## zmudz one year ago Consider the identity $$(px + (1-p)y)^2 = Ax^2 + Bxy + Cy^2.$$ Find the minimum of $$\max(A,B,C)$$ over $$0 \leq p \leq 1.$$

1. anonymous

I would expand the left side of that equation. (px + (1-p)y)^2

2. anonymous

(px + (1-p)y)^2 = (px + (1-p) y) * (px + (1-p) y) = p^2*x^2 + 2p(1-p)*xy + (1-p)^2 *y^2 = Ax^2 + Bxy + Cy^2 thus it follows A = p^2 , B = 2p(1-p), C = (1-p)^2 we want to find minimum of Max (A,B,C)

3. anonymous

this means finding the minimum of Max { p^2, 2p(1-p), (1-p)^2} for p in [0,1] plug in some values and see if you can find a pattern

4. anonymous

Using excel I made a spreadsheet and it looks like the minimum of the max of A,B,C is 0.5

5. zmudz

@jayzdd your methodology makes sense to me, but for some reason, the answer isn't 0.5 and I can't know the answer unless I give up on the problem. I have unlimited tries, though. Is there any other way to go about the problem? thanks!

6. zmudz

@jayzdd nevermind, I figured it out. It was 4/9 when p=1/3, or 2/3