Susanne
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doc#129 image of the point of tangency of the given line . Thus pencils of tangents to Q are transformed
doc#129 <formul> are two planes intersecting in a line l, tangent to Q at a point P, the two free
doc#129 tangent l<prime>. Hence, thought of as a line in a particular plane <pgr>, any tangent
doc#129 <formul> since it meets <lgr>, and hence every line of <formul> in the <formul> invariant points
doc#129 <lgr> and since it obviously meets every line of <formul> in a single point. The congruence
doc#129 <formul> which are tangent to g. Clearly, any line , l, of any bundle having one of these points
doc#129 of <formul> which is tangent to g at T. A line through two of these points, <formul> and
doc#129 (1,1) curve on Q, meets the image of any line of <formul>, which we have already found
doc#129 points. Hence its image, C<prime>, meets any line of <formul> in <formul> points. Moreover
doc#129 . Moreover, C<prime> obviously meets any line <formul> in a single point. Hence C<prime>
doc#129 observations make it clear that there exist line involutions of all orders greater than
doc#129 To do this we must first show that every line which meets g in a point P meets its image
doc#129 <formul> points cut from C by a general line , l, of the pencil correspond to the point
doc#129 k coincidences, each of which implies a line of the pencil which meets its image. However
doc#129 necessarily a cone, it follows finally that every line through a point, P, of g meets its image
doc#129 the invariant locus must have a multiple line of multiplicity either <formul> or <formul>
doc#129 first possibility requires that there be a line through P which meets g in <formul> points
doc#129 points; the second requires that there be a line through P which meets g in <formul> points
doc#129 each pencil are the multiple secant and the line joining the vertex, P, to the intersection
doc#129 secants. </p><p> Now consider an arbitrary line , l, meeting Q in two points, <formul> and