Solving system to find a curve passing through 3 points A family of curves, depending on parameters (A,B,C) has equation
$$y(t) = A / (1+B*C*t)^{1/B}$$If B=0 or B=1 we have exponential and harmonic cases.
I am looking for a curve of that family that passes through 3 known points.
It looks like a non-linear system to solve.
If the 3 known points are M1(t1,y1=y(t1)), M2(t2,y2=y(t2)) and M3(t3,y3=y(t3)).
The system looks like
$$y1 = A / (1+B*C*t1)^{1/B}$$
$$y2 = A / (1+B*C*t2)^{1/B}$$
$$y3 = A / (1+B*C*t3)^{1/B}$$
How can we solve it for A,B,C ? How many solutions exist ? 
I know some conditions exist for being able to get a solution.
 A: To make life easier, you can first decouple the parameters setting $bc=k$ $$y(t)=a (1+b c t)^{-1/b}\implies y(t)=a (1+k t)^{-1/b}$$ So, the curve going through three points, the equations are $$y_1=a (1+k t_1)^{-1/b}$$  $$y_2=a (1+k t_2)^{-1/b}$$  $$y_3=a (1+k t_3)^{-1/b}$$ We can eliminate $a$ from the first equation $$a=y_1 (1+k t_1)^{\frac{1}{b}}$$ This being replaced in the second equation gives $$b=-\frac{\log(1+kt_2)-\log(1+kt_1)}{\log(y_2)-\log(y_1)}$$ So, what is left is then the third equation in which the only unknown is $k$; for simplicity, write it as $$\log(y_3)=\log(a)-\frac 1 b\log(1+kt_3)$$ Solve it (more than likely using numerical methods  since it does not show analytical solution because highly nonlinear; I am ready to bet that Newton method would converge very fast) and when you have $k$, go back to get $b$, $c$ and $a$.
Defining $$\alpha_1=\log\left(\frac{y_3}{y_2}\right)\qquad \alpha_2=\log\left(\frac{y_1}{y_3}\right)\qquad \alpha_3=\log\left(\frac{y_2}{y_1}\right)$$
the equation to solve for $k$ just write $$f(k)=\alpha_1 \log(1+kt_1)+\alpha_2 \log(1+kt_2)+\alpha_3 \log(1+kt_3)$$ $$f'(k)= \frac{\alpha_1 t_1}{1+k t_1}+\frac{\alpha_2 t_2}{1+k t_2}+\frac{\alpha_3
   t_3}{1+k t_3}$$ $$f''(k)= -\frac{\alpha_1 t_1^2}{(1+k t_1)^2}-\frac{\alpha_2 t_2^2}{(1+k
   t_2)^2}-\frac{\alpha_3 t_3^2}{(1+k t_3)^2}$$ There is a trivial solution $k=0$ which must be discarded. The first derivative cancels  for only one  value of $k=k_*$  since $\alpha_1+\alpha_2+\alpha_3=0$.
$$k_*=\frac{(\alpha_1+\alpha_2)t_3- (\alpha_1t_1+ \alpha_2t_2)} {t_2(t_1-t_3) \alpha_1+t_1(t_2-t_3) \alpha_2}$$
So,  the first guess of Newton method should be $k_0>k_*$. However, since $f(k)$, $f'(k)$ and $f''(k)$ are available and very unexpensive to compute, Halley method could be a good method to replace Newton method.
Edit
I used your test data points $$\left(
\begin{array}{cc}
  t  &  y  \\
 0.2 & 0.5800 \\
 0.4 & 0.2300 \\
 1.1 & 0.0369
\end{array}
\right)$$ and used the last equation. The minimum is at $k_*=1.03081$;  the solution of the equation corresponds to $k=4.09869$. Newton method works very well for this problem. Starting using $k_0=2$, the iterates are $$k_1=4.20319$$ $$k_2=4.09794$$ $$k_3=4.09869$$ which is the solution for six significant figures.
A: Picking up from Claude Leibovici's answer, with $y_4=y_1$ and $t_4=t_1$ for brevity, we have   $b= P_i/Q_i$ for $i=1,2,3$ , where $P_i=\log (1+k t_{i+1})-\log (1+k t_i)$ and $Q_i=\log y_{i+1}-\log y_i.$   Now $P_3=-(P_1+P_2)$ and $Q_3=-(Q_2+Q_1).$ So we have $$P_1/Q_1=P_2/Q_2=(P_1+P_2)/(Q_1+Q_2).$$ Eliminating the denominators and simplifying, we have $$0=P_1Q_2^2+P_2Q_1^2.$$ Substituting  $P_2Q_1$ for $ P_1Q_2$ in this ,we have $$0=P_2Q_1Q_2+P_2Q_1^2= P_2Q_1(Q_2+Q_1)=-P_2Q_1Q_3.$$  Permuting the subscripts we also have $$0=-P_3Q_2Q_1=-P_1Q_3Q_2.$$ I have to leave it here as it is very late for me.
