PROBLEMS AND SOLUTIONS. 
591 
[383 
383] PROBLEMS AND SOLUTIONS. 591 
? rom an 
3 shown 
hexagon 
ne, and 
pair of 
is, the 
points A, B, C). But if the problem be indeterminate, then starting from the point 
1 and constructing the point 4, and again starting from the point 4 and making the 
very same construction, we arrive at the original point 1, that is, u must be the 
same function of u! that u' is of u; and this will be the case if b = c; hence b = c 
is the condition in order that the problem may be indeterminate. 
8. To effect the calculation, take x = 0, y = 0. z—0 for the equations of the lines 
x, y, z respectively; and let (a, /3, 7), (a', /3', 7'), (a", /3", 7") be the coordinates of the 
X, y, z, 
3n such 
be sides 
m (viz., 
of such 
points A, B, G respectively. Let 1 and 4 be given as the intersections of the line 
x = 0 with the lines y — uz= 0, y—u'z = Q, respectively; and assume that for the point 2 
we have y = 0, z—vx = 0, and for the point 3, z = 0, x — wy = 0. Then 1, C, 2 are in 
a line; as are also 2, A, 3 ; 3, B, 4; hence we obtain 
7 "u — ¡3" av — 7 , (3'w — a 
a u ¡3v 7 w 
a given 
ly; the 
in any 
ite, and 
e lines 
therefore, eliminating v and w, we have 
(ay" — a"y) y'uu' — a(3"yu — (a/3'y" — a'/3y" — a"/3'y) u — ¡3" (a'/3 — a/3') = 0. 
The required condition, therefore, is 
3 given 
ng the 
tions of 
tions of 
ninate. 
<x/3"y’ = (x/3'y" — a'/3y" — a"/3'y, or <x/3'y" — ot/3"y — a'/3y" — (x"/3'y = 0 ; 
which is linear in regard to each of the three sets (a, /3, 7), (a', /3', y), (a", /3", y"), 
separately ; that is, two of the points A, B, C being given, the locus of the remaining 
point is a line. In particular, if a = 0, (3' = 0 ; then the equation becomes a'¡3y" = 0, 
and assuming that neither a! = 0, or /3 = 0, then the equation becomes y" = 0, that is, 
its, say 
h that, 
in fact, 
:ing an 
6, 5; 
and by 
lifferent 
A, B being arbitrary points on the lines x = 0, y = 0 respectively, the locus of G is the 
line z = 0. 
9. Mr Clifford’s theorem is clearly its own reciprocal. I do not know the /precise 
analogues of his special form of the theorem; but the analogue of the more general 
theorem stated in (6) is as follows: viz., we may have in the plane n lines x, y, z, ... 
and n points A, B, G,..., such that there exist an infinity of 2n-gons whereof the 
pairs of opposite angles lie in the given lines respectively; and the pairs of opposite 
sides pass through the given points respectively; and if the n lines and n — 1 of the 
1 is on 
line z; 
hich is 
n points be assumed at pleasure, then the locus of the remaining point is a line. It 
is moreover clear by the principle of reciprocity, that if the n points and n — 1 of 
the n lines be assumed at pleasure, then the envelope of the remaining line is a 
point. 
, G, as 
means 
bonding 
4; the 
nneters 
There exists also an analogue in space; viz. we may have n lines x, y, z,... and 
n lines A, B, G, ... such that there exist an infinity of (skew) 2?i-gons whereof the 
pairs of opposite angles lie in the given lines x, y, z,... respectively; and the pairs of 
opposite sides meet in the given lines A, B, G,... respectively. It may be added, that 
if all but one of the 2 n lines x, y, z,...A, B, G... are given, then the ‘six coordi- 
course 
, z and 
nates ’ of the remaining line satisfy a certain linear equation, but I do not stop to 
explain the geometrical interpretation of this theorem.