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Difference between revisions of "1989 AIME Problems/Problem 12"

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== Problem ==
 
== Problem ==
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Let <math>ABCD^{}_{}</math> be a [[tetrahedron]] with <math>AB=41^{}_{}</math>, <math>AC=7^{}_{}</math>, <math>AD=18^{}_{}</math>, <math>BC=36^{}_{}</math>, <math>BD=27^{}_{}</math>, and <math>CD=13^{}_{}</math>, as shown in the figure. Let <math>d^{}_{}</math> be the distance between the [[midpoint]]s of [[edge]]s <math>AB^{}_{}</math> and <math>CD^{}_{}</math>. Find <math>d^{2}_{}</math>.
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Let <math>ABCD</math> be a [[tetrahedron]] with <math>AB=41</math>, <math>AC=7</math>, <math>AD=18</math>, <math>BC=36</math>, <math>BD=27</math>, and <math>CD=13</math>, as shown in the figure. Let <math>d</math> be the distance between the [[midpoint]]s of [[edge]]s <math>AB</math> and <math>CD</math>. Find <math>d^{2}</math>.
  
 
[[Image:AIME_1989_Problem_12.png]]
 
[[Image:AIME_1989_Problem_12.png]]

Revision as of 21:12, 17 July 2008

Problem

Let $ABCD$ be a tetrahedron with $AB=41$, $AC=7$, $AD=18$, $BC=36$, $BD=27$, and $CD=13$, as shown in the figure. Let $d$ be the distance between the midpoints of edges $AB$ and $CD$. Find $d^{2}$.

AIME 1989 Problem 12.png

Solution

Call the midpoint of $\overline{AB}$ $M$ and the midpoint of $\overline{CD}$ $N$. $d$ is the median of triangle $\triangle CDM$. The formula for the length of a median is $m=\sqrt{\frac{2a^2+2b^2-c^2}{4}}$, where $a$, $b$, and $c$ are the side lengths of triangle, and $c$ is the side that is bisected by median $m$. The formula is a direct result of the Law of Cosines applied twice with the angles formed by the median (Stewart's Theorem).

We first find $CM$, which is the median of $\triangle CAB$.

\[CM=\sqrt{\frac{98+2592-1681}{4}}=\frac{\sqrt{1009}}{2}\]

Now we must find $DM$, which is the median of $\triangle DAB$.

\[DM=\frac{\sqrt{425}}{2}\]

Now that we know the sides of $\triangle CDM$, we proceed to find the length of $d$.

\[d=\frac{\sqrt{548}}{2} \Longrightarrow d^2=\frac{548}{4}=\boxed{137}\]

See also

1989 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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