Carla Peri INTEGRAL GEOMETRY IN MINKOWSKI PLANE 1 ...

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114(6') n(K;Kr\K o ¥=0) = 2Il(S o +S) +L 0 L(7') n(K,PeK) = 2IlS(8') jJi(K } KnK o =*0,PeK) = 2nS o +L o Land formally coincide with Santalo's formulae for convex sets in Euclideanplane.Let K Q be a convex set of area S 0 and perimeter L 0 and let K lf... ,K n be » convex sets congruent to a convex set K of area 5 and perimeterL.We denote the set K 0 D (K x O ... C\ K n ) by K w and the area of K nby S n .From (7), exactly as in the Euclidean plane [4], we have(9) I S n dK x ... dK n = (2nS) w S 0 .More generally, if S r n denotes the area of the set of those points of K 0that are covered precisely by r sets Kj, from (7) and (8) there follows(10) [ S r n dK x ... dK n = (* ) (2n5) r (2n5 0 + L 0 L*r- r S 0where L* is the perimeter of the convex set K* obtained by reflecting Kin a point.Since the sets Kj are independent,formula (6) yields(11) / dK 1 ...dK n = ( I dK\ =(2n(5 0 +5) + L 0 L*) n .\K 0 n Kj * a} \ {*o n K i **}/Thus the average values of the areas S n and S n are(12) E(S n ) =d3) *(s;> =r(2n5) w S 0(2n(S 0 +S) + L 0 L*)"(")(2nS) r (2n5 0 + L 0 L*) n ' r S,'(2n(S 0 +5) + L 0 L*)"
1155. The isoperimetric inequality.Let K 0 ,K be two convex sets of area S and perimeter L. We denoteby m,- the measure of the set of convex sets congruent to K whose boundarieshave i points in common with dK 0 . Then formulae (6) and (4) becomem 2 + WI4 + m 6 + ... = 4115 + LL* ,2m 2 + 4m A + 6m 6 + ... = 2(L + L*)L .Note that the convex sets with i odd belong to a set of measure zero.Thus the isoperimetric inequality follows(14) L 2 -4IIS>0.This inequality has been already obtained by Guggenheimer [3] in a differentway.Where upon we shall derive a stronger inequality. Following Biberstein[1], we call anticircle of radius r the translate of the image of the isoperimetrixT in a homothety of center O.Consider a convex set K of area S and perimeter L. Let r,- be theradius of a maximal anticircle contained in K and let r e be the radius ofa minimal anticircle that contains K, then r f - < r e . Moreover let r be theradius of an anticircle K 0 such that(15) rj < r < r e .If Wj is the measure of the set of the sets congruent to K whoseboundaries have exactly i points in common with the boundary of K 0 , thenwe have, as above,m 2 + m 4 + ... = 2nCS 0 + S) + L 0 L* ,2m 2 + 4w 4 + ... = 2(L 0 + L*)Lwhere 5 0 and L 0 are the area and the perimeter of K 0 while LQ is theperimeter of the set KQ obtained by reflecting K 0 in a point.From the previous formulae there followsL 0 L 4- L*L -L 0 L*- 2Il(S 0 + S) > 0 .On the other hand, if we fix K and "move" K Q we obtain in thesame wayL 0 L + L 0 L* -L^L -2n(S 0 + 5)> 0 .
Page 1 and 2: REND. SEM. MAT.UNIVERS. POLITECN. T
Page 3 and 4: 109For the properties of these trig
Page 5 and 6: where the absolute value is used be
Page 7: 113pectively. We want to calculate
Page 11: 117REFERENCES[1] O. Biberstein, Ele

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