Carla Peri INTEGRAL GEOMETRY IN MINKOWSKI PLANE 1 ...

110provided we consider a polygonal line inscribed in T with sufficiently smallsides. Chakerian's proof may be also carried over to the unsymmetric case.Let T 0 , C 0 be the vectors joining the origin 0 of the plane to thepoints corresponding to a = 0 on the isoperimetrix T and on the indicatrixC, respectively.Let T 0 be a fixed curve of finite length L 0 . We assume that T 0 isdefined by the equations x =x(a 0 ), y = y(o 0 ) relative to the arclength o 0as parameter and to the frame (0;T 0 ,C 0 ).Consider a "moving" curve T of length L and arclength a. For"moving" curve T we mean a curve that belongs to the set of the curvescongruent to I\ Let P be a point on T determined by the coordinates(xp,y p ) relative to the frame (0; T 0 , C 0 ) and let C(a) be the unit tangentvector to F at P. We consider the frame (P; T(a), C(a)) "moving" with I\Let X = X(o), Y = Y(o) be the equations of T referred to the arclengtho as parameter and to the frame (P } T(oc), C(ot)).Using the trigonometric relations defined by Guggenheimer in [3] weobtain the following equations for T with respect to the frame (0; T 0 , C 0 )x = x,p + X(o) cm (a, 0) — Y(o) sm (0 ,ai)Then the intersection points of T 0 and r are given by solving the sys­(x(o Q )\ —x p 4- X(o)cm(a,0) — Y(o)sm(0,a)tem!y =y p + X(o)st(0,a) + Y(o) cm (0 ,a),\y(o 0 )=y p +X(o)st(0.,oi)+Y{o)cm(0,a),in the unknowns o 0 , o.Differentiation of these equations yields(dxp =x'do 0 -[X'cm{(x,0)- Y'sm(0,&)]do + [Xsm(0,a)+ Yxcm(a,0)]da\dy. p =y'do 0 -[X'st(0,a) + Y'cm(0,a)]do - [Xcm(0,a)- Yxst(0,a)] da ,where the prime symbol denote differentiation with respect to the arclengthsa 0 or a and x is the unimodular centro-affine curvature of T to the center0. For the differentiation of trigonometric functions see [3].By exterior multiplication we getdV '= dXpAdy p Ada = \-x'X'st(0,a) -x'Y'cm(0,a) ++ y'X'cm(a,0)-y'Y'sm(0,a)\do Q Ado Ada ,