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

Revision as of 18:17, 4 July 2013

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 (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 $d$ .

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

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	[[Category:Intermediate Geometry Problems]]		[[Category:Intermediate Geometry Problems]]
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1989 AIME (Problems • Answer Key • Resources)
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Difference between revisions of "1989 AIME Problems/Problem 12"

Revision as of 18:17, 4 July 2013

Problem

Solution

See also