CHAPTER V - PENCILS OF RAYS OF THE SECOND ORDER
*79. Pencil of rays of the second order defined.* If the corresponding points of two projective point-rows be joined by straight lines, a system of lines is obtained which is called a pencil of rays of the second order. This name arises from the fact, easily shown (§ 57), that at most two lines of the system may pass through any arbitrary point in the plane. For if through any point there should pass three lines of the system, then this point might be taken as the center of two projective pencils, one projecting one point-row and the other projecting the other. Since, now, these pencils have three rays of one coincident with the corresponding rays of the other, the two are identical and the two point-rows are in perspective position, which was not supposed.
[Figure 19]
FIG. 19
*80. Tangents to a circle.* To get a clear notion of this system of lines, we may first show that the tangents to a circle form a system of this kind. For take any two tangents, _u_ and _u’_, to a circle, and let _A_ and _B_ be the points of contact (Fig. 19). Let now _t_ be any third tangent with point of contact at _C_ and meeting _u_ and _u’_ in _P_ and _P’_ respectively. Join _A_, _B_, _P_, _P’_, and _C_ to _O_, the center of the circle. Tangents from any point to a circle are equal, and therefore the triangles _POA_ and _POC_ are equal, as also are the triangles _P’OB_ and _P’OC_. Therefore the angle _POP’_ is constant, being equal to half the constant angle _AOC + COB_. This being true, if we take any four harmonic points, _P__1_, _P__2_, _P__3_, _P__4_, on the line _u_, they will project to _O_ in four harmonic lines, and the tangents to the circle from these four points will meet _u’_ in four harmonic points, _P’__1_, _P’__2_, _P’__3_, _P’__4_, because the lines from these points to _O_ inclose the same angles as the lines from the points _P__1_, _P__2_, _P__3_, _P__4_ on _u_. The point-row on _u_ is therefore projective to the point-row on _u’_. Thus the tangents to a circle are seen to join corresponding points on two projective point-rows, and so, according to the definition, form a pencil of rays of the second order.
*81. Tangents to a conic.* If now this figure be projected to a point outside the plane of the circle, and any section of the resulting cone be made by a plane, we can easily see that the system of rays tangent to any conic section is a pencil of rays of the second order. The converse is also true, as we shall see later, and a pencil of rays of the second order is also a set of lines tangent to a conic section.
*82.* The point-rows _u_ and _u’_ are, themselves, lines of the system, for to the common point of the two point-rows, considered as a point of _u_, must correspond some point of _u’_, and the line joining these two corresponding points is clearly _u’_ itself. Similarly for the line _u_.
*83. Determination of the pencil.* We now show that _it is possible to assign arbitrarily three lines, __a__, __b__, and __c__, of __ the system (besides the lines __u__ and __u’__); but if these three lines are chosen, the system is completely determined._
This statement is equivalent to the following:
_Given three pairs of corresponding points in two projective point-rows, it is possible to find a point in one which corresponds to any point of the other._
We proceed, then, to the solution of the fundamental
PROBLEM. _Given three pairs of points, __AA’__, __BB’__, and __CC’__, of two projective point-rows __u__ and __u’__, to find the point __D’__ of __u’__ which corresponds to any given point __D__ of __u__._
[Figure 20]
FIG. 20
On the line _a_, joining _A_ and _A’_, take two points, _S_ and _S’_, as centers of pencils perspective to _u_ and _u’_ respectively (Fig. 20). The figure will be much simplified if we take _S_ on _BB’_ and _S’_ on _CC’_. _SA_ and _S’A’_ are corresponding rays of _S_ and _S’_, and the two pencils are therefore in perspective position. It is not difficult to see that the axis of perspectivity _m_ is the line joining _B’_ and _C_. Given any point _D_ on _u_, to find the corresponding point _D’_ on _u’_ we proceed as follows: Join _D_ to _S_ and note where the joining line meets _m_. Join this point to _S’_. This last line meets _u’_ in the desired point _D’_.
We have now in this figure six lines of the system, _a_, _b_, _c_, _d_, _u_, and _u’_. Fix now the position of _u_, _u’_, _b_, _c_, and _d_, and take four lines of the system, _a__1_, _a__2_, _a__3_, _a__4_, which meet _b_ in four harmonic points. These points project to _D_, giving four harmonic points on _m_. These again project to _D’_, giving four harmonic points on _c_. It is thus clear that the rays _a__1_, _a__2_, _a__3_, _a__4_ cut out two projective point-rows on any two lines of the system. Thus _u_ and _u’_ are not special rays, and any two rays of the system will serve as the point-rows to generate the system of lines.
