Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 66 / 1 / iv

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 givesIf the left control points are $L_0,\dots ,L_3$, endpoint positions give $L_0=P_0,L_3=H$. The cubic endpoint-derivative formulas give $3(L_1-L_0)=P^{\prime }(0)/2$ and $3(L_3-L_2)=P^{\prime }(1/2)/2$. On the right, $R_0=H,R_3=P_3$, $3(R_1-R_0)=P^{\prime }(1/2)/2$ and $3(R_3-R_2)=P^{\prime }(1)/2$. Substitution yieldsThese 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.