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1986 AIME Problems/Problem 14

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Problem

The shortest distances between an interior diagonal of a rectangular 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 of P.

Solution

Image: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 = 750.

See also

1986 AIME (ProblemsResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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