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Difference between revisions of "1985 USAMO Problems/Problem 3"

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== Solution ==
 
== Solution ==
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Suppose that <math>AB</math> is the length that is more than <math>1</math>. Let spheres with radius <math>1</math> around <math>A</math> and <math>B</math> be <math>S_A</math> and <math>S_B</math>. <math>C</math> and <math>D</math> 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 <math>AC + BC + AD + BD = 4</math>.
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In fact, <math>CD</math> must be a diameter of the circle. This maximizes the five lengths <math>AC</math>, <math>BC</math>, <math>AD</math>, <math>BD</math>, and <math>CD</math>. Thus, quadrilateral <math>ACBD</math> is a rhombus.
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Suppose that <math>\angle CAD = 2\theta</math>. Then, <math>AB + CD = 2\sin{\theta} + 2\cos{\theta}</math>. To maximize this, we must maximize <math>\sin{\theta} + \cos{\theta}</math> on the range <math>0^{\circ}</math> to <math>90^{\circ}</math>. However, note that we really only have to solve this problem on the range <math>0^{\circ}</math> to <math>45^{\circ}</math>, since <math>\theta > 45</math> is just a symmetrical function.
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For <math>\theta < 45</math>, <math>\sin{\theta} \leq \cos{\theta}</math>. We know that the derivative of <math>\sin{\theta}</math> is <math>\cos{\theta}</math>, and the derivative of <math>\cos{\theta}</math> is <math>-\sin{\theta}</math>. Thus, the derivative of <math>\sin{\theta} + \cos{\theta}</math> is <math>\cos{\theta} - \sin{\theta}</math>, which is nonnegative between <math>0^{\circ}</math> and <math>45^{\circ}</math>. Thus, we can conclude that this is an increasing function on this range.
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It must be true that <math>2\sin{\theta} \leq 1</math>, so <math>\theta \leq 30^{\circ}</math>. But, because <math>\sin{\theta} + \cos{\theta}</math> is increasing, it is maximized at <math>\theta = 30^{\circ}</math>. Thus, <math>AB = \sqrt{3}</math>, <math>CD = 1</math>, and our sum is <math>5 + \sqrt{3}</math>.
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~mathboy100
  
 
== See Also ==
 
== See Also ==

Latest revision as of 20:32, 14 December 2022

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

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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