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PROBLEMS AND SOLUTIONS. 
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2. The demonstration is as follows: We have to show that, starting from an 
arbitrary point 1 in the line x, and constructing in the prescribed manner (as shown 
successively in the figure) the points 2, 3, 4, 5, 6, the last side 61 of the hexagon 
123456 will pass through B. By the construction, we have A, 2, 3 in a line, and 
likewise 0, 4, 5; hence, by Pascal’s theorem, applied to the six points in a pair of 
lines, the points of intersection of the lines (25, 34), (SC, Ao), (A4, (72), that is, the 
points B, 6, 1, lie in a line; which is the required theorem. 
3. More generally suppose that the points A, B, C are not on the lines x, y, z, 
respectively. I remark that it is not in general possible to describe a hexagon such 
that the opposite angles lie in the lines x, y, z, respectively, and the opposite sides 
pass through the points A, B, G, respectively; but if there exists one hexagon (viz., 
a proper hexagon, not a triangle twice repeated), then there exists an infinity of such 
hexagons. 
4. In fact, if it be required to find a polygon, the angles whereof lie in given 
lines respectively, and the sides whereof pass through given points respectively; the 
problem is either indeterminate or admits of only two solutions. If therefore in any 
particular case there are three or more solutions, the problem is indeterminate, and 
has an infinity of solutions. Now, in the above-mentioned case of the three lines 
and the three points, there exist two triangles, the angles whereof lie in the given 
lines, and the sides pass through the given points; and each triangle, taking the 
angles twice over in the same order 123123, is a hexagon satisfying the conditions of 
the problem; hence, if we have besides a proper hexagon satisfying the conditions of 
the problem, there are really three solutions, and the problem is therefore indeterminate. 
5. Suppose that the three lines x, y, z, and also two of the three points, say 
the points A and B, are given; we may construct geometrically a locus, such that, 
taking for C any point of this locus, the problem shall be indeterminate: in fact, 
starting with the point 4, and constructing successively the points 3, 2; taking an 
arbitrary direction for the line 21, and constructing successively the points 1, 6, 5; 
then the intersection of the lines 21 and 54 is a position of the point C: and by 
taking any number of directions for the line 21, we obtain for each of them a different 
position of the point (7; and so construct the locus. 
6. The locus in question is, as will be shown, a line; and if the point A is on 
the line x, and the point B on the line y, then the locus of G will be the line z; 
that is, G being any point of the line z, the problem is indeterminate; which is 
Mr Clifford’s theorem. 
7. To prove this, consider the lines x, y, z, and also the points A, B, G, as 
given; the point 1 is an arbitrary point on the line x, linearly determined by means 
of a parameter u; and for every position of the point 1 we have a corresponding 
position of the point 4; let u be the corresponding parameter for the point 4; the 
series of points 1 is homographic with the series of points 4; that is, the parameters 
u, u are connected by an equation of the form auu' + bu + cu + d = 0, (where of course 
a, b, c, d are functions of the parameters which determine the given lines x, y, z and