Cube's Vertex Distance Problem
A cube has a side length of $ s, $ and its vertices are $ A = (0,0,0), $ $ B = (s,0,0), $ $ C = (s,s,0), $ $ D = (0,s,0), $ $ E = (0,0,s), $ $ F = (s,0,s), $ $ G = (s,s,s), $ and $ H = (0,s,s) $. A point $ P $ inside the cube satisfies $ PA = \sqrt{70}, $ $ PB = \sqrt{97}, $ $ PC = \sqrt{88}, $ and $ PE = \sqrt{43} $. Find the side length $ s $.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Let $ P = (x,y,z) $. Then from the given information,
\begin{align*}
x^2 + y^2 + z^2 &= 70, \quad (1) \\
(x - s)^2 + y^2 + z^2 &= 97, \quad (2) \\
(x - s)^2 + (y - s)^2 + z^2 &= 88, \quad (3) \\
x^2 + y^2 + (z - s)^2 &= 43.\quad (4)
\end{align*}Subtracting equations (1) and (2) gives us
\[-2sx + s^2 = 27,\]so $ x = \frac{s^2 - 27}{2s} $.
Subtracting equations (2) and (3) gives us
\[-2sy + s^2 = -9,\]so $ y = \frac{s^2 + 9}{2s} $.
Subtracting equations (1) and (4) gives us
\[-2sz + s^2 = -27,\]so $ z = \frac{s^2 + 27}{2s} $.
Substituting into equation (1), we get
\[\left( \frac{s^2 - 27}{2s} \right)^2 + \left( \frac{s^2 + 9}{2s} \right)^2 + \left( \frac{s^2 + 27}{2s} \right)^2 = 70.\]This simplifies to $ 3s^4 - 262s^2 + 1539 = 0, $ which factors as $ (s^2 - 81)(3s^2 - 19) = 0 $. Since $ x = \frac{s^2 - 27}{2s} $ must be positive, $ s^2 = 81 $. Hence, $ s = \boxed{9} $.
Freeflows