Seeking The Optimal Cutting Line Of Rigid Boundary Of Tight Convex Bodies In R~3

Convex bodies are the objects of study in the paper. A method of seeking the minimum projected area is given by utilizing the sufficient and necessary condition and existence of the cutting line of the convex body rigid separation. And it also proved that cutting line and projected area reach minimum at the same time. Finally, the minimum cutting line is obtained by using related knowledge of conditional-extremum in the variational method.A method of seeking the minimum projected area:The projections discussed in this paper are all on the positive quadrant set of vertical projection. Set M is a tight convex body in R3. AB is the diameter of M, make AB parallel with the xoy plane by properly rotating. Let S0is the area of P(M) at this time. Let f(P,θ) is the function of chord length between any two points of projection area P(M). f(P,θ)x and f(P,θ)y are functions of the maximum chord length in the directions of x and y respectively. In order to looking for the minimum projected area we mainly utilize rotated method, that is projected area reaches minimum when f(P,θ)x and f(P,θ)y getting minimum value in the process of rotation.Observing the change of chord length when rotating is feasible due to f(P,θ) is continuous variation as rotating proved in the second chapter of this paper. The operating steps are introduced briefly as follows:Firstly, looking for the minimum of f(P,θ)x:Set A’B’is the projection of AB in the xoy plane, CD’is the longest chord in the direction of x in P(M), and A’B’∩C’D’=m. Fixed AB, let M rotate around AB axis, each rotated angle is△θ(△θâ†'0), observe the changes of CD’.While C’,D’close to m point, the length of CD’decreases, the area ofP(M)is also decreasing. CD’achieves the minimum when C or D’no longer gets closer to point m,stop rotating and find the counter images of C, D’,they are C, D∈bdM in M.Secondly, looking for the minimum of f(P,θ)yFixed CD,let M rotate around CD axis, each rotated angle is△θ(Aθâ†'0), observe the changes of A’B’. As the first step, when the chord length of A’B’achieves minimum stop rotating and find the counter images of A’B’, they are A1B1in M, the area of P(M) is S1at this time.Thirdly, compare the value of f1(P,θ)x and f1,(P,θ)y(1) While f1(P,θ)x<f1(P,θ)y,S1is the smallest projected area.(2) While f1(P,θ)xf1(P,θ)y, repeat the steps from the one to three above, until f(P,θ)x and f(P,θ)y are no longer decreasing, the area of P(M)is S2at this time.S2is the answer. No matter how M rotates f(P,θ)x always equals f(P,θ), then M is a ball, this is proved in chapter2,the projected area is the circle surface.Normally the minimum area of P(M)can be found by rotating around at least two fixed axis. Take ellipsoid in2.3.3for example.The boundary, projected lines and planes of convex body together is considered as a cylinder. Cutting line is the intersecting line of cylinder and the convex body, which plus boundary conditions is the conditional extremum of ariational problem.The minimum value of cutting line is obtained by using that method in third chapter of this paper.