Question: A regular tetrahedron has three of its vertices at $A = (1, 1, 1)$, $B = (1, -1, -1)$, and $C = (-1, 1, -1)$. Find the integer coordinates of the fourth vertex $D$, assuming all coordinates are integers.

["Understanding the Regular Tetrahedron: Finding the Fourth Vertex with Integer Coordinates", "A regular tetrahedron is a three-dimensional shape with four equilateral triangular faces, six equal edges, and four vertices. While visualizing such symmetry helps, verifying geometric properties using coordinates allows precise determination of missing points—especially when integer coordinates are required. In this article, we explore how to find the fourth vertex $D$ of a regular tetrahedron given three vertices $A = (1, 1, 1)$, $B = (1, -1, -1)$, and $C = (-1, 1, -1)$, with the constraint that D must have integer coordinates.", "---", "### Geometry of the Problem", "Given:\n- $A = (1, 1, 1)$\n- $B = (1, -1, -1)$\n- $C = (-1, 1, -1)$", "We seek point $D = (x, y, z)$ such that the distances from $D$ to $A$, $B$, and $C$ are equal to the common edge length $s$, and also equal to $AB = BC = CA$.", "First, compute the squared distances between the known points:", "Distance $AB^2$:\n$$\n(1-1)^2 + (1 - (-1))^2 + (1 - (-1))^2 = 0 + 4 + 4 = 8\n\Rightarrow AB = \sqrt{8} = 2\sqrt{2}\n$$", "Distance $AC^2$:\n$$\n(1 - (-1))^2 + (1 - 1)^2 + (1 - (-1))^2 = 4 + 0 + 4 = 8\n$$", "Distance $BC^2$:\n$$\n(1 - (-1))^2 + (-1 - 1)^2 + (-1 - (-1))^2 = 4 + 4 + 0 = 8\n$$", "All edges are $ \sqrt{8} $, confirming triangle $ABC$ is equilateral with side length $ \sqrt{8} $. We now require $D$ such that $DA^2 = DB^2 = DC^2 = 8$ and $D$ has integer coordinates.", "---", "### Setting up Equations", "Let $D = (x, y, z) \in \mathbb{Z}^3$. Then:", "1. $DA^2 = (x - 1)^2 + (y - 1)^2 + (z - 1)^2 = 8$\n2. $DB^2 = (x - 1)^2 + (y + 1)^2 + (z + 1)^2 = 8$\n3. $DC^2 = (x + 1)^2 + (y - 1)^2 + (z + 1)^2 = 8$", "Expand each equation:", "Equation (1):\n$$\n(x - 1)^2 + (y - 1)^2 + (z - 1)^2 = x^2 - 2x + 1 + y^2 - 2y + 1 + z^2 - 2z + 1 = 8\n\Rightarrow x^2 + y^2 + z^2 - 2x - 2y - 2z + 3 = 8\n\Rightarrow x^2 + y^2 + z^2 - 2x - 2y - 2z = 5 \quad (1)\n$$", "Equation (2):\n$$\n(x - 1)^2 + (y + 1)^2 + (z + 1)^2 = x^2 - 2x + 1 + y^2 + 2y + 1 + z^2 + 2z + 1 = 8\n\Rightarrow x^2 + y^2 + z^2 - 2x + 2y + 2z + 3 = 8\n\Rightarrow x^2 + y^2 + z^2 - 2x + 2y + 2z = 5 \quad (2)\n$$", "Equation (3):\n$$\n(x + 1)^2 + (y - 1)^2 + (z + 1)^2 = x^2 + 2x + 1 + y^2 - 2y + 1 + z^2 + 2z + 1 = 8\n\Rightarrow x^2 + y^2 + z^2 + 2x - 2y + 2z + 3 = 8\n\Rightarrow x^2 + y^2 + z^2 + 2x - 2y + 2z = 5 \quad (3)\n$$", "Now subtract equations to eliminate the quadratic term.", "Subtract (1) from (2):\n$$\n[ -2x + 2y + 2z ] - [ -2x - 2y - 2z ] = 5 - 5\n\Rightarrow ( -2x + 2y + 2z ) - ( -2x - 2y - 2z ) = 0\n\Rightarrow 4y + 4z = 0 \Rightarrow y + z = 0 \quad (a)\n$$", "Subtract (2) from (3):\n$$\n[ +2x - 2y + 2z ] - [ -2x + 2y + 2z ] = 0\n\Rightarrow 4x - 4y = 0 \Rightarrow x = y \quad (b)\n$$", "From (a) and (b): $z = -y = -x$. So:\n$$\ny = x, \quad z = -x\n\Rightarrow D = (x, x, -x)\n$$", "Now substitute into equation (1):", "$$\nx^2 + x^2 + (-x)^2 - 2x - 2x - 2(-x) = 5\n\Rightarrow 3x^2 - 4x + 2x = 5\n\Rightarrow 3x^2 - 2x = 5\n\Rightarrow 3x^2 - 2x - 5 = 0\n$$", "Solve the quadratic:\n$$\nx = \frac{2 \pm \sqrt{(-2)^2 + 4 \cdot 3 \cdot 5}}{2 \cdot 3} = \frac{2 \pm \sqrt{4 + 60}}{6} = \frac{2 \pm \sqrt{64}}{6} = \frac{2 \pm 8}{6}\n$$", "So:\n- $x = \frac{10}{6} = \frac{5}{3}$ → not integer\n- $x = \frac{-6}{6} = -1$ → integer", "Thus, $x = -1$, so $y = -1$, $z = -(-1) = 1$", "Hence, $D = (-1, -1, 1)$", "---", "### Verification", "Compute distances from $D = (-1, -1, 1)$ to $A$, $B$, $C$:", "To $A = (1,1,1)$:\n$$\n(-1-1)^2 + (-1-1)^2 + (1-1)^2 = 4 + 4 + 0 = 8 \Rightarrow \sqrt{8}\n$$", "To $B = (1,-1,-1)$:\n$$\n(-1-1)^2 + (-1+1)^2 + (1+1)^2 = 4 + 0 + 4 = 8\n$$", "To $C = (-1,1,-1)$:\n$$\n(-1+1)^2 + (-1-1)^2 + (1+1)^2 = 0 + 4 + 4 = 8\n$$", "All match. $D$ has integer coordinates.", "---", "### Why This is Special: Integer Symmetry in 3D Space", "The tetrahedron formed by $ (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1) $ exhibits high symmetry—all coordinates are integers with values in $ {-1, 1 } $, forming a regular tetrahedron centered at the origin. This configuration is rare and represents one of the few regular tetrahedra embedded in $\mathbb{Z}^3$ with integer vertices.", "This problem illustrates a powerful technique: using vector algebra and symmetry constraints to solve geometric problems with integer solutions, commonly used in discrete geometry, crystallography, and computational geometry.", "---", "Conclusion\nThe only integer-coordinate point $D$ forming a regular tetrahedron with $A$, $B$, and $C$ is $(-1, -1, 1)$. This solution demonstrates how algebraic constraints on distances and symmetry yield precise, unique results—especially within the discrete lattice of integer coordinates.", "---", "Key Terms for SEO:\nRegular tetrahedron, integer coordinates, 3D geometry, distance formula, symmetry in 3D, dihedral angles, lattice points, vector geometry, symmetric polyhedra, integer solutions in Euclidean space", "---", "For further exploration, consider how such integer-vertex regular tetrahedra generalize or how symmetries relate to group theory—the foundations of crystallography in mathematics and materials science."]