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

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== Solution ==
 
== Solution ==
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Call the midpoint of <math>\overline{AB}</math> <math>M</math> and the midpoint of <math>\overline{CD}</math> <math>N</math>. <math>d</math> is the [[median]] of triangle <math>\triangle CDM</math>. The formula for the length of a median is <math>m=\sqrt{\frac{2a^2+2b^2-c^2}{4}}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are the side lengths of triangle, and <math>c</math> is the side that is bisected by median <math>m</math>. The formula is a direct result of the [[Law of Cosines]] applied twice with the angles formed by the median.  
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Call the midpoint of <math>\overline{AB}</math> <math>M</math> and the midpoint of <math>\overline{CD}</math> <math>N</math>. <math>d</math> is the [[median]] of triangle <math>\triangle CDM</math>. The formula for the length of a median is <math>m=\sqrt{\frac{2a^2+2b^2-c^2}{4}}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are the side lengths of triangle, and <math>c</math> is the side that is bisected by median <math>m</math>. 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 <math>CM</math>, which is the median of <math>\triangle CAB</math>.
 
We first find <math>CM</math>, which is the median of <math>\triangle CAB</math>.

Revision as of 23:17, 26 January 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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