Real constants a, b, c are such that there is exactly one square all of whose vertices lie on the cubic curve y = x3 + ax2 + bx + c. Prove that the square has sides of length 721/4.
First off, with a fair bit of algebra you can translate the cubic to its center of rotation to the origin y=x^3+px where p=(a^2-3b)/3.
I'll assume the square is also centered at the origin, since I see no other way have a square's vertices all lie on the cubic.
Now by symmetry if one vertex is of the square is at (x,x^3+px) the opposite one is at (-x,-x^3-px) and the others are 90 degree rotations: (x^3+px,-x) and (-x^3-px,x)
Now either of these new points must lie on the cubic so we can substitute and get the degree 9 polynomial equation:
x=(x^3+px)^3+p(x^3+px)
One solution is x=0 so we can divide out x and simplify this to the degree 8:
1=x^2(x^2+p)^3+p(x^2+p)
This equation has either 0, 4, or 8 solutions. We can also note this is even and make the substitution z=x^2
1=z(z+p)^3+p(z+p)
This can have 0, 2, or 4 solutions. We want this quartic to have 2 solutions. Then everything can be subbed back to find the solution. That's where I'm stuck for now.
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Posted by Jer
on 2024-04-12 15:23:04 |
A start
