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        riddles >> putnam exam (pure math) >> Find All Reals Such That...
        
(Message started by: THUDandBLUNDER on Dec 6^th, 2004, 7:27am)

Title: Find All Reals Such That...
Post by THUDandBLUNDER on Dec 6^th, 2004, 7:27am

Determine all real numbers a > 0 for which there exists a nonnegative continuous function f(x) defined on [0,a]
with the property that the region R = {(x,y) ; 0 [smiley=eqslantless.gif] x [smiley=eqslantless.gif] a, 0 [smiley=eqslantless.gif] y [smiley=eqslantless.gif] f(x)}
has perimeter k units and area k square units for some real number k.

Title: Re: Find All Reals Such That...
Post by The Man on Dec 6^th, 2004, 11:17am

Not sure if this is right, but I say no values of a exist... Since a is positive, the perimeter of the region, k, is positive. But, a region of fixed perimeter with maximum area is a circle. So, area<=pi*(radius)^2=pi*(k/(2*pi))^2=(k^2) / (4*pi)...
So, k^2 <= (k^2) / (4*pi)
This is a contradiction for positive values of k.

Title: Re: Find All Reals Such That...
Post by Icarus on Dec 6^th, 2004, 4:32pm

You took a wrong turn in the middle. The area of the region is to be k, not k². And of course k [le] k²/4[pi] only requires that k [ge] 4[pi]

Title: Re: Find All Reals Such That...
Post by Icarus on Dec 6^th, 2004, 6:25pm

But your idea still holds merit. I believe (not sure - my brain is too fogged right now to remember), that the maximum area for fixed perimeter where one side must be flat is a semicircle. This means that k must be [le] the area of a semicircle of perimeter k. Assuming that my foggy brain has not spasmed somewhere, this means that ::[hide]1+8/[pi] <= a[/hide]::. However, I do not have a proof that all such a are possible. For a>2, f(x)=b works for the right b.

Title: Re: Find All Reals Such That...
Post by blah blah on Dec 30^th, 2004, 4:50pm

It does not work for a<=2. The proof is astonishingly simple. Let x0 be the point on the interval where f attains its max, which exists by extreme value thm. Let Q=(x0, f(x0)). Clearly, the perimeter is greater than the distance from the origin to Q + the distance from Q to (a, 0), which is in turn greater than 2*f(x0). Believe it or not, this is tight enough since the area must not exceed a*f(x0) <= 2f(x0) for a<=2.

Title: Re: Find All Reals Such That...
Post by Icarus on Dec 30^th, 2004, 6:18pm

Excellent insight! Since f(x)=2a/(a-2) works for a>2, the answer is all a>2.