1985 USAMO Problems/Problem 3

Problem

Let $A,B,C,D$ denote four points in space such that at most one of the distances $AB,AC,AD,BC,BD,CD$ is greater than $1$ . Determine the maximum value of the sum of the six distances.

Solution

Suppose that $AB$ is the length that is more than $1$ . Let spheres with radius $1$ around $A$ and $B$ be $S_A$ and $S_B$ . $C$ and $D$ 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 $AC + BC + AD + BD = 4$ .

In fact, $CD$ must be a diameter of the circle. This maximizes the five lengths $AC$ , $BC$ , $AD$ , $BD$ , and $CD$ . Thus, quadrilateral $ACBD$ is a rhombus.

Suppose that $\angle CAD = 2\theta$ . Then, $AB + CD = 2\sin{\theta} + 2\cos{\theta}$ . To maximize this, we must maximize $\sin{\theta} + \cos{\theta}$ on the range $0^{\circ}$ to $90^{\circ}$ . However, note that we really only have to solve this problem on the range $0^{\circ}$ to $45^{\circ}$ , since $\theta > 45$ is just a symmetrical function.

For $\theta < 45$ , $\sin{\theta} \leq \cos{\theta}$ . We know that the derivative of $\sin{\theta}$ is $\cos{\theta}$ , and the derivative of $\cos{\theta}$ is $-\sin{\theta}$ . Thus, the derivative of $\sin{\theta} + \cos{\theta}$ is $\cos{\theta} - \sin{\theta}$ , which is nonnegative between $0^{\circ}$ and $45^{\circ}$ . Thus, we can conclude that this is an increasing function on this range.

It must be true that $2\sin{\theta} \leq 1$ , so $\theta \leq 30^{\circ}$ . But, because $\sin{\theta} + \cos{\theta}$ is increasing, it is maximized at $\theta = 30^{\circ}$ . Thus, $AB = \sqrt{3}$ , $CD = 1$ , and our sum is $5 + \sqrt{3}$ .

~mathboy100

1985 USAMO (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5
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1985 USAMO Problems/Problem 3

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