Discovering and proving geometric inequalities by CAS (pdf)

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2 2 2 2 2 2 2 GROEBNER_BASIS((x-a) + y - b ,(x-u) + (y-v) - c , u + 2 2 2 2 2 2 2 2 2 2 2 2 2 v - d , x + y - e ,(u-a) + v - f , a + b + c + d - e - 2 f - k, [b, c, d, e, f, a, x, y, u, v, k]) the polynomial x 2 + y 2 − 2xu + u 2 − 2yv + v 2 − 2xa +2ua + a 2 − k. This polynomial can be expressed in the form of the sum of squares k =(x − u − a) 2 +(y − v) 2 . (8) Substitution of k from (8) into (7) gives the following identity a 2 + b 2 + c 2 + d 2 − e 2 − f 2 =(x − u − a) 2 +(y − v) 2 , (9) which implies the inequality (6). The equality in (6) is attained if and only if ABCD is a parallelogram as we could see from the parallelogram law. Instead of (6) we can write (9) which describes the situation better including the case of equality. By (9) we rediscovered the theorem which is ascribed to L. Euler [8]: Given a skew quadrilateral ABCD. DenoteP, Q the midpoints of diagonals AC and BD respectively. Then |AB| 2 + |BC| 2 + |CD| 2 + |DA| 2 −|AC| 2 −|BD| 2 =4|PQ| 2 . (10) A close inspection shows that the relation (10) is equal to (9). Figure 7: In a skew quadrilateral |AB| 2 +|BC| 2 +|CD| 2 +|DA| 2 = |AC| 2 +|BD| 2 +4|PQ| 2 holds General case The inequality (6) is a special case of the following theorem which was published in 1980 by L. Gerber [6]: Let Π=P0,P1,...,Pn−1 be a closed skew n-gon in the Euclidean space E N . Then n−1 k=0 |PkPk+2| 2 n−1 2 π ≤ 4cos n k=0 |PkPk+1| 2 , (11)
with the equality if and only if Π is a planar affine-regular n-gon. For n = 4 we get the relation (6) since an affine image of a square is a parallelogram. Another generalization of the inequality (11) was published in 1990 by P. Pech, see [11]. It is as follows: Let Π=P0,P1,...,Pn−1 be a closed skew n-gon in the Euclidean space E N and let Pn+j = Pj, for j =0, 1,... Then for all p =0, 1,...,n− 1 n−1 2 π sin |PkPk+p| n k=0 2 ≤ sin 2 (p π n ) n−1 k=0 with the equality if and only if Π is a planar affine-regular n-gon. For p = 2 we get from (12) the inequality (11). Euler’s inequality |PkPk+1| 2 , (12) In 1765 L. Euler [5] published the relation (13) which expresses the distance d of the incenter and circumcenter of an arbitrary triangle where r is the circumradius and p is the inradius. d = r(r − 2p) , (13) In [10] it is given that H. Wieleitner [14] mentions, that the relation (13) was published Figure 8: Euler’s relation: d = r(r − 2p) even earlier by W. Chapple from England in 1746. The following inequality, which is a consequence of (13), is r ≥ 2p , (14) called Euler’s inequality [4]. It is obvious that the equality in (14) is attained iff a triangle is equilateral, because only in this case the circumcenter and incenter coincide and d =0 in (13). We will prove the Euler’s inequality (14) by computer. We will proceed in a similar way as W. Koepf [7] without introducing a coordinate system. Let ABC be an arbitrary triangle with the lengths of sides a, b, c. Denote by r and p its
Page 1 and 2: Introduction DISCOVERING AND PROVIN
Page 3 and 4: Given a parallelogram with the leng
Page 5 and 6: Figure 4: In a trapezoid b 2 + d 2
Page 7: Inequality between diagonals of a q
Page 11: References Figure 9: A classical pr

Discovering and proving geometric inequalities by CAS (pdf)

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