Suppose $A$, $B$ $C$ are three non collinear points (A challenging problem) Please help me. I didn't get any idea to solve this challenging problem:
Suppose $A, B, C$ are three non-collinear points corresponding to complex numbers $$z_0 = ai, \;z_1 = \frac{1}{2}+bi,\; z_2 = 1+ci$$ ($a, b$ and $c$ being real numbers), respectively. Prove that the curve $$z = z_0 \cos^4 t + 2 z_1 \cos^2 t \cdot \sin^2 t + z_2 \sin^4 t \qquad (t \in \mathbb R)$$ shares a single common point with the line bisecting $AB$ and parallel to $AC$ and $\Delta ABC$, and find this point.
 A: *

*Note that "the line bisecting $AB$ and parallel to $AC$" is simply the line that connects the bisectors of $AB$ and $BC$.

*Change variable to $u=\sin^2 t$.

*As @Dustan Levenstein remarks, the complex numbers here are just being used as a real vector space; their multiplicative structure is irrelevant.

*Note that the coefficients of $z_0$, $z_1$, $z_2$ sum to $1$, so you can add the same vector to all three points and the curve will translate similarly.

*Thus, by an appropriate affine transformation you can switch to a nicer coordinate system where the coordinates are $A(0,0)$, $B(1,2)$, $C(2,0)$, without changing the expression for the curve.

*You should now be able to prove that the $y$-coordinate reaches $1$ exactly once while $u$ goes from $0$ to $1$. This shows that the curve meets the line in a single point.

*By symmetry, this point can only be the point halfway between the midpoints of $AB$ and $BC$.

*This allows to you find the point in the original complex coordinates.

