This is the solution I came up with during the test: w.l.o.g we can assume f(0)=0 (or consider the function f(x)-f(0)) and we can also multiply a nonzero constant to f in order to make f(1)=1 First we take $ a=0$ in the condition , we get that if distinct nonzero real numbers b,c,d satisfying $ \frac {1}{b} + \frac {1}{d} = \frac {2}{c}$, then $ \frac {1}{f(b)} + \frac {1}{f(d)} = \frac {2}{f(c)}$. So it reminds me to consider a new function g(x) defined as $ g(x)=\frac {1}{f(\frac {1}{x})}$ ($ g: R/{0}\to R/{0}$). It is easy to check that g is also satisfying the original condition and since if distinct nonzero real numbers b,c,d satisfying $ \frac {1}{b} + \frac {1}{d} = \frac {2}{c}$, then $ \frac {1}{f(b)} + \frac {1}{f(d)} = \frac {2}{f(c)}$, we get that for distinct nonzero real numbers x,y,z, if $ x+y=2z$, then $ g(x)+g(y)=2g(z)$. i.e. for distinct nonzero real numbers x and y, if $ x+y \not = 0$, then $ g(x)+g(y)=2g(\frac {x+y}{2})$ since f(1)=1, we have g(1)=1.using $ g(x)+g(y)=2g(\frac {x+y}{2})$ we can easily get for any nonzero rational number x, $ g(x)=x$. Then it suffices to prove g is monotone. my method is a little tricky, I think. we prove a lemma: for nonzero real numbers x>y>z, if $ \frac {1}{x} + \frac {1}{y} \not = \frac {1}{z}$, then $ g(x)-g(z)$ and $ g(y)-g(z)$ are both positive or negative. lemma's proof: in the condition we take $ a=x$, $ c=y$, $ b=z+\sqrt {(x-z)(y-z)}$, $ d=z-\sqrt {(x-z)(y-z)}$. According to $ x>y>z$, we have $ a>b>c>d$.It is obvious that a and c is not 0.if b or d is 0, then $ z^2 = (x-z)(y-z)$, which leads to $ \frac {1}{x} + \frac {1}{y} = \frac {1}{z}$, a contradiction.So a,b,c,d are distinct nonzero real number,i.e. they satisfies the condition in th original problem. Thus we get $ \frac {f(a) - f(b)}{f(b) - f(c)} + \frac {f(a) - f(d)}{f(d) - f(c)} = 0$ It equals that $ \frac {g(a) - g(b)}{g(b) - g(c)} + \frac {g(a) - g(d)}{g(d) - g(c)} = 0$ i.e. $ {[2g(a)-g(b)-g(d)]}{[2g(c)-g(b)-g(d)]} = {{g(b)-g(d)}^2}$ since $ {g(b)+g(d)}=2g(\frac {b+d}{2})$ we get $ {[g(a)-g(\frac {b+d}{2})]}{[g(c)-g(\frac {b+d}{2})]} = \frac {[g(b)-g(d)]^2}{4}$ So $ [g(x)-g(z)][g(y)-g(z)] = \frac {[g(b)-g(d)]^2}{4} \ge 0$, the lemma is proved To finish the answer, we assume there exist a real number $ x$ such that $ g(x) \not = x$ case 1: $ g(x)>x$, then we can find a rational number y satisfying $ x<y<g(x)$ and a rational number z satisfying $ g(z)>g(x)$ and $ \frac {1}{z} + \frac {1}{y} \not = \frac {1}{x}$. So g(z)=z and g(y)=y. we have z>y>x but $ g(z)-g(x)$ is positive while $ g(y)-g(x)$ is negative, a contradiction. case 2: $ g(x)<x$ is similar to get a contradiction. so the hypothesis of the existence of such an x is wrong.i.e. g(x)=x for all nonzero x.Accoding to the definition of g we get f(x)=x for nonzero real number x. Also f(0)=0, so f(x)=x for all real x, and the original f must be a linear function. Q.E.D.

linboll wrote: we have g(1)=1.using $ g(x) + g(y) = 2g(\frac {x + y}{2})$ we can easily get for any nonzero rational number x, $ g(x) = x$. Sorry .I don't know how to prove $ g(x)=x \forall x \in Q$ : .We can not substitute $ x=y$.Could anyone explain please?

