A William Lowell Putnam Problem and its Solution

Problem:

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

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

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

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

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

Prove that A \bigcap B = C \bigcap D

(1987 William Lowell Putnam Mathematics Competition)

Solution:

Let z= x+iy. The equations defining A and B are the real and imaginary parts of the equation z^{2}=z^{-1}+3i, and similarly the equations defining C and D are the real and imaginary parts of z^{3}-3iz=1. Hence, for all real x and y, we have (x,y) \in A \bigcap B if and only if z^{2}=z^{-1}+3i. This is equivalent to z^{3}-3iz=1, that is, (x, y) \in C \bigcap D.

Thus, A \bigcap B = C \bigcap D.

Isn’t that an elegant solution? What do you think?

Nalin Pithwa

PS: Solution published in “Complex Numbers from A to …Z” by Titu Andreescu and Dorin Andrica

 

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