If we take the points in different order, the value of the anharmonic ratio will not necessarily remain the same.  The twenty-four different ways of writing them will, however, give not more than six different values for the anharmonic ratio, for by writing out the fractions which define them we can find that (ABCD) = (BADC) = (CDAB) = (DCBA).  If we write (ABCD) = a, it is not difficult to show that the six values are

[formula]

      The proof of this we leave to the student.

If A, B, C, D are four harmonic points (see Fig. 6, p. 22), and a quadrilateral _KLMN_ is constructed such that _KL_ and _MN_ pass through _A_, _KN_ and _LM_ through _C_, _LN_ through _B_, and _KM_ through _D_, then, projecting _A_, _B_, _C_, _D_ from _L_ upon _KM_, we have _(ABCD) = (KOMD)_, where _O_ is the intersection of _KM_ with _LN_.  But, projecting again the points _K_, _O_, _M_, _D_ from _N_ back upon the line _AB_, we have _(KOMD) = (CBAD)_.  From this we have

_(ABCD) = (CBAD),_

or

[formula]

whence a = 0 or a = 2.  But it is easy to see that a = 0 implies that two of the four points coincide.  For four harmonic points, therefore, the six values of the anharmonic ratio reduce to three, namely, 2, [formula], and -1.  Incidentally we see that if an interchange of any two points in an anharmonic ratio does not change its value, then the four points are harmonic.

[Figure 49]

FIG. 49

Many theorems of projective geometry are succinctly stated in terms of anharmonic ratios.  Thus, the anharmonic ratio of any four elements of a form is equal to the anharmonic ratio of the corresponding four elements in any form projectively related to it.  The anharmonic ratio of the lines joining any four fixed points on a conic to a variable fifthpoint on the conic is constant.  The locus of points from which four points in a plane are seen along four rays of constant anharmonic ratio is a conic through the four points. We leave these theorems for the student, who may also justify the following solution of the problem:  Given three points and a certain anharmonic ratio, to find a fourth point which shall have with the given three the given anharmonic ratio. Let A, B, D be the three given points (Fig. 49).  On any convenient line through A take two points B’ and D’ such that AB’/AD’ is equal to the given anharmonic ratio.  Join BB’ and DD’ and let the two lines meet in S.  Draw through S a parallel to AB’.  This line will meet AB in the required point C.

    2 Pappus, Mathematicae Collectiones, vii, 129.

    3 J. Verneri, Libellus super vigintiduobus elementis conicis, etc.
      1522.

    4 Kepler, Ad Vitellionem paralipomena quibus astronomiae pars optica
      traditur. 1604.