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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 midpoints of edges <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]]
  
 
== Solution ==
 
== Solution ==
−
Call the midpoint of AB M and the midpoint of CD N. d 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 a, b, and c are the side lengths of triangle, and c is the side that is bisected by median m.
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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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We first find CM, which is the median of <math>\triangle CAB</math>.
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We first find <math>CM</math>, which is the median of <math>\triangle CAB</math>.
  
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<math>CM=\sqrt{\frac{98+2592-1681}{4}}=\frac{\sqrt{1009}}{2}</math>
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<cmath>CM=\sqrt{\frac{98+2592-1681}{4}}=\frac{\sqrt{1009}}{2}</cmath>
  
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Now we must find DM, which is the median of <math>\triangle DAB</math>.
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Now we must find <math>DM</math>, which is the median of <math>\triangle DAB</math>.
  
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<math>DM=\frac{\sqrt{425}}{2}</math>
+
<cmath>DM=\frac{\sqrt{425}}{2}</cmath>
  
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Now that we know the sides of <math>\triangle CDM</math>, we proceed to find the length of d.
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Now that we know the sides of <math>\triangle CDM</math>, we proceed to find the length of <math>d</math>.
−
 
 
−
<math>d=\frac{\sqrt{548}}{2}</math>
 
−
 
 
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<math>d^2=\frac{548}{4}=\boxed{137}</math>
 
  
 +
<cmath>d=\frac{\sqrt{548}}{2} \Longrightarrow d^2=\frac{548}{4}=\boxed{137}</cmath>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1989|num-b=11|num-a=13}}
 
{{AIME box|year=1989|num-b=11|num-a=13}}
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 +
[[Category:Intermediate Geometry Problems]]

Revision as of 13:07, 25 November 2007

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.

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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