1989 AIME Problems/Problem 12: Difference between revisions
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== Solution == | == Solution == | ||
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. | 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 22:17, 26 January 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 | ||
