Just in case someone finds this question using google at some point, and is also curious about the solution for the quadratic case, its solution is similar:

$A(t) = p_2$

$B(t) = (1-t)^2p_1 + 2t(1-t)p_2 + t^2p_3$

$C(t) = sA(t) + (1-s)B(t) = up_1 + (1-u)p_2$

This requires solving:

$$sp_2 + (1-s)((1-t)^2p_1 + 2t(1-t)p_2 + t^2p_3) - up_1 + (1-u)p_2 = 0$$

which, expressed in terms of the control points, is:

$$((1-s)(1-t)^2 - u)p_1 + (s+2t(1-s)(1-t))p_2 + ((1-s)t^2 + u - 1)p_3 = 0$$

If we want these coefficients to become identically zero, we solve the following two identies for *s*:

$$u = (1-s)(1-t)^2 ∧ u = -((1-s)t^2 - 1)$$

which means solving:

$$(1-s)(1-t)^2 + ((1-s)t^2 - 1) = 0$$

which gives us the following expressions for *s* and *u* (after substituting *s* into either of the identities for *u* and solving):

$$s = \frac{2t^2 - 2t}{2t^2 - 2t + 1}$$

$$u = \frac{(t-1)^2}{2t^2 - 2t + 1}$$

(Also note that there are no solutions for curves of order 4 and higher; unlike for quadratic and cubic curves, the ratio between the two distances is not a fixed value for higher order curves, unfortunately)