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

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== Solution ==
 
== 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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<center>[[Image:AIME_1986_Problem_14_sol.png]]</center>
  
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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.
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In the above diagram, we focus on the line that appears closest and is parallel to <math>BC</math>. All the blue lines are perpendicular lines to <math>BC</math> and their other points are on <math>AB</math>, 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 <math>AC</math>, and have the same lengths as their corresponding blue lines. So we want to find the shortest distance between <math>AC</math> and that corner, which is <math>\frac {wl}{\sqrt {w^2 + l^2}}</math>.
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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.
 
  
 
So we have:
 
So we have:
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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>
 
<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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We invert the above equations to get a system of linear equations in <math>\frac {1}{h^2}</math>, <math>\frac {1}{l^2}</math>, and <math>\frac {1}{w^2}</math>:
 
<cmath>\frac {1}{l^2} + \frac {1}{w^2} = \frac {45}{900}</cmath>
 
<cmath>\frac {1}{l^2} + \frac {1}{w^2} = \frac {45}{900}</cmath>
 
<cmath>\frac {1}{h^2} + \frac {1}{w^2} = \frac {13}{900}</cmath>
 
<cmath>\frac {1}{h^2} + \frac {1}{w^2} = \frac {13}{900}</cmath>
 
<cmath>\frac {1}{h^2} + \frac {1}{l^2} = \frac {40}{900}</cmath>
 
<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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We see that <math>h = 15</math>, <math>l = 5</math>, <math>w = 10</math>. Therefore <math>V = 5 \cdot 10 \cdot 15 = \boxed{750}</math>
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 12:13, 22 July 2020

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

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 $V = 5 \cdot 10 \cdot 15 = \boxed{750}$

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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