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Algebra Level 3

Given that $x$ and $y$ are real numbers satisfying $x+y = H(x,y)$ , where $H(x,y)$ denotes the harmonic mean of $x$ and $y$ , how many ordered pairs $(x,y)$ exist?

More than 4, but finitely many 4 No ordered pairs satisfy 1 3 2 This is a yet-to-be-solved Millenium problem, just like the Riemann Hypothesis Infinitely many ordered pairs exist

1 solution

Manuel Kahayon
Jun 16, 2016

$\large x+y = \frac{2}{\frac{1}{x}+ \frac{1}{y}} = \frac{2}{\frac{x+y}{xy}} = \frac{2xy}{x+y}$ .

Multiplying both sides by $x+y$ gives us $(x+y)^2 = 2xy$ , $x^2+2xy+y^2=2xy$ , $x^2+y^2 = 0$

Since all perfect squares are greater than or equal to zero, and can only be equal to zero if the original numbers themselves are zero, this implies that $x=y=0$

But, then, the harmonic mean becomes undefined, since $\frac{2}{\frac{1}{0}+\frac{1}{0}}$ becomes undefined since $\frac{1}{0}$ is undefined.

Moral Lesson: Never forget to check your answer, and never answer based on "instinct"

Or you could write it as:

$(x+y)=\pm\sqrt{ 2xy}$ $\implies t+\dfrac 1t=\pm \sqrt 2\in(-2,2)$ $\small{\left(t=\sqrt{\dfrac xy}\right)}$ While by AM-GM, we know $t+\dfrac 1t$ can never attain values in $(-2,2)$ hence no solution.

Rishabh Jain - 4 years, 12 months ago

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