1989 AIME Problems/Problem 12: Difference between revisions
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== Problem == | == Problem == | ||
Let <math>ABCD | 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 
be a tetrahedron with 
, 
, 
, 
, 
, and 
, as shown in the figure. Let 
be the distance between the midpoints of edges 
and 
. Find 
.
Solution
Call the midpoint of 
and the midpoint of 
. is the median of triangle 
. The formula for the length of a median is 
, where 
, 
, and 
are the side lengths of triangle, and is the side that is bisected by median 
. 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 
, which is the median of 
.
![\[CM=\sqrt{\frac{98+2592-1681}{4}}=\frac{\sqrt{1009}}{2}\]](http://latex.artofproblemsolving.com/a/b/e/abe130a49cd7411828cc3f705e9efd3a0aaef928.png)
Now we must find 
, which is the median of 
.
![\[DM=\frac{\sqrt{425}}{2}\]](http://latex.artofproblemsolving.com/9/8/0/9807aa4b02f58eff7b178a99e6f1cc48495a1695.png)
Now that we know the sides of , we proceed to find the length of .
![\[d=\frac{\sqrt{548}}{2} \Longrightarrow d^2=\frac{548}{4}=\boxed{137}\]](http://latex.artofproblemsolving.com/7/5/0/75091e008ce263f5c200534e4a659e819a5504b6.png)
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 | ||
