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> Yeah, that looks like it's going to be messy.

I agree with Olin: the problem starts with three variables and three conditions so, no matter how you look at it, you will end up with a system of equations.

However, you can take certain shortcuts (using your 1st picture): 

$$\begin{align}
R_{23}&=R_3||(R_2+R_o) \tag{1} \\\\
R_{13}&=R_3||(R_1+R_i) \tag{2} \\\\
R_i&=R_1+R_{23} \tag{3} \\\\
R_o&=R_2+R_{13} \tag{4}
\end{align}$$

Then, the input source always sees its \$R_i\$ is series with the equivalent \$R_i\$, while the output voltage is always \$\frac{k}2\$ over \$R_o||R_o^{eq}\$, thus:

$$\begin{align}
I_1&=\dfrac{1}{2R_i} \tag{5} \\\\
I_2&=\dfrac{k}{2R_o} \tag{6} \\\\
I_3&=I_1-I_2=\dfrac12\left(\dfrac{1}{R_i}-\dfrac{k}{R_o}\right) \tag{7}
\end{align}$$

Since the biggest unknown is \$V_x\$, calculate it with (3) in mind:

$$V_x=\dfrac12\dfrac{R_{23}}{R_{23}+R_1}=\dfrac12\dfrac{R_i-R_1}{R_i-R_1+R_1}=\dfrac12\dfrac{R_i-R_1}{R_i}=\dfrac12\left(1-\dfrac{R_1}{R_i}\right) \tag{8}$$

Now calculate \$R_2\$ and \$R_3\$ based on (6) and (7):

$$\begin{align}
R_2&=\dfrac{V_x-\dfrac{k}{2}}{\dfrac{1}{2R_i}}=\left(\dfrac{2V_x}{k}-1\right)R_o=\left(\dfrac{1-\dfrac{R_1}{R_i}}{k}-1\right)R_o=\dfrac{\big((1-k)R_i-R_1\big)R_o}{kR_i} \tag{9} \\\\
R_3&=\dfrac{V_x}{\dfrac12\left(\dfrac{1}{R_i}-\dfrac{k}{R_o}\right)}=\dfrac{2V_xR_iR_o}{R_o-kR_i}=\dfrac{\left(1-\dfrac{R_1}{R_i}\right)R_iR_o}{R_o-kR_i}=\dfrac{(R_i-R_1)R_o}{R_o-kR_i} \tag{10}
\end{align}$$

Since \$R_i\$, \$k\$, and \$R_o\$ are given, both (9) and (10) are dependent on \$R_1\$, only. Now, use (2) and (4) to derive the expression for \$R_1\$:

$$R_o\stackrel{(2,4)}{=}R_2+\dfrac{(R_1+R_i)R_3}{R_1+R_3+R_i} \tag{11}$$

After expanding and collecting the terms:

$$R_1=\dfrac{(R_o-R_2)R_i-(R_2+R_i-R_o)R_3}{R_2+R_3-R_o} \tag{12}$$

Substituting (9) and (10) in (12) takes a few lines of simplifications to give:

$$R_1=\dfrac{k^2R_i^2-(2k-1)R_i^2R_o}{R_o-k^2R_i} \tag{I}$$

At this point, if you want to keep things simple(-ish) then use (13) for \$R_1\$ and then, sequentially, calculate (9) and (10) through substitution. Otherwise, you'll need a few more lines of simplifications to give:

$$\begin{align}
R_2&=\dfrac{R_o^2+(k^2-2k)R_iR_o}{R_o-k^2R_i} \tag{II} \\\\
R_3&=\dfrac{2kR_iR_o}{R_o-k^2R_i} \tag{III} \\\
\end{align}$$

I find it interesting that all of them are divided by \$R_o-k^2R_i\$. Maybe there is something to it, hopefully a simplification but, now, my eyes are getting crossed by the amount of MathJax so, I'll take a break. I'll double check later on but, for now, it looks like I haven't made any mistakes.

A SPICE test with your values confirms both approaches (blue text is active, upper sources output the resistor values):

![confirmation](https://electrical.codidact.com/uploads/XZK4NR82QMVTSKeMPQ6pZAbd)