So all edges must be $\sqrt2$. Set up equations for $AD = BD = CD = \sqrt2$: - Imagemakers
All Edges Must Be $ \sqrt{2} $: Setting Up a Coordinate Geometry Challenge
All Edges Must Be $ \sqrt{2} $: Setting Up a Coordinate Geometry Challenge
In the elegant world of Euclidean geometry, the lengths of segments define shape, symmetry, and balance. A particularly intriguing challenge arises when every edge of a geometric figure is constrained to be $ \sqrt{2} $, a length rich in simplicity and algebraic beauty. Here, we explore how to construct a figure—such as a tetrahedron or a triangle—where $ AD = BD = CD = \sqrt{2} $, revealing the mathematical backbone that ensures all edges maintain this exact length.
This article derives the essential equations that enforce $ AD = BD = CD = \sqrt{2} $, laying the foundation for precise geometric modeling. Whether used in spatial reasoning, 3D modeling, or olympiad geometry, understanding these equations unlocks powerful problem-solving tools.
Understanding the Context
Foundation: Distance in the Coordinate Plane
Let’s define the coordinates of point $ D $ as $ (0, 0, 0) $—a natural origin that simplifies the derivation. When $ AD = BD = CD = \sqrt{2} $, the Euclidean distance formula becomes our key instrument. For any point $ A = (x_A, y_A, z_A) $, the distance from $ D $ is:
$$
AD = \sqrt{(x_A - 0)^2 + (y_A - 0)^2 + (z_A - 0)^2} = \sqrt{x_A^2 + y_A^2 + z_A^2} = \sqrt{2}
$$
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Key Insights
Squaring both sides yields:
$$
x_A^2 + y_A^2 + z_A^2 = 2 \ ag{1}
$$
The same logic applies to points $ B $ and $ C $. Thus, every vertex on the configuration must lie on the sphere centered at $ D $:
$$
x^2 + y^2 + z^2 = 2
$$
Enforcing Equal Distances: $ AD = BD = \sqrt{2} $
For $ AD = BD $, take two unknown points:
$ A = (x_A, y_A, z_A) $,
$ B = (x_B, y_B, z_B) $.
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We require:
$$
x_A^2 + y_A^2 + z_A^2 = x_B^2 + y_B^2 + z_B^2 = 2 \ ag{2}
$$
Subtracting these equations ensures symmetry:
$$
(x_A^2 - x_B^2) + (y_A^2 - y_B^2) + (z_A^2 - z_B^2) = 0
$$
or
$$
(x_A - x_B)(x_A + x_B) + (y_A - y_B)(y_A + y_B) + (z_A - z_B)(z_A + z_B) = 0 \ ag{3}
$$
This equation describes a geometric midpoint or perpendicularity condition, highlight- ing how distances constrain relative positions.
Extending to a Three-Dimensional Configuration: $ AD = CD = \sqrt{2} $
Now include point $ C = (x_C, y_C, z_C) $, so:
$$
x_C^2 + y_C^2 + z_C^2 = 2 \ ag{4}
$$
The equal distances imply:
$$
AD^2 = CD^2 \Rightarrow x_A^2 + y_A^2 + z_A^2 = x_C^2 + y_C^2 + z_C^2
$$
Equating equations (1) and (4) ensures $ A $ and $ C $ lie on the same sphere centered at $ D $.
To enforce $ BD = \sqrt{2} $, repeat for $ B $ and $ C $:
$$
x_B^2 + y_B^2 + z_B^2 = x_C^2 + y_C^2 + z_C^2 \ ag{5}
$$
Combine with earlier differences: the symmetry across $ A, B, C $ suggests they form an equilateral triangle in some plane, all at unit-distance (scaled) from $ D $.
