An Elementary Course in Synthetic Projective Geometry — John Shaqi
An Elementary Course in Synthetic Projective GeometryLehmer, Derrick Norman
Science
An Elementary Course in Synthetic Projective Geometry
Lehmer, Derrick Norman
Geometry, Projective
*62. Determination of the locus.* We now show that _it is possible to
assign arbitrarily the position of three points, __A__, __B__, and __C__,
on the locus (besides the points __S__ and __S’__); but, these three
points being chosen, the locus is completely determined._
*63.* This statement is equivalent to the following:
_Given three pairs of corresponding rays in two projective pencils, it is
possible to find a ray of one which corresponds to any ray of the other._
*64.* We proceed, then, to the solution of the fundamental
PROBLEM: _Given three pairs of rays, __aa’__, __bb’__, and __cc’__, of two
protective pencils, __S__ and __S’__, to find the ray __d’__ of __S’__
which corresponds to any ray __d__ of __S__._
[Figure 12]
FIG. 12
Call _A_ the intersection of _aa’_, _B_ the intersection of _bb’_, and _C_
the intersection of _cc’_ (Fig. 12). Join _AB_ by the line _u_, and _AC_
by the line _u’_. Consider _u_ as a point-row perspective to _S_, and _u’_
as a point-row perspective to _S’_. _u_ and _u’_ are projectively related
to each other, since _S_ and _S’_ are, by hypothesis, so related. But
their point of intersection _A_ is a self-corresponding point, since _a_
and _a’_ were supposed to be corresponding rays. It follows (§ 52) that
_u_ and _u’_ are in perspective position, and that lines through
corresponding points all pass through a point _M_, the center of
perspectivity, the position of which will be determined by any two such
lines. But the intersection of _a_ with _u_ and the intersection of _c’_
with _u’_ are corresponding points on _u_ and _u’_, and the line joining
them is clearly _c_ itself. Similarly, _b’_ joins two corresponding points
on _u_ and _u’_, and so the center _M_ of perspectivity of _u_ and _u’_ is
the intersection of _c_ and _b’_. To find _d’_ in _S’_ corresponding to a
given line _d_ of _S_ we note the point _L_ where _d_ meets _u_. Join _L_
to _M_ and get the point _N_ where this line meets _u’_. _L_ and _N_ are
corresponding points on _u_ and _u’_, and _d’_ must therefore pass through
_N_. The intersection _P_ of _d_ and _d’_ is thus another point on the
locus. In the same manner any number of other points may be obtained.
*65.* The lines _u_ and _u’_ might have been drawn in any direction
through _A_ (avoiding, of course, the line _a_ for _u_ and the line _a’_
for _u’_), and the center of perspectivity _M_ would be easily obtainable;
but the above construction furnishes a simple and instructive figure. An
equally simple one is obtained by taking _a’_ for _u_ and _a_ for _u’_.
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