1975 USAMO Problems/Problem 2
Contents
Problem
Let denote four points in space and the distance between and , and so on. Show that
Solutions
Solution 1
If we project points onto the plane parallel to and , and stay the same but all decrease, making the inequality sharper. Thus, it suffices to prove the inequality when are coplanar:
Let . We wish to prove that . Let us fix and the length and let vary on the circle centered at with radius . If we find the minimum value of , which is the only variable quantity, and prove that it is larger than , we will be done.
First, we express in terms of , using the Law of Cosines: is a function of , so we take the derivative with respect to and obtain that takes a minimum when
Define and :
Solution 2
Let
It is clear that every other case can be reduced to this. Then, with the distance formula and expanding,
which rearranges to the desired inequality.
Solution 3
Because the distances are all squared, we must only prove the result in one dimension, and then we can just add up the three individual inequalities for the , , and dimension to get the desired result. Let , , , and be the positions of , , , and respectively. Then we must show that,
So we are done.
Solution 4 (Vector bash)
Let , , , correspond to the position vectors of points A, B, C, and D, respectively, with respect to an arbitrary origin O. Let us also for simplicity define , where is the magnitude of vector . Because squares are non-negative, is non-negative for all vectors . Thus, Because dot product is linear, we expand to obtain from which we add to both sides, rearrange, and complete the square to get As and likewise for the others, which is what we wanted to prove.
NOTES:
1. Equality holds when the vector equality holds, which occurs when A, B, C, and D are the vertices of a (planar) parallelogram, in that order.
2. The algebra employed here is almost identical to the algebra used in Solution 3, which means that Solution 3 is just a simplification of this solution.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1975 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.