Difference between revisions of "1989 AIME Problems/Problem 12"
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== 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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| + | We first find CM, 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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| + | Now we must find DM, which is the median of <math>\triangle DAB</math>. | ||
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| + | <math>DM=\frac{\sqrt{425}}{2}</math> | ||
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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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| + | <math>d=\frac{\sqrt{548}}{2}</math> | ||
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| + | <math>d^2=\frac{548}{4}=\boxed{137}</math> | ||
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== See also == | == See also == | ||
{{AIME box|year=1989|num-b=11|num-a=13}} | {{AIME box|year=1989|num-b=11|num-a=13}} | ||
Revision as of 08:47, 12 November 2007
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 AB M and the midpoint of CD N. d is the median of triangle 
. The formula for the length of a median is 
, where a, b, and c are the side lengths of triangle, and c is the side that is bisected by median m.
We first find CM, which is the median of 
.
Now we must find DM, which is the median of 
.
Now that we know the sides of , we proceed to find the length of d.
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 | ||
