= Solution
A cubic <polynomial> on either half is determined by its endpoint positions and endpoint <derivatives>. Use normalized parameters $s=2t$ on the first half and $s=2t-1$ on the second, so that <derivatives> with respect to $s$ are half the <derivatives> with respect to $t$. Evaluation of the <Bernstein basis> at the split gives
$$
H=P(1/2)=\frac{P_0+3P_1+3P_2+P_3}{8},\qquad P'(1/2)=\frac34(-P_0-P_1+P_2+P_3).
$$
If the left <control points> are $L_0,\ldots,L_3$, endpoint positions give $L_0=P_0,L_3=H$. The cubic endpoint-derivative formulas give $3(L_1-L_0)=P'(0)/2$ and $3(L_3-L_2)=P'(1/2)/2$. On the right, $R_0=H,R_3=P_3$, $3(R_1-R_0)=P'(1/2)/2$ and $3(R_3-R_2)=P'(1)/2$. Substitution yields
$$
\boxed{\begin{aligned}
(L_0,L_1,L_2,L_3)&=\left(P_0,\frac{P_0+P_1}{2},\frac{P_0+2P_1+P_2}{4},H\right),\\
(R_0,R_1,R_2,R_3)&=\left(H,\frac{P_1+2P_2+P_3}{4},\frac{P_2+P_3}{2},P_3\right).
\end{aligned}}
$$
These are exact restricted <Bézier curves>, since they match all four endpoint Hermite data of the respective cubics.
The construction is also the midpoint case of <De Casteljau's algorithm>. Form $A_j=(P_j+P_{j+1})/2$ for $j=0,1,2$, then $B_0=(A_0+A_1)/2$, $B_1=(A_1+A_2)/2$, and $H=(B_0+B_1)/2$. The two polygons are $(P_0,A_0,B_0,H)$ and $(H,B_1,A_2,P_3)$. Their common tangent edges satisfy $H-B_0=B_1-H$, as required for equal half-intervals. Thus the triangular midpoint construction simultaneously evaluates the split point and supplies the two exact representations.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-66-bezier-halves.png]
{title=Midpoint construction and the two exact cubic Bézier subcurves}
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