These curious cases would, of course, become objects of research, from the great facilities they afforded for the solutions of the pioblents to which they belonged, and the elegance which they introduced into them ; and partaking in some measure of the nature of problems, as well as of theorems, they formed an intermediate class of proposi tions of great importance, to which, when enunciated in a peculiar manner, the name of porisms was attached.
As an example of the manner in which a porism might be discovered, we shall consider the following problem.
A circle ABC, Plate CCCCLVII. (Fig. 1.) a straight line DE, and a point F, being given in position, to find a point G in the sraight line DE, such that GF, the line drawn from it to the given point, shall be equal to GB, the line drawn from it touching the given circle.
Suppose the point G to be found, and GB to be drawn touching the circle ABC in B; let H be the centre of the circle ABC; join FIB, and let HD be perpendicular to 1)E; and from D draw DL, touching the circle ABC in L, and join HE. Also, from the centre G, with the dis tance GB or GF, describe the circle BKF, meeting HD in the points K and K'.
It is plain that the lines HD and DL are given in posi tion and in magnitude. Also, because GB touches the circle ABC, HBG is a right angle, and G is the centre of the circle BKF ; therefore HB touches the circle BKF, and consequently the square of HB or of 11L is equal to the rectangle K'HK. But the rectangle K'IIK, together with the square of DK, is equal to the square of DH, be cause KK' is bisected in 1); therefore the squares of 111_, and DK are also equal to the square of DII. But the squares of III. and LD are equal to the square of DII ; wherefore, the square of DK is equal to the square of DL, and the line DK to the line DL. But DL is given in mag nitude, therefore DK is given in magnitude, and K is therefore a given point. For the same reason, K' is a given point, and the point F being also given by hypothe sis, the circle BKF is given in position. The point G, therefore, the centre of the circle BKF, is given, which was to be found.
Ilence this construction : Having drawn HD perpen dicular to DE, and DL touching the circle ABC, make DK and DK' each equal to DL, and find by the centre of a circle described through the points K, F and K', that is, let l'K' be joined, and bisected at right angles by the line MN, which meets DE in G; G twill be the point required, or it will be such a point, that if GB be drawn from it, touching the circle ABC, and GF to the given point, GB and BF will be equal to one another.
In this instance, we have a problem which admits in general but of one solution, since only one circle can pass through three given points; yet if the point F which is given should coincide with either of the two points K or K' which are found, it is evident that an infinite number of circles can pass through two given points, and their cen tres will be situated on a right line perpendicular to the middle point of the line which joins them : in this case, then, the problem becomes indeterminate, since any point in that Iine will satisfy the conditions.
The indeterminate case is thus enunciated as a porism : A circle ABC being given by position, and also a straight line DE, which does not cut the circle, a point K may be found such, that if G be any point whatever in the line given, the straight line drawn from G to the point K shall be equal to the straight line drawn from G touching the circle ABC. This is in fact the 66th proposition in Dr. Simson's restoration, slightly altered in its statement.
As another instance of a problem leading to a porism, we will give one which appears to have led to the inven tion of the second porism in the treatise of Simson.
A circle A BC, (Plate CCCCLXVII. Fig. 2) and two points 1) and E, in a diameter of it being given, to find a point F in the circumference of the given circle, from which, if straight lines be drawn to the given points E and D, these straight lines shall have to one another the given ratio of a to p.
Suppose the problem resolved, and that F is found, so that FE has to I'D the given ratio of a to P. Produce El: any how to B, bisect the angle EFD by the line FL, and the angle DFB by the line FM.
Then because the angle EFD is bisected by FL, EL is to LD as EF is to I'D, that is in a given ratio; and as ED is given, each of the segments EL, LD is given, and also the point L.