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110    T.-H. Lˆ e and A. Galligo
                           Corollary 15. To detect an oval of the double locus in the box [0; 1] × [0; 1],itis
                           sufficient to calculate the critical points of the parameterization. This provides an
                           economy of calculation.

                           Example 16. a =3,b =5,c =2,d = −4,e =1,f = −2.

                           The surface is defined by the parameterization:
                                     ⎧
                                     ⎨ x = t(1 − u) +6tu(1 − u) − 8(1 − t)u(1 − u)
                                                 2
                                       y =(1 − t)u +10tu(1 − u)+2(1 − t)u(1 − u)
                                                 2
                                     ⎩
                                       z =(1 − t)(1 − u) +4tu(1 − u) − 4(1 − t)u(1 − u)
                                                      2
                           We have that:
                             C(t, u) = 4(234476t u − 446468t u − 215277tu + 42550u + 212531t 2
                                              2 2
                                                          2
                                                                      2
                                                                                2
                                          + 496138tu − 131320u − 281358t + 92806)
                                       M x
                                            =(−9u +9u − 4)t − 86u +35u − 1+54u    3
                                                                   2
                                                   2
                                      C(t, u)
                                       M y
                                            =(11 + 11u − 22u)t − 7+7u − 17u +17u
                                                                             2
                                                       2
                                                                       3
                                      C(t, u)
                                        M z
                                             =(2u +9u − 4)t − 85u + 125u − 43u − 1
                                                                  3
                                                                         2
                                                  2
                                      C(t, u)
                           Therefore the three affine critical points are the roots of the followed system:

                                                73u − 180u + 148u − 39 = 0
                                                   3
                                                          2
                                                      146     221    85
                                                  t =    u −     u +
                                                          2
                                                      11      11     11
                           The system above has only one root (t 0 ≈ 0.72427400,u 0 ≈ 0.5442518227) in the
                           box [0; 1] × [0; 1], i.e the surface has only one critical point E =(t 0 ,u 0 ) in the box
                           [0; 1] × [0; 1]. Therefore the surface has the pinch point P = Φ(E).
                              We calculate the points of C(t, u) on the borders, i.e. the intersections of C(t, u)
                           and of the lines (u =0), (u =1), (t =0), (t =1) in the parameter space. We
                           obtain 4 points: (t 1 ≈ 0.7002714999,u 1 =0), (t 2 ≈ 0.6235730217,u 2 =0),
                           (t 3 =1,u 3 ≈ 0.4402786871), (t 4 =1,u 4 ≈ 0.8820099327).
                              Now, we look for the points on C(t, u) in the box [0; 1]×[0; 1] that correspond to 4
                           points (t i ,u i ) ,i =1,..., 4 in order to detect the segment of C(t, u) corresponding to
                           the self-intersections of the surface. (Two points on C(t, u) are called corresponding
                           if their images by Φ are coincident on the singular locus of the surface. We also note
                           that two points in the parameter space satisfying this condition lie on C(t, u).Wesee
                           that a critical point does not have any corresponding point except itself). Hence, for
                           i =1,..., 4, to find a point (t, u) ∈ [0; 1] corresponding to the point (t i ,u i ),we
                                                              2
                           resolve the system Φ(t, u)= Φ(t i ,u i ) in variables (t, u) in [0; 1] . We obtain only
                                                                                2
                           the point (t ≈ 0.6632643380,u ≈ 0.1700120872) which correspond to the point
                           (t 4 ,u 4 ). We denote these two points by B and A.
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