1985 USAMO Problems/Problem 3
Problem
Let denote four points in space such that at most one of the distances is greater than . Determine the maximum value of the sum of the six distances.
Solution
Suppose that is the length that is more than . Let spheres with radius around and be and . and must be in the intersection of these spheres, and they must be on the circle created by the intersection to maximize the distance. We have .
In fact, must be a diameter of the circle. This maximizes the five lengths , , , , and . Thus, quadrilateral is a rhombus.
Suppose that . Then, . To maximize this, we must maximize on the range to . However, note that we really only have to solve this problem on the range to , since is just a symmetrical function.
For , . We know that the derivative of is , and the derivative of is . Thus, the derivative of is , which is nonnegative between and . Thus, we can conclude that this is an increasing function on this range.
It must be true that , so . But, because is increasing, it is maximized at . Thus, , , and our sum is .
~mathboy100
See Also
1985 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
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