1989 AIME Problems/Problem 12

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 $d^{2}_{}$ .

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.

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 $d$ .

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

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
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1989 AIME Problems/Problem 12

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