F(x)= x^4-4mx²+2, find m, so that f(x) has three max/min points which form

a triangle whose barycenter is at the origin?
so y'= 4x³-8mx =0,
x=0, or x=± √(2m)
let A(0,2, B(√(2m), f(√(2m)) and C( -√(2m), f(-√(2m))
How do I locate the barycenter which satisfies the question? thanks for helping.

f'(x) = 0 = 4x³ - 8mx = 4x(x²-2m)

The three max/min points are:
[0, f(0)] = (0, 2)
[√(2m), f(√(2m))] = (√(2m), 4m²-8m² + 2) = (√(2m), -4m²+2)
[-√(2m), f(-√(2m))] = (-√(2m), -4m²-8m²+2) = (-√(2m), -4m²+2)

Coordinates of the barycenter of a triangle are the arithmetic mean of the coordinates of the vertices of the triangle.
x = [0 + √(2m) - √(2m)]/3 = 0
y = (2 - 4m² - 4m²+4)/3 = (6-8m²)/3 = 0

(6-8m²)/3 = 0
3-4m² = 0
4m² = 3
m² = 3/4
m = √3/2 (discard the negative root since m > 0 to avoid complex coefficients in the function definition)

f(x) = x^4 - 2√3x² + 2

So the triangle is ABC. Barycenter or the centroid of a triangle is the intersection point of its medians.

1). Find mid-point of AB then you can find the linear function that passes through point C and AB/2.
2). Do the same method for either AC or BC and point B or point A.
3). Once you get two linear functions, let both of them intersect at origin to find m-value.

** You only need to find two linear functions, no need to find the third one**

f(√(2m)) = (4m^2)-4m(2m)+2 = -4m^2+2
f(-√(2m)) = -4m^2+2

The barycenter of triangle ABC is (A+B+C)/3.

(A+B+C)/3 = 0

==> (0,2) + (√(2m), f(√(2m)) + ( -√(2m), f(-√(2m)) = 0

==> 2 + f(√(2m)) + f(-√(2m)) = 0

==> 2 + 2*( -4m^2+2) = 0
==> 3 + -4m^2 = 0
==> m = sqrt(3/4)