Discovering and proving geometric inequalities by CAS (pdf)

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and y − v =0, i.e., that A − D = B − C. Formal verification Use R::=Q[xyuvabcdef]; I:=Ideal((x-a)^2+y^2-b^2,(x-u)^2+(y-v)^2-c^2,u^2+v^2-d^2,x^2 +y^2-e^2,(u-a)^2+v^2-f^2,a^2+b^2+c^2+d^2-e^2-f^2); NF((x-u-a)^2+(y-v)^2,I); confirms that normal form equals 0 and a quadrilateral which fulfils a 2 +b 2 +c 2 +d 2 = e 2 +f 2 is a parallelogram. The statement is true. Remark: If we would enter x − u − a =0andy − v =0, instead of the ”equivalent” condition (x − u − a) 2 +(y − v) 2 =0, which also characterizes a parallelogram, we have failed. We get Use R ::= Q[xyuvabcdefts]; I:=Ideal((x-a)^2+y^2-b^2,(x-u)^2+(y-v)^2-c^2,u^2+v^2-d^2, x^2+y^2-e^2,(u-a)^2+v^2-f^2,a^2+b^2+c^2+d^2-e^2-f^2,(x-u-a)t-1); NF(1,I); the result 1. What is the difference between the conditions (x − u − a) 2 +(y − v) 2 =0 and x − u − a =0∧ y − v =0? ’The only difference‘ consists in the fact that whereas in a real case both conditions are the same, in the case of complex numbers is (x − u − a) 2 +(y − v) 2 =(x − u − a + i(y − v))(x − u − a − i(y − v)), i.e., we get two conditions x − u − a + i(y − v) =0 ∨ x − u − a − i(y − v) =0! This example should be instructive. We have always to consider such conditions which the system requires (and mostly yields). Any ”changes” of these conditions, despite they describe the reality properly, mostly lead to the failure. Solving inequalities We will study some geometric inequalities discovered and proved by means of Buchberger’s algorithm for computing Groebner bases of ideals. However we should realize that we are working in an algebraic closed field, in our case in the field of complex numbers which, as known, can not be ordered. Hence we can not use the signs > or
Inequality between diagonals of a quadrilateral The following statement is a generalization of the parallelogram law on an arbitrary quadrilateral. It holds: Given a quadrilateral ABCD with lengths of sides a, b, c, d and diagonals e, f. Then a 2 + b 2 + c 2 + d 2 ≥ e 2 + f 2 . (6) The sign of equality in (6) is attained if and only if ABCD is a parallelogram. Let a = |AB|,b = |BC|,c = |CD|,d = |DA|,e = |AC|,f = |BD| and A =[0, 0],B = [a, 0],C =[x, y],D =[u, v]. Then Figure 6: In a quadrilateral ABCD the inequality a 2 + b 2 + c 2 + d 2 ≥ e 2 + f 2 holds b = |BC|⇔h1 :(x− a) 2 + y2 = b2 , c = |CD|⇔h2 :(x− u) 2 +(y − v) 2 = c2 , d = |DA| ⇔h3 : u2 + v2 = d2 , e = |AC| ⇔h4 : x2 + y2 = e2 , f = |BD|⇔h5 :(u− a) 2 + v2 = f 2 . To prove (6), we will try to express the difference (a2 + b2 + c2 + d2 ) − (e2 + f 2 )insuch a form from which non - negativity of (a2 + b2 + c2 + d2 ) − (e2 + f 2 ) would follow. We introduce a new variable k and put hence let h6 :(a 2 + b 2 + c 2 + d 2 ) − (e 2 + f 2 ) − k =0. (a 2 + b 2 + c 2 + d 2 ) − (e 2 + f 2 )=k, (7) In the ideal I =(h1,h2,...,h6) we eliminate dependent variables b, c, d, e, f to express k in terms of independent variables x, y, u, v, a. We get in CoCoA using the command Elim Use R::=Q[xyuvabcdefk]; J:=Ideal((x-a)^2+y^2-b^2,(x-u)^2+(y-v)^2-c^2,u^2+v^2-d^2, x^2+y^2-e^2,(u-a)^2+v^2-f^2,a^2+b^2+c^2+d^2-e^2-f^2-k); Elim(b..f,J); or in Derive using the command GROEBNER BASIS
Page 1 and 2: Introduction DISCOVERING AND PROVIN
Page 3 and 4: Given a parallelogram with the leng
Page 5: Figure 4: In a trapezoid b 2 + d 2
Page 9 and 10: with the equality if and only if Π
Page 11: References Figure 9: A classical pr

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