Tuyển tập đề thi vô địch bất đẳng thức thế giới P6 , tài liệu tham khảo, tài liệu gồm các bài toán bất đẳng thức cực khó, với bài tập hay và cách giải được sưu tầm trên thế giói các bạn có đào sâu kiến thức toán về mảng này, Tai liệu được viết bằng tiêng anh. Chúc các bạn học tốt. | SOLUTIONS 401 respect to the same axis is a segment of length here we assume that a point is segment of zero length . Therefore the perimeter of H is equal to P P and the number of H s sides can take any value from the largest n1 or n2 to n1 n2 depending on for how many axes both basic sets of F and G are sides and not vertices simultaneously. . We will prove a more general statement. Recall that cardinality of a set is for a finite set the number of its element. Ramsey s theorem. Let p q and r be positive integers such that p q r. Then there exists a number N N p q r with the following property if r-tuples from a set S of cardinality N are divided at random into two nonintersecting families a and ft then either there exists a p-tuple of elements from S all subsets of cardinality r of which are contained in a or there exists a q-tuple all subsets of cardinality r of which are contained in ft. The desired statement follows easily from Ramsey s theorem. Indeed let N N p 5 4 and family a consist of quadruples of elements of an N-element set of points whose convex hulls are quadrilaterals. Then there exists a subset of n elements of the given set of points the convex hulls of any its four-elements subset being quadrilaterals because there is no five-element subset such that the convex hulls of any four-element subsets of which are triangles see Problem . It remains to make use of the result of Problem . Now let us prove Ramsey s theorem. It is easy to verify that for N p q 1 N r q r and N p r r one can take numbers p q 1 q and p respectively. Now let us prove that if p r and q r then for N p q r one can take numbers N pi q1 r 1 1 where pi N p 1 q r and q1 N p q 1 r . Indeed let us delete from the N p q r -element set S one element and divide the r 1 -element subsets of the obtained set S into two families family a resp. ft consists of subsets whose union with the deleted element enters a resp. ft . Then either 1 there exists a p1-element subset of S all r
