So all edges must be $\sqrt2$. Set up equations for $AD = BD = CD = \sqrt2$: - inBeat
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:
$$
AD = \sqrt{(x_A - 0)^2 + (y_A - 0)^2 + (z_A - 0)^2} = \sqrt{x_A^2 + y_A^2 + z_A^2} = \sqrt{2}
$$
Image Gallery
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) $.
🔗 Related Articles You Might Like:
📰 molly mae documentary usa 📰 is trump deporting people with green cards 📰 is trump alive 📰 Apple Music Dolby Atmos 9364256 📰 How Much Water Are You Supposed To Drink Everyday 2009417 📰 Explore The Most Beloved Christian Cafe Its Secret Is Worth Visiting Every Day 8639454 📰 How The State Farm Growth Fund Is Dominating 2024Dont Miss This Breakthrough 4533476 📰 You Wont Believe What Happened When Our Friend Group Got Turned Inside Out 4480915 📰 Why Germanys Crazy Games Are Hitting Internet Explosionsgamers Need To See This 735379 📰 Why This Simple Trick Makes Inserting A Degree Symbol In Word A Breeze 2123514 📰 Stand Offish Meaning 6316672 📰 Business Card Credit Cards 7262545 📰 Burger King Whopper Wednesday 1757484 📰 Adjusted Current Earnings 3369924 📰 5 Download Faster Save Better The Best Tools For Making Pdfs In Minutes 125459 📰 Clcu Online The Ultimate Tool Thats Taking Productivity By Storm 1485683 📰 You Wont Believe How Easy It Is To Upgrade Your Windshield Wipers Forever 6588684 📰 Baseball Insiders Finally Explain What Ops Actually Means On The Field 1020070Final Thoughts
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 $.
