1987 Putnam A1

Curves $A,B,C$ and $D$ are defined in the plane as follows:

$$

\begin{aligned}

A &= \left\{ (x,y): x^2-y^2 = \frac{x}{x^2+y^2} \right\}, \\

B &= \left\{ (x,y): 2xy + \frac{y}{x^2+y^2} = 3 \right\}, \\

C &= \left\{ (x,y): x^3-3xy^2+3y=1 \right\}, \\

D &= \left\{ (x,y): 3x^2 y - 3x - y^3 = 0\right\}.

\end{aligned}

$$

Prove that $A \cap B = C \cap D$.

曲线 $A、B、C$ 和 $D$ 在平面中定义如下：

$$

\开始{对齐}

A &= \left\{ (x,y): x^2-y^2 = \frac{x}{x^2+y^2} \right\}, \\

B &= \left\{ (x,y): 2xy + \frac{y}{x^2+y^2} = 3 \right\}, \\

C &= \left\{ (x,y): x^3-3xy^2+3y=1 \right\}, \\

D &= \left\{ (x,y): 3x^2 y - 3x - y^3 = 0\right\}。

\结束{对齐}

$$

证明 $A \cap B = C \cap D$。