Let the fourth vertex be $(x, y, z)$, and recall that in a regular tetrahedron, all edges have equal length.

["Let the Fourth Vertex Be $(x, y, z)$: Finding the Missing Corner of a Regular Tetrahedron", "In the elegant world of geometry, the regular tetrahedron stands out as a symbol of perfect symmetry and uniformity. One of its defining properties is that all six edges are of equal length—making it one of nature’s most harmonious 3D shapes. Whenever tasked with determining the coordinates of the fourth vertex $(x, y, z)$ of a regular tetrahedron given three vertices, the challenge lies in preserving equal distances across all edges. This article explains how to compute $(x, y, z)$ using spatial logic and distance formulas.", "### Understanding the Regular Tetrahedron Structure", "A regular tetrahedron consists of four equilateral triangular faces, with each vertex connected to the other three by edges of identical length. Given three vertices $A$, $B$, and $C$, we seek a fourth point $D = (x, y, z)$ such that the distances $AD = BD = CD = AB = BC = CA$ are all equal.", "Let the coordinates of the three given points be $A = (x_1, y_1, z_1)$, $B = (x_2, y_2, z_2)$, and $C = (x_3, y_3, z_3)$. The edge length $L$ is the distance between any two of these three points—say, the distance $AB$:", "$$\nL = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\n$$", "To find $D = (x, y, z)$, we set up three equations ensuring $AD = BD = CD = L$. This translates to:", "$$\n(x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2 = L^2 \quad (1)\n$$\n$$\n(x - x_2)^2 + (y - y_2)^2 + (z - z_2)^2 = L^2 \quad (2)\n$$\n$$\n(x - x_3)^2 + (y - y_3)^2 + (z - z_3)^2 = L^2 \quad (3)\n$$", "Subtracting equation (1) from (2) and (1) from (3) eliminates the $L^2$ term, yielding two linear equations in $x, y, z$. Subtracting (2) from (3) gives a third linear equation. This system reduces the problem to two independent linear equations—representing the perpendicular bisector planes of the edges—intersecting along a line: the set of possible locations for the fourth vertex.", "### Geometric Insight: The Circumcentric Condition", "The point $D$ must lie at the unique position equidistant and symmetric relative to the triangle $ABC$, lying along the line perpendicular to the plane of $ABC$ passing through its center (the circumcenter). For a regular tetrahedron, among two possible symmetric positions (above and below the plane), only one will yield all edges equal to $L$.", "Let $G$ be the centroid (and circumcenter) of triangle $ABC$. Then $D$ lies along the line perpendicular to the plane $ABC$, a distance $h$ from $G$, where:", "$$\nh = \sqrt{L^2 - R^2}\n$$", "Here, $R$ is the circumradius of triangle $ABC$, which for an equilateral triangle of side $L$ is:", "$$\nR = \frac{L}{\sqrt{3}}\n$$", "Thus,", "$$\nh = \sqrt{L^2 - \left(\frac{L}{\sqrt{3}}\right)^2} = \sqrt{L^2 - \frac{L^2}{3}} = \sqrt{\frac{2L^2}{3}} = L \sqrt{\frac{2}{3}}\n$$", "Hence, $D$ lies at a distance $h = L\sqrt{2/3}$ from the plane of $ABC$, along the normal vector perpendicular to the triangle’s plane.", "### Computing $(x, y, z)$ Explicitly", "Assume without loss of generality that $A, B, C$ are placed in a convenient orientation to simplify calculations—such as placing $A = (0, 0, 0)$, $B = (L, 0, 0)$, and $C = \left(\frac{L}{2}, \frac{L\sqrt{3}}{2}, 0\right)$ to form an equilateral triangle in the $xy$-plane.", "The centroid $G$ has coordinates:", "$$\nG = \left( \frac{0 + L + \frac{L}{2}}{3}, \frac{0 + 0 + \frac{L\sqrt{3}}{2}}{3}, 0 \right) = \left( \frac{L}{2}, \frac{L\sqrt{3}}{6}, 0 \right)\n$$", "The normal vector to the plane (and hence direction from $G$ to $D$) is proportional to $(0, 0, 1)$ since the triangle lies in the $xy$-plane. Thus, $D$ has coordinates:", "$$\nD = \left( \frac{L}{2}, \frac{L\sqrt{3}}{6}, \pm L\sqrt{\frac{2}{3}} \right)\n$$", "Therefore, the fourth vertex $(x, y, z)$ satisfies:", "$$\nx = \frac{L}{2}, \quad y = \frac{L\sqrt{3}}{6}, \quad z = \pm L\sqrt{\frac{2}{3}}\n$$", "### Summary: Role of $(x, y, z)$ in a Regular Tetrahedron", "Setting $D = (x, y, z)$ means solving a system constrained by equal edge lengths—translating geometric symmetry into algebraic conditions. The solution lies precisely at a height $L\sqrt{2/3}$ along the perpendicular axis through the equilateral triangle’s circumcenter. This makes $(x, y, z)$ not merely a coordinate, but a key to unlocking the tetrahedron’s spatial harmony.", "Whether constructing in 3D modeling, exploring crystallography, or solving olympiad geometry problems, computing the fourth vertex as $(x, y, z)$ ensures perfect regularity—proving once again that symmetry governs the most elegant structures.", "---", "Keywords: regular tetrahedron, fourth vertex $(x, y, z)$, equal edge length, geometric construction, 3D coordinates, symmetry, distance formula, circumcenter, height of tetrahedron, $L\sqrt{2/3}$, equilateral triangle in 3D."]