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Math Problems - Riyuran - 2010-11-26
If a and b are the sides of the rectangle and r is the radius of the circle, then:
r = sqrt(a^2+b^2)/2
Then the area of the circle is:
pi*r^2 = pi*(a^2+b^2)/4
Replacing b with 200/a:
Area = (pi/4)(a^2+(200/a)^2) = (pi/4)(a^2+40000/a^2)
Differentiating area wrt a:
(pi/4)(2a-80000/a^3)
Setting this equal to zero, we can get rid of the pi/4 and are left with:
2a-80000/a^3 = 0
Getting a common denominator & solving for a:
(2a^4-80000)/a^3 = 0
2a^4-80000 = 0
a = sqrt(200)
Which makes sense because the area of the circle would be minimized when radius is minimized. And radius is minimized when a = b (i.e. a square).
Edit: Whoops. Somehow I didn't see the post above mine. O.o He's right too!
EDIT2: Anyone wanna help me out with a math problem I got stuck on?
Find all integers n with the property that
![[Image: 32vf5jo.png]](http://mathurl.com/32vf5jo.png)
is also an integer.
I spent like 2 hours trying to algebraically manipulate it and practically got nowhere. I then used my calculator to find out that n=49 is a solution. I haven't tried finding any other solutions, as "I used my calculator" is not a mathematical proof at all.
Math Problems - Hazzy - 2010-11-26
29=6; 49=4; 181=2;
Three solutions. Who needs a proof?
This problem is a mother fucker. I hate it.
Math Problems - Riyuran - 2010-11-26
Excel hoar. L>Proof, not brute force! 
EDIT: Figured it out with Wolfram and Hazzy's help.
Set the expression equal to b.
Used a ton of algebraic manipulation (moved terms around, squared both sides twice) to get rid of the sqrts, then solved for n.
Got n = (b^2)/4 + 720/b^2.
Plugged in integer values for b and saw which ones outputted integer n values.