Hostname: page-component-78c5997874-xbtfd Total loading time: 0 Render date: 2024-11-14T07:24:14.823Z Has data issue: false hasContentIssue false
  • English
  • Français

Oblique Pedals

Published online by Cambridge University Press:  03 November 2016

E. H. Neville
Get access Rights & Permissions [Opens in a new window]

Extract

The elementary theorem that, if the triangles ΩAB, ΩPQ are directly similar, so also are the triangles ΩAP, ΩBQ, can be elevated into a general principle:

Given a point Ω and any number of points P1, P2,.. lying on a curve Γ, let Q1, Q2, … be points such that the triangles ΩP1Q1, ΩP2Q2, .. are all directly similar. Then Q1, Q2, … lie on a curve similar to Γ, obtainable by rotating Γ through the angle PrΩQr and applying the scale-factor ΩQr/ΩPr.

Type
Research Article
Information
The Mathematical Gazette , Volume 38 , Issue 325 , September 1954 , pp. 166 - 171
Copyright
Copyright © Mathematical Association 1954

Access options

Get access to the full version of this content by using one of the access options below. (Log in options will check for institutional or personal access. Content may require purchase if you do not have access.)

References

Page 166 of note * “The general isogonal circle,” Math. Gaz., vol. XXXIV (1950), p. 281. I have adopted or adapted Mr. Wilson’s notation for most of the points involved. It is unfortunate that in his figure the angle PFB is obviously obtuse, since the strain on the imagination in equating it to PDC is rendered acute.

Page 167 of note * Modern Geometry of the Triangle, a booklet which Collidge does not use; Gallatly does not distinguish cearly between the theorem and its converse, but is evidently acquainted with both.

Page 167 of note † “An theorem about isogonal conjugates,” Amer. math, Mthly., vol. XX, p. 251.

Page 168 of note * This argument is preferable to the argument CD/CM is equal to CP/CQ and is therefore constant, which uses a ratio of lengths that are not parallel. Similarly, Mr. Wilson’s expression of the ratio BX/CX as b. BP BQ/c.CP CQ can, strictly speaking, prove only that either AX, BY, CZ are concurrent or X, Y, Z are collinear, and it is a nuisance to have to dispose of the second alternative; that is why I suggest below a different proof.

Page 170 of note * To anticipate a charge of circular reasoning, let me remind resders that to prove that, in the usual notation, the locus Y is the auxiliary circle, we have only to show that the tringale SYX is similar to SPM.