China Mathematical Competition (Shaanxi)

Let $z=x+yi$ ($x,y\in \mathbb{R}$), then $$x+yi=a{\cos }^{4}t\cdot i+2\left(\frac{1}{2}+bi\right){\cos }^{2}t\cdot {\sin }^{2}t+(1+ci){\sin }^{4}t.$$ Separating real and imaginary parts, we get $$\begin{array}{rl}x& ={\cos }^{2}t\cdot {\sin }^{2}t+{\sin }^{4}t={\sin }^{2}t,\\ y& =a(1-x{)}^2+2b(1-x)x+c{x}^2\\ & (0\le x\le 1).\end{array}$$

That is, $y=(a+c-2b)x^2+2(b-a)x+a$ ($0\le x\le 1$) $$a(0\le x\le 1)\text{\hspace{1em}(1)}$$ Since $A$, $B$, $C$ are non-collinear, $a+c-2b\ne 0$. So Equation (1) is the segment of a parabola (see the diagram). Furthermore, the midpoints of $AB$ and $BC$ are $D(\frac{1}{4},\frac{a+b}{2})$ and $E(\frac{3}{4},\frac{b+c}{2})$, respectively. So the equation of line $DE$ is $y=(c-a)x+\frac{1}{4}(3a+2b-c)$. (2) Solving Equations (1) and (2) simultaneously, we get $(a+c-2b)(x-\frac{1}{2}{)}^2=0$. Then $x=\frac{1}{2}$, since $a+c-2b\ne 0$. So the parabola and line $DE$ have one and only one common point $P(\frac{1}{2},\frac{a+c+2b}{4})$. Notice that $\frac{1}{4}<\frac{1}{2}<\frac{3}{4}$, so point $P$ is on the segment $DE$ and satisfies Equation (1), as required.

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Alternative solution.

We can solve the problem using the method of complex numbers directly. Let $D$, $E$ be the midpoints of $AB$, $CB$, respectively. Then the complex numbers corresponding to $D$, $E$ are $\frac{1}{2}(z_0+z_1)=\frac{1}{4}+\frac{a+b}{2}i$, $\frac{1}{2}(z_1+z_2)=\frac{3}{4}+\frac{b+c}{2}i$, respectively. So, complex number $z$ corresponding to a point on the segment $DE$ satisfies $$z=\lambda \left(\frac{1}{4}+\frac{a+b}{2}i\right)+(1-\lambda )\left(\frac{3}{4}+\frac{b+c}{2}i\right),0\le \lambda \le 1.$$

Substitute the above expression into the equation of the curve $$z=z_0{\cos }^{4}t+2z_1{\cos }^{2}t\cdot {\sin }^{2}t+z_2{\sin }^{4}t,$$ and separate the real and imaginary parts from both sides to give the following two equations, $$\{\begin{array}{l}\frac{3}{4}-\frac{\lambda }{2}={\sin }^{2}t{\cos }^{2}t+{\sin }^{4}t,\\ \frac{1}{2}[\lambda a+b(1-\lambda )c]=a{\cos }^{4}t+2b{\sin }^{2}t{\cos }^{2}t+c{\sin }^{4}t.\end{array}$$ Eliminating $\lambda$ from the equations, we get $$\begin{array}{rl}\frac{3}{4}(a-c)+\frac{b+c}{2}& =a{\cos }^{4}t+(2b+a-c){\sin }^{2}t{\cos }^{2}t+a{\sin }^{4}t\\ & =a(1-2{\sin }^{2}t{\cos }^{2}t)+(2b+a-c){\sin }^{2}t{\cos }^{2}t\\ & =a+(2b-a-c){\sin }^{2}t{\cos }^{2}t.\end{array}$$ Then $(2b-a-c)({\sin }^{2}t{\cos }^{2}t-\frac{1}{4})=0$. Since $A$, $B$, $C$ are non-collinear, we know that $z_1\ne \frac{1}{2}(z_0+z_2)$. So $2b-a-c\ne 0$. Then ${\sin }^{2}t{\cos }^{2}t={\sin }^{2}t(1-{\sin }^{2}t)=\frac{1}{4}$, that is, $({\sin }^{2}t-\frac{1}{2}{)}^2=0$. Then we have $\frac{3}{4}-\frac{\lambda }{2}=\frac{1}{4}+(\frac{1}{2}{)}^2=\frac{1}{2}$, so $\lambda =\frac{1}{2}\in [0,1]$. That means that the curve and the line $DE$ have one and only one common point, and the complex number corresponding to this common point is $$z=\frac{1}{2}\left(\frac{1}{4}+\frac{a+b}{2}i\right)+\frac{1}{2}\left(\frac{3}{4}+\frac{b+c}{2}i\right)=\frac{1}{2}+\frac{a+c+2b}{4}i.$$