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Difference between revisions of "1986 AIME Problems/Problem 14"

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(solution by minsoens)
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== Problem ==
 
== Problem ==
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The shortest distances between an interior diagonal of a rectangular parallelepiped, <math>\displaystyle P</math>, and the edges it does not meet are <math>\displaystyle 2\sqrt{5}</math>, <math>\displaystyle \frac{30}{\sqrt{13}}</math>, and <math>\displaystyle \frac{15}{\sqrt{10}}</math>. Determine the volume of <math>\displaystyle P</math>.
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The shortest distances between an interior [[diagonal]] of a rectangular [[parallelepiped]], <math>P</math>, and the edges it does not meet are <math>2\sqrt{5}</math>, <math>\frac{30}{\sqrt{13}}</math>, and <math>\frac{15}{\sqrt{10}}</math>. Determine the [[volume]] of <math>P</math>.
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== Solution ==
 
== Solution ==
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{{solution}}
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If you draw a [[right triangle]] with the space diagonal as the [[hypotenuse]], one side as a leg, and the corresponding face diagonal as the other leg, then you will notice that the minimum distance from the triangle to another side parallel to the first leg of the triangle is what the problem asks for.
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In other words, draw a space diagonal of any face that shares a vertex with the space diagonal. It is only necessary to find the min. distance to another vertex of that face.
 +
 
 +
Using similar triangles and [[Pythagorean Theorem|Pythagoras]], the distance equation is:
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<math>\frac {xy}{\sqrt {x^2 + y^2}}</math>, where x and y are any of the sides.
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So we have:
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<cmath>\frac {lw}{\sqrt {l^2 + w^2}} = \frac {10}{\sqrt {5}}</cmath>
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<cmath>\frac {hw}{\sqrt {h^2 + w^2}} = \frac {30}{\sqrt {13}}</cmath>
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<cmath>\frac {hl}{\sqrt {h^2 + l^2}} = \frac {15}{\sqrt {10}}</cmath>
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 +
Notice the familiar roots: <math>\sqrt {5}</math>, <math>\sqrt {13}</math>, <math>\sqrt {10}</math>, which are <math>\sqrt {1^2 + 2^2}</math>, <math>\sqrt {2^2 + 3^2}</math>, <math>\sqrt {1^2 + 3^2}</math>, respectively. (This would give us the guess that the sides are of the ratio 1:2:3, but let's provide the complete solution.)
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<cmath>\frac {l^2w^2}{l^2 + w^2} = \frac {1}{\frac {1}{l^2} + \frac {1}{w^2}} = 20</cmath>
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<cmath>\frac {h^2w^2}{h^2 + w^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{w^2}} = \frac {900}{13}</cmath>
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<cmath>\frac {h^2l^2}{h^2 + l^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{l^2}} = \frac {45}{2}</cmath>
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 +
We solve the above equations for <math>\frac {1}{h^2}</math>, <math>\frac {1}{l^2}</math>, and <math>\frac {1}{w^2}</math>:
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<cmath>\frac {1}{l^2} + \frac {1}{w^2} = \frac {45}{900}</cmath>
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<cmath>\frac {1}{h^2} + \frac {1}{w^2} = \frac {13}{900}</cmath>
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<cmath>\frac {1}{h^2} + \frac {1}{l^2} = \frac {40}{900}</cmath>
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 +
<math>h = 15</math>, <math>l = 5</math>, <math>w = 10</math>, as expected, so <math>V = 5 \cdot 10 \cdot 15 = 750</math>.
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== See also ==
 
== See also ==
 
{{AIME box|year=1986|num-b=13|num-a=15}}
 
{{AIME box|year=1986|num-b=13|num-a=15}}
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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[[Category:Intermediate Geometry Problems]]
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* [[Mathematics competition resources]]
 

Revision as of 19:14, 29 September 2007

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

If you draw a right triangle with the space diagonal as the hypotenuse, one side as a leg, and the corresponding face diagonal as the other leg, then you will notice that the minimum distance from the triangle to another side parallel to the first leg of the triangle is what the problem asks for.

In other words, draw a space diagonal of any face that shares a vertex with the space diagonal. It is only necessary to find the min. distance to another vertex of that face.

Using similar triangles and Pythagoras, the distance equation is: $\frac {xy}{\sqrt {x^2 + y^2}}$, where x and y are any of the sides.

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 solve the above equations for $\frac {1}{h^2}$, $\frac {1}{l^2}$, and $\frac {1}{w^2}$: \[\frac {1}{l^2} + \frac {1}{w^2} = \frac {45}{900}\] \[\frac {1}{h^2} + \frac {1}{w^2} = \frac {13}{900}\] \[\frac {1}{h^2} + \frac {1}{l^2} = \frac {40}{900}\]

$h = 15$, $l = 5$, $w = 10$, as expected, so $V = 5 \cdot 10 \cdot 15 = 750$.

See also

1986 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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