The shortest distances between an interior [diagonal](https://artofproblemsolving.com/wiki/index.php/Diagonal) of a rectangular [parallelepiped](https://artofproblemsolving.com/wiki/index.php/Parallelepiped), $P$, and the edges it does not meet are $2\sqrt{5}$, $\frac{30}{\sqrt{13}}$, and $\frac{15}{\sqrt{10}}$. Determine the [volume](https://artofproblemsolving.com/wiki/index.php/Volume) of $P$.
[AIME 1986 Problem 14 sol.png](https://artofproblemsolving.com/wiki/index.php/File:AIME_1986_Problem_14_sol.png)
In the above diagram, we focus on the line that appears closest and is parallel to $BC$. All the blue lines are perpendicular lines to $BC$ and their other points are on $AB$, the main diagonal. The green lines are projections of the blue lines onto the bottom face; all of the green lines originate in the corner and reach out to $AC$, and have the same lengths as their corresponding blue lines. So we want to find the shortest distance between $AC$ and that corner, which is $\frac {wl}{\sqrt {w^2 + l^2}}$.
So we have:
\[\frac {lw}{\sqrt {l^2 + w^2}} = \frac {10}{\sqrt {5}}\]
\[\frac {hw}{\sqrt {h^2 + w^2}} = \frac {30}{\sqrt {13}}\]
\[\frac {hl}{\sqrt {h^2 + l^2}} = \frac {15}{\sqrt {10}}\]
Notice the familiar roots: $\sqrt {5}$, $\sqrt {13}$, $\sqrt {10}$, which are $\sqrt {1^2 + 2^2}$, $\sqrt {2^2 + 3^2}$, $\sqrt {1^2 + 3^2}$, respectively. (This would give us the guess that the sides are of the ratio 1:2:3, but let's provide the complete solution.)
\[\frac {l^2w^2}{l^2 + w^2} = \frac {1}{\frac {1}{l^2} + \frac {1}{w^2}} = 20\]
\[\frac {h^2w^2}{h^2 + w^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{w^2}} = \frac {900}{13}\]
\[\frac {h^2l^2}{h^2 + l^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{l^2}} = \frac {45}{2}\]
We invert the above equations to get a system of linear equations in $\frac {1}{h^2}$, $\frac {1}{l^2}$, and $\frac {1}{w^2}$:
\[\frac {1}{l^2} + \frac {1}{w^2} = \frac {45}{900}\]
\[\frac {1}{h^2} + \frac {1}{w^2} = \frac {13}{900}\]
\[\frac {1}{h^2} + \frac {1}{l^2} = \frac {40}{900}\]
We see that $h = 15$, $l = 5$, $w = 10$. Therefore $V = 5 \cdot 10 \cdot 15 = \boxed{750}$