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Home » Average per hour: 190/3 = 63.333... - Veritas Home Health

Average per hour: 190/3 = 63.333... - Veritas Home Health

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 $AC = \sqrt{(7-1)^2 + (8-2)^2 + (9-3)^2} = \sqrt{36 + 36 + 36} = \sqrt{108} = 6\sqrt{3}$

📌 $BC = \sqrt{(7-4)^2 + (8-5)^2 + (9-6)^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}$

📌 We observe that $AB = BC = 3\sqrt{3}$, but $AC = 6\sqrt{3}$, so the three points are not equidistant. Thus, they cannot form a face of a regular tetrahedron. Therefore, the given points cannot be vertices of a regular tetrahedron with integer coordinates.

📌 However, if we reinterpret the question as seeking a fourth point $D = (x, y, z)$ such that all six edges $AB, AC, AD, BC, BD, CD$ are equal and the coordinates are integers, we proceed by solving the system:

📌 Let $AB = BC = CD = DA = DB = DC = s$. Then we solve the system of equations:

📌 Subtracting the first equation from the second and the second from the third gives linear equations in $x, y, z$. Subtracting the first from the second:

📌 [(x - 4)^2 - (x - 1)^2] + [(y - 5)^2 - (y - 2)^2] + [(z - 6)^2 - (z - 3)^2] = 0

📌 -6x -6y -6z + 63 = 0 \Rightarrow x + y + z = 10.5

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\boxed{240^\circ} \quad \text{و} \quad \boxed{300^\circ}
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