On Packing And Covering Of Triangles With Squares

Let C,Dn（n = 1,2,...）be plane convex polygons.We say that the sequence {Dn}permits a covering of C if C C（?）∪ Dn.We say that {Dn} can be packed into C if C（?）∪Dn and for arbitrary i,j ∈ {1,2,· · · },i≠ j,int Di∩ int Dj = φ.In particular,a covering or packing of C with {Dn} is called parallel if there is a side of C such that each Dn has a side parallel to this side of C.Denote by A（C）the area of a plane convex body C.Let C,D be plane convex bodies,D1,D2,...be the homothetic copies of D.Let f（C,D）=min {f:any sequence {Dn}permits a parallel covering of C provided the total area of Dn is not less than f·A（C）}.Let p（C,D）= max {p:any sequence {Dn} permits a parallel packing of C provided the total area of Dn is not exceed p · A（C）}In chapter 1 we consider the parallel covering of isosceles triangles with sequences of squares and get the following results:Any sequence of squares permits a parallel covering of an isosceles triangle with base length 1 and with height2^1/2/2 provided the total area of the squares is not less than 1.Any sequence of squares permits a parallel covering of an isosceles triangle with base length 1 and with height2^1/2 provided the total area of the squares is not less than 2.Let Th be an isosceles triangle with base length 1 and with height h.Suppose that a side of a square S is parallel to the base of Th.if h<2^1/2/2 then f（Th,S）=2/h;2^1/2/2≤ h<1,then f（Th,S）= 4h;if1≤h<2^1/2,then f（Th,S）= 4/h;if h≥2^1/2,then f（Th,S）= 2h.Let T = ABC be an arbitrary triangle with base AB = 1,with height h and with the ∠CBA ≤ ∠CAB≤ And let Th = ABC’ be the isosceles triangle with base length 1 and with height h.If a side of a square S is parallel to the base of T.Then f（T,S）= f（Th,S）.In chapter 2 we consider the parallel packing of isosceles triangles with sequences of squares and get the following results:Any sequence of squares can be parallel packed into an isosceles triangle with base length 1 and with height2^1/2/2 provided the total area of the squares does not exceed 3-22^1/2.Let T’1 = ABC be an arbitrary triangle with base AB = 1,with height2^1/2/2,and with the∠CBA ≤∠CAB≤π/2 a side of a square S is parallel to the base of T’1,then p（T’1,S）=62^1/2-8.Any sequence of squares can be parallel packed into an isosceles triangle with base length 1 and with height 1 provided the total area of the squares does not exceed 2/9Let T’2 = ABC be an arbitrary triangle with base AB = 1,with height 1,and with the ∠CBA≤∠CAB≤π/2.If a side of a square S is parallel to the base of T’2,then p（T’2,S）= 4/9.