, 0), (6, 1, 0), (5, 2, 0), ... (0, 7, 0)$ ; $(6, 0, 1), (5, 1, 1), (4, 2, 1), ... (0, 6, 1)$ ; $(5, 0, 2), (4, 1, 2), (3, 2, 2)$ ; ... $(0, 5, 2)$ , and so on until $(0, 0, 7)$ . (Algebraically, these are all the unit cubes in which $x_a + y_a + z_a = 7$ , since each of these unit cubes contain the coordinates $(x_a + 1, y_a + 1, z_a + 1)$ , and these coordinates are on the plane $x + y + z = 10$ because $(x_a + 1) + (y_a + 1) + (z_a + 1) = x_a + y_a + z_a + 3 = 7 + 3 = 10$ .) There are $8 + 7 + ... + 1 = \frac{8(8 + 1)}{2} = 36$ of them.

The plane would also intersect the unit cubes directly in front of those, which are as follows:

These include the unit cubes $(8, 0, 0), (7, 1, 0), (6, 2, 0), ... (0, 8, 0)$ ; $(7, 0, 1), (6, 1, 1), (5, 2, 1), ... (0, 7, 1)$ ; $(6, 0, 2), (5, 1, 2), (4, 2, 2)$ ; ... $(0, 6, 2)$ , and so on until $(0, 0, 8)$ . (Algebraically, these are all the unit cubes in which $x_a + y_a + z_a = 8$ , since each of these unit cubes contain the coordinates $(x_a + \frac{2}{3}, y_a + \frac{2}{3}, z_a + \frac{2}{3})$ , and these coordinates are on the plane $x + y + z = 10$ because $(x_a + \frac{2}{3}) + (y_a + \frac{2}{3}) + (z_a + \frac{2}{3}) = x_a + y_a + z_a + 2 = 8 + 2 = 10$ .) There are $9 + 8 + 7 + ... + 1 = \frac{9(9 + 1)}{2} = 45$ of them.

The plane would also intersect the unit cubes directly in front of those, which are as follows:

These include the unit cubes $(9, 0, 0), (8, 1, 0), (7, 2, 0), ... (0, 9, 0)$ ; $(8, 0, 1), (7, 1, 1), (6, 2, 1), ... (0, 8, 1)$ ; $(7, 0, 2), (6, 1, 2), (5, 2, 2)$ ; ... $(0, 7, 2)$ , and so on until $(0, 0, 9)$ . (Algebraically, these are all the unit cubes in which $x_a + y_a + z_a = 9$ , since each of these unit cubes contain the coordinates $(x_a + \frac{1}{3}, y_a + \frac{1}{3}, z_a + \frac{1}{3})$ , and these coordinates are on the plane $x + y + z = 10$ because $(x_a + \frac{1}{3}) + (y_a + \frac{1}{3}) + (z_a + \frac{1}{3}) = x_a + y_a + z_a + 2 = 8 + 1 = 10$ .) There are $10 + 9 + 8 + ... + 1 = \frac{10(10 + 1)}{2} = 55$ of them.

The plane would also intersect the bottom back vertex of the unit cubes directly in front of those, which are as follows:

These include the unit cubes $(10, 0, 0), (9, 1, 0), (8, 2, 0), ... (0, 10, 0)$ ; $(9, 0, 1), (8, 1, 1), (7, 2, 1), ... (0, 9, 1)$ ; $(8, 0, 2), (7, 1, 2), (6, 2, 2)$ ; ... $(0, 8, 2)$ , and so on until $(0, 0, 10)$ . (Algebraically, these are all the unit cubes in which $x_a + y_a + z_a = 10$ .) There are $9 + 10 + 9 + 8 + 7 + ... + 2 = 9 + \frac{10(10 + 1)}{2} - 1 = 63$ of them.

Therefore, there are a total of $36 + 45 + 55 + 63 = \boxed{199}$ unit cubes that intersect the plane $x + y + z = 10$ .

I had been checking this with the Z3 SMT Solver with the SBV package . I am getting 199 solutions. Here is my code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

import Data.SBV
import Control.Monad

checkCube a b c = do
  [x, y, z] <- mapM sReal ["x", "y", "z"]
  constrain $ x + y + z .== 10
  constrain $ x `inRange` (fromIntegral a, (fromIntegral a) + 1)
  constrain $ y `inRange` (fromIntegral b, (fromIntegral b) + 1)
  constrain $ z `inRange` (fromIntegral c, (fromIntegral c) + 1)

countCubes = filterM (\(x, y, z) -> isSatisfiable $ checkCube x y z) $ [ (x, y, z) | x <- [0..9], y <- [0..9], z <- [0..9]]

main :: IO ()
main = (length <$> countCubes) >>= print