An Olympiad math problem

[jay_shark]

Given any 7 real numbers , prove that there are two of them x and y such that

0<=(x-y)/(1+xy) <=1/sqrt3

This is a difficult math problem which was the last question on an old olympaid . The solution to this problem is extremely clever and the inequality becomes automatic . See if anyone can figure this out .

don't think too hard :P

[Darryl_P]

Seems pretty straightforward...just a case of the iterated pigeon-hole principle.

From 7 reals there must be 4 on either the interval (-infinity, 0] or [0, infinity)

WLOG we can assume the latter, for if it were the former we could just multiply everything by -1 and the result would be the same.

From these 4 there must be 2 on one of the following intervals:

[0, 1/sqrt3)
[1/sqrt3, 3/sqrt3)
[3/sqrt3, infinity)

Looking at each case individually the algebra falls into place fairly easily.

[arahant]

[ QUOTE ]
Seems pretty straightforward...just a case of the iterated pigeon-hole principle.

From 7 reals there must be 4 on either the interval (-infinity, 0] or [0, infinity)

WLOG we can assume the latter, for if it were the former we could just multiply everything by -1 and the result would be the same.

From these 4 there must be 2 on one of the following intervals:

[0, 1/sqrt3)
[1/sqrt3, 3/sqrt3)
[3/sqrt3, infinity)

Looking at each case individually the algebra falls into place fairly easily.

[/ QUOTE ]

Ah, pigeonhole, how i miss thee...
Been so long since I did this stuff.

[jay_shark]

There is a more elegant way of doing this .

Notice than (x-y)/(1+xy) is similar to tanx-tany/(1+tanxtany)=tan(x-y)

From the pigeonhole principle if you divide 180 into 6 equal segments then two of seven numbers are within 30 degrees of eachother .

[Darryl_P]

n1

[Enrique]

Putting my solution before reading the other posts:

x-y / (1+xy) makes one think immediately of angles.

Tan(a - b) = (Tan(a) - Tan(b))/(1 + Tan(a)Tan(b)).

We can think of the seven real numbers as slopes (tangents).

x-y / (1 + xy) would be the tangent of the angle between two lines. But we have 180 degrees. We can split it in 6 regions of 30 degrees.
Two lines are in one of those regions, hence there angle is at most 30 degrees, hence their slop is at most tan(30) = 1/sqrt(3).

Just pick the bigger slope to make it >=0.

[Enrique]

[ QUOTE ]
From these 4 there must be 2 on one of the following intervals:

[0, 1/sqrt3)
[1/sqrt3, 3/sqrt3)
[3/sqrt3, infinity)

Looking at each case individually the algebra falls into place fairly easily.

[/ QUOTE ]

How do you the algebra for the second interval?
I can prove fairly easily that if two elements are in that interval then (x-y) / (1+xy) <= 3/2 * (1/sqrt(3)). I notice that reaching the upper limit is impossible, but I didn't find quick easy ways to improve on the inequality.
Although I guess it cn be quickly shown with tangents (your intervals pretty much are 0,30 degrees, 30-60 degrees and 60-90 degrees).

[Darryl_P]

[ QUOTE ]
How do you the algebra for the second interval?

[/ QUOTE ]

Just do a change of variable:

Let a = (sqrt3)*x - 2
let b = (sqrt3)*y - 2

and rewrite the inequality in terms of a and b, noting that a and b must be between -1 and 1 in this case.