How can we find a function $g(x)$ such that, for any given value of $x$, the point $(x, g(x))$ is the arithmetic average of some two points on the graph of $f(x) = \frac{1}{x}$, where the distance between these two points is exactly $1$?
At first, I attempted to approach the problem by creating a circle with a center at $(x, \frac{1}{x})$ and a radius of $1$, hoping to find the intersection point. However, this approach quickly became overly complicated.
For example: for the point $(a, f(a))$, I derived the following equations: $(x - a)^2 + \left(\frac{1}{x} - \frac{1}{a}\right)^2 = 1$. Unfortunately, due to the nature of these equations (4th degree), expressing $x$ in terms of $a$ is chanalging.
Let $(a,1/a)$ and $(b,1/b)$ be two points on the graph of $f(x)=1/x$ such that the distance between them is $1$. Then $$ (a-b)^2+\left(\frac{1}{a}-\frac{1}{b}\right)^2=1. \tag{1} $$ If $(x,y)$ is the arithmetic average of those points, then $$ \begin{cases} x=\frac{a+b}{2}, \\ y=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{x}{ab}, \end{cases} \quad\implies\quad \begin{cases} a+b=2x, \\ ab=\frac{x}{y}. \tag{2} \end{cases} $$ Let's now rewrite $(1)$ in terms of $a+b$ and $ab$: \begin{align} (1)\implies&(a-b)^2\left(1+\frac{1}{a^2b^2}\right)=1 \\ \implies& [(a+b)^2-4ab]\left(1+\frac{1}{a^2b^2}\right)=1 \\ \overset{(2)}{\implies}&\left(4x^2-\frac{4x}{y}\right)\left(1+\frac{y^2}{x^2}\right)=1 \\ \implies&\left(1-\frac{1}{xy}\right)\left(x^2+y^2\right)=\frac{1}{4}. \tag{3} \end{align} Eq. $(3)$ defines implicitly the function $y=g(x)$ with the desired property.