Let $f(x)$ be a function with the property that $f(x)+f\left(\frac{x-1}{3 x-2}\right)=x$ for all real numbers $x$ other $\operatorname{than} \frac{2}{3}$. What is the sum $f(0)+f(1)+f(2) ?$
With $x=1, \frac{x-1}{3 x-2}=0$, and so the given identity implies $f(1)+f(0)=1$. Since we need to find $f(0)+f(1)+f(2)$, it remains to find the value of $f(2)$.

With $x=2, \frac{x-1}{3 x-2}=\frac{2-1}{6-2}=\frac{1}{4}$. Applying the identity gives $f(2)+f\left(\frac{1}{4}\right)=2$.

With $x=\frac{1}{4}, \frac{x-1}{3 x-2}=\frac{\frac{1}{4}-1}{\frac{3}{4}-2}=\frac{1-4}{3-8}=\frac{3}{5}$. Applying the identity gives $f\left(\frac{1}{4}\right)+f\left(\frac{3}{5}\right)=\frac{1}{4}$.

With $x=\frac{3}{5}, \frac{x-1}{3 x-2}=\frac{\frac{3}{5}-1}{\frac{9}{5}-2}=\frac{3-5}{9-10}=2$. Applying the identity gives $f\left(\frac{3}{5}\right)+f(2)=\frac{3}{5}$.

Let $a=f(2), b=f\left(\frac{1}{4}\right)$, and $c=f\left(\frac{3}{5}\right)$. We have shown that the following three equations hold:

$$
\begin{aligned}
& a+b=2 \\
& b+c=\frac{1}{4} \\
& a+c=\frac{3}{5}
\end{aligned}
$$

Adding the first and third equations gives $2 a+(b+c)=2+\frac{3}{5}=\frac{13}{5}$.

Since $b+c=\frac{1}{4}$, this means $2 a=\frac{13}{5}-\frac{1}{4}=\frac{47}{20}$, and so $f(2)=a=\frac{47}{40}$.

We already know that $f(0)+f(1)=1$, so $f(0)+f(1)+f(2)=1+\frac{47}{40}=\frac{87}{40}$.

Remark: It can be shown that the given condition implies that $f(x)=\frac{-9 x^{3}+6 x^{2}+x-1}{2(-3 x+1)(3 x-2)}$ for all $x$ except $x=\frac{1}{3}$ and $=\frac{2}{3}$ and that $f\left(\frac{1}{3}\right)$ and $f\left(\frac{2}{3}\right)$ are cannot be uniquely determined. However, the values of $f(0), f(1)$, and $f(2)$ are uniquely determined, which is all that was needed for this problem.

## Relay Problems

(Note: Where possible, the solutions to parts (b) and (c) of each relay are written as if the value of $t$ is not initially known, and then $t$ is substituted at the end.)