*84. Brianchon’s theorem.* From the figure also appears a fundamental theorem due to Brianchon:
_If __1__, __2__, __3__, __4__, __5__, __6__ are any six rays of a pencil of the second order, then the lines __l = (12, 45)__, __m = (23, 56)__, __n = (34, 61)__ all pass through a point._
[Figure 21]
FIG. 21
*85.* To make the notation fit the figure (Fig. 21), make _a=1_, _b = 2_, _u’ = 3_, _d = 4_, _u = 5_, _c = 6_; or, interchanging two of the lines, _a = 1_, _c = 2_, _u = 3_, _d = 4_, _u’ = 5_, _b = 6_. Thus, by different namings of the lines, it appears that not more than 60 different _Brianchon points_ are possible. If we call 12 and 45 opposite vertices of a circumscribed hexagon, then Brianchon’s theorem may be stated as follows:
_The three lines joining the three pairs of opposite vertices of a hexagon circumscribed about a conic meet in a point._
*86. Construction of the pencil by Brianchon’s theorem.* Brianchon’s theorem furnishes a ready method of determining a sixth line of the pencil of rays of the second order when five are given. Thus, select a point in line 1 and suppose that line 6 is to pass through it. Then _l = (12, 45)_, _n = (34, 61)_, and the line _m = (23, 56)_ must pass through _(l, n)_. Then _(23, ln)_ meets 5 in a point of the required sixth line.
[Figure 22]
FIG. 22
*87. Point of contact of a tangent to a conic.* If the line 2 approach as a limiting position the line 1, then the intersection _(1, 2)_ approaches as a limiting position the point of contact of 1 with the conic. This suggests an easy way to construct the point of contact of any tangent with the conic. Thus (Fig. 22), given the lines 1, 2, 3, 4, 5 to construct the point of contact of _1=6_. Draw _l = (12,45)_, _m =(23,56)_; then _(34, lm)_ meets 1 in the required point of contact _T_.
[Figure 23]
FIG. 23
*88. Circumscribed quadrilateral.* If two pairs of lines in Brianchon’s hexagon coalesce, we have a theorem concerning a quadrilateral circumscribed about a conic. It is easily found to be (Fig. 23)
_The four lines joining the two opposite pairs of vertices and the two opposite points of contact of a quadrilateral circumscribed about a conic all meet in a point._ The consequences of this theorem will be deduced later.
[Figure 24]
FIG. 24
*89. Circumscribed triangle.* The hexagon may further degenerate into a triangle, giving the theorem (Fig. 24) _The lines joining the vertices to the points of contact of the opposite sides of a triangle circumscribed about a conic all meet in a point._
*90.* Brianchon’s theorem may also be used to solve the following problems:
_Given four tangents and the point of contact on any one of them, to construct other tangents to a conic. Given three tangents and the points of contact of any two of them, to construct other tangents to a conic._
*91. Harmonic tangents.* We have seen that a variable tangent cuts out on any two fixed tangents projective point-rows. It follows that if four tangents cut a fifth in four harmonic points, they must cut every tangent in four harmonic points. It is possible, therefore, to make the following definition:
_Four tangents to a conic are said to be harmonic when they meet every other tangent in four harmonic points._
*92. Projectivity and perspectivity.* This definition suggests the possibility of defining a projective correspondence between the elements of a pencil of rays of the second order and the elements of any form heretofore discussed. In particular, the points on a tangent are said to be _perspectively related_ to the tangents of a conic when each point lies on the tangent which corresponds to it. These notions are of importance in the higher developments of the subject.
[Figure 25]
FIG. 25
*93.* Brianchon’s theorem may also be applied to a degenerate conic made up of two points and the lines through them. Thus(Fig. 25),
_If __a__, __b__, __c__ are three lines through a point __S__, and __a’__, __b’__, __c’__ are three lines through another point __S’__, then the lines __l = (ab’, a’b)__, __m = (bc’, b’c)__, and __n = (ca’, c’a)__ all meet in a point._
*94. Law of duality.* The observant student will not have failed to note the remarkable similarity between the theorems of this chapter and those of the preceding. He will have noted that points have replaced lines and lines have replaced points; that points on a curve have been replaced by tangents to a curve; that pencils have been replaced by point-rows, and that a conic considered as made up of a succession of points has been replaced by a conic considered as generated by a moving tangent line. The theory upon which this wonderful _law of duality_ is based will be developed in the next chapter.
PROBLEMS
1. Given four lines in the plane, to construct another which shall meet them in four harmonic points.
2. Where are all such lines found?
3. Given any five lines in the plane, construct on each the point of contact with the conic tangent to them all.
4. Given four lines and the point of contact on one, to construct the conic. ("To construct the conic" means here to draw as many other tangents as may be desired.)
5. Given three lines and the point of contact on two of them, to construct the conic.
6. Given four lines and the line at infinity, to construct the conic.
7. Given three lines and the line at infinity, together with the point of contact at infinity, to construct the conic.
8. Given three lines, two of which are asymptotes, to construct the conic.
9. Given five tangents to a conic, to draw a tangent which shall be parallel to any one of them.
10. The lines _a_, _b_, _c_ are drawn parallel to each other. The lines _a’_, _b’_, _c’_ are also drawn parallel to each other. Show why the lines (_ab’_, _a’b_), (_bc’_, _b’c_), (_ca’_, _c’a_) meet in a point. (In problems 6 to 10 inclusive, parallel lines are to be drawn.)