KDS wrote: linboll wrote: we have g(1)=1.using $ g(x) + g(y) = 2g(\frac {x + y}{2})$ we can easily get for any nonzero rational number x, $ g(x) = x$. Sorry.I don't know how to prove $ g(x)=x \forall x \in Q$ : .We can not substitute $ x=y$.Could anyone explain please? I think this is wrong. In particular $g(x)$ can be any function of a kind $g(x)=kx+b$ where $k+b = 1 (x \not = 0)$ It seems, it's possible to prove that $g(x)$ should be only of that kind. In this version, we are to check (?) if functions of a kind $\frac{1}{k \frac{1}{x} +b}$ may do. Put it into original so we may check that only linear functions are suitable. Does anyone know how to solve it with less efforts?

linboll wrote: we can easily get for any nonzero rational number x, $ g(x)=x$. Then it suffices to prove g is monotone. We can tweak linboll's solution slightly to get a correct solution. Indeed, if we have $g(x)+g(y)=2g(\frac{x+y}{2})$ then starting with 2 known values of $g$ we can get a dense subset $D\subset\mathbb{R}^*$ in which $g$ is a linear function, where $\mathbb{R}*=\mathbb{R}\backslash \{0\}$. My recommendation is to take $g(\sqrt{2}+1)=a$, and work from there (with $g(1)=1$), since the subset constructed will exclude 0 (because $\sqrt{2}$ is irrational). Following which, we can finish with his lemma that $g$ is increasing, swapping out $\mathbb{Q}$ for $D$. Conclude first that $g$ is linear, then reason that if $g(x)\neq x$ then we can pick a value of $x$ such that $g(x)=0$, which shouldn't be possible. Thus $g(x)=x$ for all $x$, and we are done.

The problem is same as problem N. 138. of the KöMaL magazine (April 1997.) http://db.komal.hu/KomalHU/feladat.phtml?id=56447

linboll wrote: Thus we get $ \frac {f(a) - f(b)}{f(b) - f(c)} + \frac {f(a) - f(d)}{f(d) - f(c)} = 0$ It equals that $ \frac {g(a) - g(b)}{g(b) - g(c)} + \frac {g(a) - g(d)}{g(d) - g(c)} = 0$ how could you get that?

It is clear that $f $ must be injective. Postcomposing with Möbius transformation $\implies f(0)\to 0,f(1)\to 1,f (\infty)\to\infty. $ Assume that $f(0)=0,f(1)=1,f(\infty)=\infty. $ Claim- $f $ is a ring homomorphism. Proof- Note that $\left [0,2x: x,\infty\right] $ is a harmonic range, so $f(2x)=2f(x).$ We know that $\left[a,b:\frac {(a+b)}{2},\infty\right]$ is a harmonic range. $f\left(\frac {(a+b)}{2}\right)=\frac{f(a)+f(b)}{2}\implies f(a+b)=2f\left (\frac{(a+b)}{2}\right)=f(a)+f(b). $ So $f $ is additive.We can prove that $f(t)=t\,\forall t\,\text {in the prime subfield}$ either $\mathbb Q $ or $\mathbb F_p. $ Using this we prove $f(-x)=-f(x)\,\forall x$ We know $\left[0,-b:\frac {(b^2-b)}{2},\frac {(1-b)}{2}\right]$ is a harmonic range, as $\left [0,b:\frac {(b^2+b)}{2},\frac {(1+b)}{2}\right]. $ Since, $f\left(\frac {(1-b)}{2}\right)=\frac {(1-f(b))}{2},f\left(\frac{(b^2-b)}{2}\right)=\frac {\left(f(b)^2+f(b)\right)}{2}. $ Adding the properties yields $f(b^2)=b^2. $ It is not known if any of these are $=, $ that is possible if $b=0,\pm 1. $ So $f(x^2)=f(x)^2\,\forall x. $ Thus, $f ((a+b)^2)=f(a^2)+2f(ab)+f(b^2)=(f(a)+f(b))^2=f(a)^2+2f(a)(b)+f(b)^2\implies f(ab)=f(a)f(b)\,\forall a,b. $ Hence, $f $ is additive and multiplicative. $f $ is a ring homomorphism. Note- Homogeneous coordinates for $P^1 (k) $ should be in the form $[a,b] $ for $a,b\in K. $ Since, we don't know notation for harmonic range. Let $[x,y: z,w] $ be the 4 points of the harmonic range. Cause $P^1 (k)=k\cup\infty. $