1979 AHSME Problems/Problem 23
Problem 23
The edges of a regular tetrahedron with vertices , and each have length one. Find the least possible distance between a pair of points and , where is on edge and is on edge .
Solution 1
Note that the distance will be minimized when is the midpoint of and is the midpoint of .
To find this distance, consider triangle . is the midpoint of , so . Additionally, since is the altitude of equilateral , .
Next, we need to find in order to find by the Law of Cosines. To do so, drop down onto to get the point .
is congruent to , since , , and are collinear. Therefore, we can just find .
Note that is a right triangle with as a right angle.
As given by the problem, .
Note that is the centroid of equilateral . Additionally, since is equilateral, is also the orthocenter. Due to this, the distance from to is of the altitude of . Therefore, .
Since ,
Simplifying, . Therefore,
Solution by treetor10145
Solution 2 (less overkill)
Notice, like above said, that is the midpoint of and is the midpoint of .
To find the length of , first draw in lines and . Notice that is an altitude of . We find that (since is equilateral), and . Use the properties of 30-60-90 triangles to get . Since is an altitude of a congruent equilateral triangle, .
Notice that is isosceles with . Also, since is the midpoint of base , we can conclude that is an altitude. We can use Pythagorean theorem to get the following (taking into consideration ):
-WannabeCharmander
See also
1979 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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