Let’s find $(x, y, z)$ such that its distance to each of the three points is also $\sqrt{2}$:

["# Finding Points $(x, y, z)$ Equidistant $\sqrt{2}$ from Three Given Points: A Geometric Exploration", "When working with 3D coordinate geometry, one fascinating problem is identifying a point $(x, y, z)$ that maintains a fixed distance from multiple reference points. A particularly insightful challenge involves finding all points at a distance of $\sqrt{2}$ from three non-collinear points in space. Whether you’re tackling this for competitive math, computer graphics, or spatial reasoning, understanding the intersection of spheres centered at these points offers clarity and precision.", "## The Problem: Points at Fixed Distance from a Triangle", "Suppose we are given three points in 3D space—let’s call them $A$, $B$, and $C$—that form a triangle. We seek all points $(x, y, z)$ such that the Euclidean distance from $(x, y, z)$ to each of $A$, $B$, and $C$ equals $\sqrt{2}$. Mathematically, this translates to solving the system:", "$$\n\begin{aligned}\n(x - x_A)^2 + (y - y_A)^2 + (z - z_A)^2 &= 2 \\n(x - x_B)^2 + (y - y_B)^2 + (z - z_B)^2 &= 2 \\n(x - x_C)^2 + (y - y_C)^2 + (z - z_C)^2 &= 2 \\n\end{aligned}\n$$", "where $A = (x_A, y_A, z_A)$, $B = (x_B, y_B, z_B)$, and $C = (x_C, y_C, z_C)$.", "## Why This Is More Complex than 2D", "In 2D, three circles (of radius $\sqrt{2}$ centered at each point) may meet at zero, one, or two points. In 3D, spheres defined by these equations extend throughout space, allowing up to eight intersection points in ideal geometric configurations. However, practical solutions often yield fewer due to symmetry or the triangle’s relative positioning.", "### Critical Condition: The Triangle’s Geometry", "For solutions to exist at all, the circumsphere (the sphere passing through all three points) must have a radius no greater than $\sqrt{2}$. If the circumradius $R$ of triangle $ABC$ exceeds $\sqrt{2}$, no such point exists. Importantly, if the triangle lies flat (coplanar) with side lengths not compatible with radius $\sqrt{2}$, solutions may be empty or degenerate.", "## Step-by-Step Solution Strategy", "### Step 1: Translate Points to Origin-Managed Coordinates", "To simplify, shift coordinate system so one point is at the origin. Let $A = (0,0,0)$, $B = \vec{b} = (b_x, b_y, b_z)$, $C = \vec{c} = (c_x, c_y, c_z)$. Now solve:\n$$\n\begin{aligned}\nx^2 + y^2 + z^2 &= 2 \\n(x - b_x)^2 + (y - b_y)^2 + (z - b_z)^2 &= 2 \\n(x - c_x)^2 + (y - c_y)^2 + (z - c_z)^2 &= 2 \\n\end{aligned}\n$$", "### Step 2: Subtract Equations to Eliminate Quadratic Terms", "Subtracting first equation from the others gives two linear equations—each representing a plane where solutions lie:", "$$\n\begin{aligned}\n-2b_x x - 2b_y y - 2b_z z &= 0 \quad \Rightarrow\quad b_x x + b_y y + b_z z = 0 \quad &\ ext{(Plane 1)} \\n-2c_x x - 2c_y y - 2c_z z &= 0 \quad \Rightarrow\quad c_x x + c_y y + c_z z = 0 \quad &\ ext{(Plane 2)} \\n\end{aligned}\n$$", "These two planes intersect along a line. The true intersection (points satisfying all three distance constraints) lies where the sphere intersections align, typically at the line of intersection projected onto the sphere.", "### Step 3: Solve the Linear System with Substitution", "Using Plane 1: $b_x x + b_y y + b_z z = 0$, express one variable in terms of the others (e.g., $z = -\frac{b_x x + b_y y}{b_z}$, assuming $b_z <br/>\ne 0$). Substitute into the sphere equation:\n$$\nx^2 + y^2 + \left(-\frac{b_x x + b_y y}{b_z}\right)^2 = 2\n$$\nThis yields a quadratic in $x$ and $y$, defining a conic section in the plane—potentially a circle (intersection of sphere and plane) or line segment.", "### Step 4: Use Symmetry or Vector Geometry to Simplify", "For symmetric triangles—say, equilateral, isosceles, or aligned with axes—explicit solutions simplify. For example, place $A = (0,0,0)$, $B = (1,1,0)$, $C = (0,1,1)$ (a non-degenerate obtuse triangle). Solving analytically gives:\n$$\nx^2 + y^2 + z^2 = 2\n$$\n$$\n(x - 1)^2 + (y - 1)^2 + z^2 = 2\n$$\n$$\nx^2 + (y - 1)^2 + (z - 1)^2 = 2\n$$", "Subtract first and second:\n$ -2x - 2y + 1 = 0 \Rightarrow x + y = \frac{1}{2} $\nSubtract first and third:\n$ -2y - 2z + 1 = 0 \Rightarrow y + z = \frac{1}{2} $", "Now substitute $x = \frac{1}{2} - y$, $z = \frac{1}{2} - y$ into the first equation:\n$$\n\left(\frac{1}{2} - y\right)^2 + y^2 + \left(\frac{1}{2} - y\right)^2 = 2\n\Rightarrow 2\left(\frac{1}{4} - y + y^2\right) + y^2 = 2\n\Rightarrow \frac{1}{2} - 2y + 2y^2 + y^2 = 2\n\Rightarrow 3y^2 - 2y - \frac{3}{2} = 0\n$$\nSolving:\n$$\ny = \frac{2 \pm \sqrt{4 + 18}}{6} = \frac{2 \pm \sqrt{22}}{6} = \frac{1 \pm \sqrt{22}/2}{3}\n$$\nCorresponding $x, z$ values derived similarly yield specific triplets.", "### Step 5: Consider All Possible Configurations", "Depending on triangle geometry—acute, obtuse, right—the number and nature of solutions vary. Solutions may appear complex, but they lie along the radical axis intersection of three spheres. Tools like symbolic computation (Mathematica, SageMath) efficiently determine these points numerically or algebraically.", "## Real-W-world Applications", "Finding equidistant points with fixed radius is vital in:\n- Error correction in spatial networks\n- Optimal placement in robotics path planning\n- 3D modeling and mesh reconstruction\n- Interpolation and simulation in computer graphics", "## Conclusion", "Locating $(x, y, z)$ such that its distance to three given 3D points is $\sqrt{2}$ involves solving a system of nonlinear equations rooted in sphere geometry. By transforming into a linear system via plane intersections and substituting into the sphere equation, we reduce complexity and uncover solution sets—often small but rich in geometric insight. Whether symmetric or arbitrary, this method equips mathematicians and engineers to locate robust reference points in spatial design.", "Exploring such problems deepens your grasp of multivariate geometry and opens pathways to advanced applications in computational design and spatial analytics.", "---", "## Key Takeaways", "- Solve three distance equations as a system of quadratic/linear constraints.\n- Use coordinate shifts and planar subtraction to simplify.\n- Symmetric configurations yield explicit solutions.\n- Solutions reflect geometric relationships at the spheres’ intersection.\n- Practical tools assist verification in complex cases.", "Keywords: $ (x, y, z) $, distance $\sqrt{2}$, spheres, 3D geometry, intersecting planes, radical axis, spatial point location, equidistant points."]