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- W2079297260 abstract "Pl(1, 0), P1 (- 1, 0); P2(o, 0), P2 (-a, 0); P3(r, 0), P3 (-r, 0). We know that the harmonic conjugate of (o, 0) with respect to (1, 0), (-1, 0) is (r-1, 0) and since a2 -1, we have a-1 . i.e. P2, P' are harmonic conjugates with respect to P1, P'; so are P3, P3'. The harmonic conjugate of P2 with respect to P3, P3 iS [7-- (-Tr+ (or-r)1)-', 0] by formula 2.9 of [I]. We can easily check that (o- -r)-1 =-(o-r)/2 and (o + r)-1 -(+ r)/2. Using these relations we get the coordinates of the harmonic conjugate of P2 to be (-o, 0) i.e. P2, P2' are indeed harmonic conjugates with respect to P3, P3', which proves the first part of our theorem. Let us now prove the converse. Without loss of generality we can take the given line as y = 0, 0 as (0, 0) and P1 as (1, 0). Let P2 and P3 be (a, 0) and (,B, 0). P', P2 and P3 will then be (-1, 0), (-a, 0) and (-j, 0). Since (a-', 0) is the harmonic conjugate of (a, 0) with respect to (1, 0), (-1, 0), we must have a-'= -a or a2= -1. Similarly 2=2 -1." @default.
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- W2079297260 date "1968-04-01" @default.
- W2079297260 modified "2023-10-18" @default.
- W2079297260 title "The quaternion group in plane geometry" @default.
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- W2079297260 doi "https://doi.org/10.1090/s0002-9939-1968-0234346-9" @default.
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