Question

Consider the identity
\((px + (1-p)y)^2 = Ax^2 + Bxy + Cy^2.\)

Find the minimum of \(\max(A,B,C)\) over \(0 \leq p \leq 1.\)

Answer 1

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

Answer 2

(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)

Answer 3

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

Answer 4

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

Answer 5

@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!

Answer 6

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