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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 can determine s(t):

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

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):

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

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

(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)

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:

$A(t) = p_2$

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

$C(t) = sA(t) + (1-s)B(t) = up_1 + (1-u)p_3$

This requires solving:

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

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 can determine s(t):

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

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):

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

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

(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)

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)

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 can determine s(t):

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

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):

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

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

(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)

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 gives us the following expressions for s and u:

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

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

(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)

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)