36 1. Various Ways of Representing Surfaces and Examples a Ia b Ib c = Ic a Ia b Ib Figure 1.22. An orientation preserving isometry with no fixed points is a translation. Rotations are entirely determined by the centre of rotation and the angle of rotation, and so are specified by three parameters. Case 2 : An orientation preserving isometry I with no fixed points is a translation. The easiest way to see that is to use the complex algebraic description. Writing Iz = az + b with |a| = 1, we observe that if a = 1, we can solve az + b = z to find a fixed point for I. Since no such point exists, we have a = 1, hence I : z → z + b is a translation. One can also make a purely synthetic argument for this case we show that the intervals [a, Ia] and [b, Ib] must be parallel and of equal length for every a, b. Indeed, if they fail to be parallel for some a, b, then their perpendicular bisectors intersect in some point c, as shown in Figure 1.22. Since [a, Ia, c] and [b, Ib, c] are isosceles triangles, we have d(a, c) = d(Ia, c) and d(b, c) = d(Ib, c), hence Ic = c since I preserves orientations. But I has no fixed point, and so [a, Ia] and [b, Ib] must be parallel since I is an isometry, d(Ia, Ib) = d(a, b), and hence the quadrilateral [a, Ia, Ib, b] is a parallelogram. It follows that the intervals [a, Ia] are all parallel and of equal length, and so I is a translation. We only require two parameters to specify a translation since the space of translations is two-dimensional, almost every orientation preserving isometry is a rotation, and hence has a fixed point. Case 3 : An orientation reversing isometry which possesses a fixed point is a reflection. Say Ix = x, and fix y = x. Let be the line bisecting the angle formed by the points y, x, Iy. Using the same

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