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9‐Point and 7‐Point Configurations in 3‐Space
Author(s) -
Croft H. T.
Publication year - 1963
Publication title -
proceedings of the london mathematical society
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 1.899
H-Index - 65
eISSN - 1460-244X
pISSN - 0024-6115
DOI - 10.1112/plms/s3-13.1.384-s
Subject(s) - mathematics , combinatorics , locus (genetics) , point (geometry) , intersection (aeronautics) , perpendicular , geometry , biochemistry , chemistry , engineering , gene , aerospace engineering
DR G. L. WATSON has kindly pointed out to me that my proof of Lemma 22 of the above article is unclear and wrong. The first paragraph from ‘Let X be another point of the P ‐set’ down to ‘ X lies on Π′ ’ must be replaced by the following: Let X be another point of the P ‐set. Let us suppose first that X does not lie on Π′, the common perpendicular bisector of P 1 P 4 , P 2 P 3 . Then XP 1 ≠ XP 4 and XP 2 ≠ XP 3 . By consideration of Δ XP 1 P 4 , we must have either XP 1 = P 1 P 4 = b or XP 4 = P 1 P 4 = b . Because of the symmetry of the figure, we may assume the former, without loss of generality. Again, by consideration of Δ XP 1 P 2 , we see that either XP 2 = P 1 P 2 = a or else XP 2 = XP 1 = b . We consider these cases separately. Case (i). XP 2 = a . By consideration of Δ XP 2 P 4 , we obtain XP 4 = a , for we have already supposed XP 4 ≠ XP 1 . We now have XP 1 = b , XP 2 = XP 4 = a . Hence X is a point of intersection of two circle‐loci in the plane that perpendicularly bisects P 2 P 4 , one being the locus of points distant b from P 1 , the other that of points distant a from P 2 (or, equivalently, P 4 ). But clearly these circles touch at P 3 , and hence no further such point X can exist. Case (ii). XP 2 = b . By consideration of Δ XP 2 P 3 , we obtain XP 3 = a , for we have already supposed XP 2 ≠ XP 2 . We now have XP 3 = a , XP 1 = XP 2 = b . Hence X is a point of intersection of two circle‐loci in the plane that perpendicularly bisects P 1 P 2 , one being that locus of points distant a from P 3 , the other that of points distant b from P 1 (or P 2 ). As before, these circles are seen to touch, at P 4 , and hence no further such point X can exist. Thus no point X of the set (distant from the P 's) can exist that is not on Π′. We must suppose then that X lies on Π′.