# 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