Here's a nice problem to get your teeth into if you have nothing better to do around Christmas...
Given any seven distinct, real numbers. Prove that there are two among them, say a and b, such that
0 < (a-b) < (1+ab)*sqrt(3).
Merry Christmas!
2006-12-22 04:55:55 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
A wee mistake in there... you don't need 7 distinct reals, six is enough.
Hint: Move the (1+ab) factor as a denominator to the iddle, and you might recognize the form of a trig subtraction...
2006-12-24 01:46:46 · update #1
The expression (a-b)/(1+ab) can be restated as Tan(A-B), where a = Tan(A) and b = Tan(B). So, if A-B < 60 degrees, then Tan(A-B) < Sqrt(3). If I have 6 angles with values 0 to 360, then at least 2 of them will differ by less than 60 degrees, as evident by drawing a 6-sided polygon formed by the angles. QED, and thanks for the holiday puzzler, I liked it.
2006-12-26 03:58:02 · answer #1 · answered by Scythian1950 7 · 0⤊ 0⤋
I was kinda guessing trig with the square root of three but I haven't yet made that work. However, I think straight inequalities will work. For instance:
Suppose there are two numbers at least 1/sqrt(3). Let a=max of these and b=min of these, then a-b
Suppose there are two numbers between 0 and 1/sqrt(3). Again let a be the largest and b be the smallest. Then a-b<=a<=1/sqrt(3)
If there are six numbers at least three have the same sign. Amongst any three (say) positive numbers, at least two pair up on one side of 1/sqrt(3), either at least two no smaller or at least no larger.
The inequality can be re-arranged for negative numbers: a-b=(-b)-(-a) and (1+ab)sqrt(3)= (1+ (-a)(-b)) sqrt(3). So if the three that "team up" on the same side of zero are negative numbers, you reverse the smallest/largest definitions for a and b above.
I am going to try your hint and see what comes up.
2006-12-24 18:08:42 · answer #2 · answered by a_math_guy 5 · 0⤊ 0⤋
what?
2006-12-22 13:30:22 · answer #3 · answered by Anonymous · 1⤊ 1⤋