A MECHANICAL PROOF TO A GEOMETRIC INEQUALITY OF ZIRAKZADEH THROUGH RECTANGULAR PARTITION OF POLYHEDRA

In this paper is concerned with a machine proof to the following geometric inequality,which was presented by A.Zirakzadeh in 1964:given any triangle ABC and three points P,Q,R,which divide the perimeter of ABC into three equal parts,the perimeter of PQR is equal to or greater than the one half of that of ABC.By representing the Zirakzadeh inequality as radical problem of form f(u,v,w) =(m_1)~(1/2)+(m_2)~(1/2) +(m_3)~(1/2) -3(m_4)~(1/2)≥0,where m_1,m_2,m_3,m_4 are cubic polynomials of u,u,w and(u,v,w) belongs to a polyhedron P in R~3 with 14 vertices,21 edges and 9 faces,we first construct a cube B~* =[-0.1,0.1]~3C P and a polynomial g(u,v,w) of total degree 8 so that f(u,v,w)≥g(u,v,w) and g(u,v,w)≥0 in B~*,then we prove two lemmas on error analysis for testing the positive definiteness of a continuous differentiable function on a rectangular region or a polytope through its values at the vertices,and finally we show the inequality f(u,v,w)0 for(u,v,w)∈P \ B~* by partitioning the polyhedron into a finitely many small rectangular sets which edges are 1/(1280) or larger and the values of f(u,v,w) at the vertices of the rectangular sets exceed the test numbers for positive definiteness.The total computation time for the third step is 159.124 hours on personal computers.We also establish a mechanical method for computing the lower and upper bounds of polynomials and the test numbers of radical expressions over polyhedra in R~3.