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.
We need to make y=x^3+px and -x=y^3+py a system with 5 real solutions. See
https://www.desmos.com/calculator/ucqo8gwnzy
to see this is equivalent to
0=x^{4}+3px^{3}+3p^{2}x^{2}+(p^{3}+p)x+p^{2}+1
having 2 double roots.
Letting a be the distance left and right of the centerline gives the factoring seen on
https://www.desmos.com/calculator/9dvizui4xk
From which we can solve for a and p
https://www.desmos.com/calculator/piunhgmr2v
a=sqrt(3/2), p=-sqrt(8)
Then plug them back in and solve the quartic (easy since we have the factored form.)
The square root of this solves the octic. And we have one corner of the square.
See lines 7 and 8
https://www.desmos.com/calculator/tlu5ubqauo
Finally apply the distance formula to find the length of half the square's diagonal. Square this then double to get the square's area. (lines 12&13) and simplify.
The area is sqrt(72) so the side length is 72^(1/4)
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Posted by Jer
on 2024-04-14 12:34:25 |
The finish
