Solution: Let the fourth vertex be $D = (x, y, z)$. In a regular tetrahedron, all six edges must be equal. First, compute the distance between known points: - Aurero
Title: Solving for Vertex $D = (x, y, z)$ in a Regular Tetrahedron: Equal Edge Lengths Explained
Title: Solving for Vertex $D = (x, y, z)$ in a Regular Tetrahedron: Equal Edge Lengths Explained
Introduction
Creating a regular tetrahedron in 3D space requires all six edges to be equal in length—this presents a classic geometric challenge. In this article, we explore a key solution approach: solving for the unknown fourth vertex $D = (x, y, z)$ when several vertices are already defined. By computing distances between known points and enforcing uniform edge lengths, we establish equations that determine the precise coordinates of $D$, ensuring symmetry and regularity. Let’s dive into the details.
Understanding the Context
Understanding a Regular Tetrahedron
A regular tetrahedron is a polyhedron with four equilateral triangular faces, six equal edges, and four vertices, where every pair of vertices is the same distance apart. For any three vertices known, the fourth vertex must satisfy three equal distance constraints to each of the known points—this forms a system of equations that fully determines $D = (x, y, z)$.
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Key Insights
Geometric Setup and Coordinate Strategy
To simplify calculations, a strategic placement of vertices helps. Without loss of generality, place three points along a triangular base in the $xy$-plane for symmetry:
- Let $A = (0, 0, 0)$
- $B = (a, 0, 0)$
- $C = \left(\frac{a}{2}, \frac{a\sqrt{3}}{2}, 0\right)$,
where $a$ is the edge length. This triangle is equilateral with side length $a$.
Computing Distances to Determine Vertex $D = (x, y, z)$
Since the tetrahedron is regular, the unknown vertex $D = (x, y, z)$ must satisfy:
$$
|AD| = |BD| = |CD| = a
$$
This gives us three equations:
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Distance from $A = (0,0,0)$:
$$
\sqrt{x^2 + y^2 + z^2} = a \quad \Rightarrow \quad x^2 + y^2 + z^2 = a^2 \ ag{1}
$$ -
Distance from $B = (a, 0, 0)$:
$$
\sqrt{(x - a)^2 + y^2 + z^2} = a \quad \Rightarrow \quad (x - a)^2 + y^2 + z^2 = a^2 \ ag{2}
$$ -
Distance from $C = \left(\frac{a}{2}, \frac{a\sqrt{3}}{2}, 0\right)$:
$$
\sqrt{\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{a\sqrt{3}}{2}\right)^2 + z^2} = a \quad \Rightarrow \quad \left(x - \frac{a}{2}\right)^2 + \left(y - \frac{a\sqrt{3}}{2}\right)^2 + z^2 = a^2 \ ag{3}
$$
Solving the System of Equations
Subtract (1) from (2):
$$
(x - a)^2 + y^2 + z^2 - (x^2 + y^2 + z^2) = 0
\Rightarrow x^2 - 2ax + a^2 - x^2 = 0 \Rightarrow -2ax + a^2 = 0
\Rightarrow x = \frac{a}{2}
$$
Now subtract (1) from (3):
$$
\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{a\sqrt{3}}{2}\right)^2 + z^2 - (x^2 + y^2 + z^2) = 0
$$
Substitute $x = \frac{a}{2}$:
$$
0 - 2\left(\frac{a}{2}\right)\cdot\frac{a}{2} + \frac{a^2}{4} + y^2 - 2y\cdot\frac{a\sqrt{3}}{2} + \frac{3a^2}{4} = 0
\Rightarrow -\frac{a^2}{2} + \frac{a^2}{4} + y^2 - a\sqrt{3}\, y + \frac{3a^2}{4} = 0
\Rightarrow \left(-\frac{1}{2} + \frac{1}{4} + \frac{3}{4}\right)a^2 + y^2 - a\sqrt{3}\, y = 0
\Rightarrow a^2 + y^2 - a\sqrt{3}\, y = 0
$$
Solve the quadratic in $y$:
$$
y^2 - a\sqrt{3}\, y + a^2 = 0
\Rightarrow y = \frac{a\sqrt{3} \pm \sqrt{3a^2 - 4a^2}}{2} = \frac{a\sqrt{3} \pm \sqrt{-a^2}}{2}
$$
Wait—this suggests an error in sign. Rechecking the algebra, the correct expansion yields:
From:
$$
-\frac{a^2}{2} + \frac{a^2}{4} + \frac{3a^2}{4} + y^2 - a\sqrt{3}\, y = 0
\Rightarrow \left(-\frac{1}{2} + 1\right)a^2 + y^2 - a\sqrt{3}\, y = 0
\Rightarrow \frac{1}{2}a^2 + y^2 - a\sqrt{3}\, y = 0
\Rightarrow y^2 - a\sqrt{3}\, y + \frac{1}{2}a^2 = 0
$$
Discriminant:
$$
\Delta = (a\sqrt{3})^2 - 4 \cdot 1 \cdot \frac{1}{2}a^2 = 3a^2 - 2a^2 = a^2 > 0
\Rightarrow y = \frac{a\sqrt{3} \pm a}{2}
$$
Thus:
- $y = \frac{a(\sqrt{3} + 1)}{2}$ or $y = \frac{a(\sqrt{3} - 1)}{2}$
Now substitute $x = \frac{a}{2}$, $y$, and solve for $z$ using equation (1):
From (1):
$$
\left(\frac{a}{2}\right)^2 + y^2 + z^2 = a^2 \Rightarrow \frac{a^2}{4} + y^2 + z^2 = a^2
\Rightarrow z^2 = a^2 - \frac{a^2}{4} - y^2 = \frac{3a^2}{4} - y^2
$$
Using $y = \frac{a(\sqrt{3} - 1)}{2}$:
$$
y^2 = \frac{a^2}{4} \cdot ( (\sqrt{3} - 1)^2 ) = \frac{a^2}{4} (3 - 2\sqrt{3} + 1) = \frac{a^2}{4} (4 - 2\sqrt{3}) = \frac{a^2}{2}(2 - \sqrt{3})
$$
Then:
$$
z^2 = \frac{3a^2}{4} - \frac{a^2}{2}(2 - \sqrt{3}) = a^2\left( \frac{3}{4} - 1 + \frac{\sqrt{3}}{2} \right) = a^2\left( -\frac{1}{4} + \frac{\sqrt{3}}{2} \right) = a^2 \left( \frac{2\sqrt{3} - 1}{4} \right)
$$
Thus:
$$
z = \pm a \sqrt{ \frac{2\sqrt{3} - 1}{4} }
$$
