Question: A regular tetrahedron has three of its vertices at $(0,0,0)$, $(1,1,0)$, and $(1,0,1)$. Find the coordinates of the fourth vertex, assuming all coordinates are integers.

["Finding the Fourth Vertex of a Regular Tetrahedron: A Geometric Solution", "When working with regular tetrahedrons in three-dimensional space, the challenge often lies in determining unknown vertex coordinates based on symmetry and equal edge lengths. This article explores a precise method to find the fourth vertex of a regular tetrahedron given three vertices at $(0,0,0)$, $(1,1,0)$, and $(1,0,1)$, with the condition that all coordinates are integers.", "### Understanding the Regular Tetrahedron", "A regular tetrahedron is a polyhedron with four equilateral triangular faces, four vertices, and six equal edge lengths. Given three vertices $A(0,0,0)$, $B(1,1,0)$, and $C(1,0,1)$, we seek point $D(x,y,z)$ such that the distances $AB = BC = CA = AD = BD = CD$, and $D$ has integer coordinates.", "### Step 1: Compute the Common Edge Length", "First, compute the squared distance between known points to determine the edge length squared:", "- $AB^2 = (1-0)^2 + (1-0)^2 + (0-0)^2 = 1 + 1 = 2$\n- $AC^2 = (1-0)^2 + (0-0)^2 + (1-0)^2 = 1 + 1 = 2$\n- $BC^2 = (1-1)^2 + (0-1)^2 + (1-0)^2 = 0 + 1 + 1 = 2$", "So all edges among $A$, $B$, and $C$ are of length $\sqrt{2}$. Therefore, the unknown point $D(x,y,z)$ must satisfy:", "$$\nAD^2 = BD^2 = CD^2 = 2\n$$", "### Step 2: Set Up Distance Equations", "Using the squared distance formula:", "1. $AD^2 = x^2 + y^2 + z^2 = 2$\n2. $BD^2 = (x-1)^2 + (y-1)^2 + z^2 = 2$\n3. $CD^2 = (x-1)^2 + y^2 + (z-1)^2 = 2$", "### Step 3: Simplify the System", "Expand equations 2 and 3:", "Equation 2:\n$$\n(x-1)^2 + (y-1)^2 + z^2 = x^2 - 2x + 1 + y^2 - 2y + 1 + z^2 = 2\n$$\nSubstitute $x^2 + y^2 + z^2 = 2$:\n$$\n2 - 2x - 2y + 2 = 2 \Rightarrow -2x - 2y + 4 = 2 \Rightarrow -2x - 2y = -2 \Rightarrow x + y = 1 \quad \ ext{(Eq.) (a)}\n$$", "Equation 3:\n$$\n(x-1)^2 + y^2 + (z-1)^2 = x^2 - 2x + 1 + y^2 + z^2 - 2z + 1 = 2\n$$\nAgain, substitute $x^2 + y^2 + z^2 = 2$:\n$$\n2 - 2x - 2z + 2 = 2 \Rightarrow -2x - 2z + 4 = 2 \Rightarrow -2x - 2z = -2 \Rightarrow x + z = 1 \quad \ ext{(Eq.) (b)}\n$$", "### Step 4: Solve the Linear System", "From (a): $y = 1 - x$\nFrom (b): $z = 1 - x$", "Substitute into the first equation:", "$$\nx^2 + (1 - x)^2 + (1 - x)^2 = 2\n$$\nSimplify:\n$$\nx^2 + (1 - 2x + x^2) + (1 - 2x + x^2) = 2 \\nx^2 + 1 - 2x + x^2 + 1 - 2x + x^2 = 2 \\n3x^2 - 4x + 2 = 2 \\n3x^2 - 4x = 0 \\nx(3x - 4) = 0\n$$", "Solutions: $x = 0$ or $x = \frac{4}{3}$", "Since we require integer coordinates, $x = 0$ is valid; $x = \frac{4}{3}$ yields non-integer values.", "Then:\n- $x = 0$\n- $y = 1 - 0 = 1$\n- $z = 1 - 0 = 1$", "So $D = (0,1,1)$", "### Step 5: Verify All Distances", "Check distances from $D(0,1,1)$ to $A$, $B$, $C$:", "- $AD^2 = 0^2 + 1^2 + 1^2 = 2$\n- $BD^2 = (0-1)^2 + (1-1)^2 + (1-0)^2 = 1 + 0 + 1 = 2$\n- $CD^2 = (0-1)^2 + (1-0)^2 + (1-1)^2 = 1 + 1 + 0 = 2$", "All match $AB^2 = BC^2 = CA^2 = 2$, confirming the tetrahedron is regular.", "### Conclusion", "The only integer-coordinate point completing the regular tetrahedron with vertices at $(0,0,0)$, $(1,1,0)$, and $(1,0,1)$ is:", "$$\n\boxed{(0,1,1)}\n$$", "This solution exemplifies how symmetry and algebraic constraints guide precise geometric constructions—especially valuable in discrete geometry and computational applications where integer solutions are required.", "---", "Keywords: regular tetrahedron, integer coordinates, 3D geometry, vertex calculation, distance formula, algebraic geometry, integer solution, $ (x,y,z) $, $ (0,1,1) $"]