% This paper has been transcribed in Plain TeX by
% David R. Wilkins
% School of Mathematics, Trinity College, Dublin 2, Ireland
% (dwilkins@maths.tcd.ie)
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% Trinity College, 2000.

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\centerline{\Largebf ON ``GAUCHE'' POLYGONS IN CENTRAL}

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\centerline{\Largebf SURFACES OF THE SECOND ORDER}

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\centerline{\Largebf By}

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\centerline{\Largebf William Rowan Hamilton}

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\centerline{\largerm (Proceedings of the Royal Irish Academy,
   4 (1850), pp.\ 541--557.)}

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\centerline{\largerm Edited by David R. Wilkins}

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\centerline{\largerm 2000}

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\centerline{\largeit On ``Gauche'' Polygons in Central Surfaces
of the Second Order.}

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\centerline{{\largeit By\/}
{\largerm Sir} {\largesc William R. Hamilton.}}

\bigskip

\centerline{Communicated May~13, 1850.}

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\centerline{[{\it Proceedings of the Royal Irish Academy},
vol.~4 (1850), pp.\ 541--557.]}

\bigskip

Sir William Rowan Hamilton gave an account of some geometrical
reasonings, tending to explain and confirm certain results to
which he had been previously conducted by the method of
quaternions, respecting the inscription of gauche polygons in
central surfaces of the second order.

\bigbreak

1.
It is a very well known property of the conic sections, that if
three of the four sides of a plane quadrilateral inscribed in a
given plane conic be cut by a rectilinear transversal in three
given points, the fourth side of the same variable quadrilateral
is cut by the same fixed right line in a fourth point likewise
fixed.  And whether we refer to the relation of involution
discovered by Desargues, or employ other principles, it is easy
to extend this property to {\it surfaces\/} of the second order,
so far as the inscription in them of {\it plane\/} quadrilaterals
is concerned.  If then we merely wish to pass from one
point~${\sc p}$ to another point~${\sc r}$ of such a surface,
under the condition that some other point~${\sc q}$ of the same
surface shall exist, such that the two successive and rectilinear
chords, ${\sc p} {\sc q}$ and ${\sc q} {\sc r}$, shall pass
respectively through some two given {\it guide-points}, ${\sc a}$
and ${\sc b}$, internal or external to the surface; we are
allowed to {\it substitute}, for this pair of guide-points,
{\it another pair}, such as ${\sc b}'$ and ${\sc a}'$, situated
{\it on the same straight line\/} ${\sc a} {\sc b}$; and may
choose {\it one\/} of these two new points {\it anywhere\/} upon
that line, provided that the {\it other\/} be then suitably
chosen.  In fact, if ${\sc c}$ and ${\sc c}'$ be the two (real or
imaginary) points in which the surface is crossed by the given
transversal~${\sc a} {\sc b}$, we have only to take care that the
three pairs of points ${\sc a} {\sc a}'$, ${\sc b} {\sc b}'$,
${\sc c} {\sc c}'$, shall be in involution.  And it is important
to observe, that in order to determine one of the new
guide-points, ${\sc b}'$ and ${\sc a}'$, when the other is given,
it is by no means necessary to employ the points
${\sc c}$,~${\sc c}'$, of intersection of the transversal with
the surface, which may be as often imaginary as real.  We have
only to {\it assume\/} at pleasure a point~${\sc p}$ upon the
given surface; to draw from it the chords
${\sc p} {\sc a} {\sc q}$, ${\sc q} {\sc b} {\sc r}$;
and then if ${\sc a}'$ be given, and ${\sc b}'$ sought, to draw
the two new chords
${\sc r} {\sc a}' {\sc s}$, ${\sc s} {\sc b}' {\sc p}$;
or else if ${\sc a}'$ is to be found from ${\sc b}'$, to draw the
chords ${\sc p} {\sc b}' {\sc s}$, ${\sc s} {\sc a}' {\sc r}$.
For example, if we choose to throw off the new
guide-point~${\sc b}'$ to infinity, or to make it a
{\it guide-star}, in the direction of the given
line~${\sc a} {\sc b}$, we have only to draw, from the assumed
initial and superficial point~${\sc p}$, a rectilinear
chord~${\sc p} {\sc s}$ of the surface, which shall be parallel
to ${\sc a} {\sc b}$, and then to join ${\sc s} {\sc r}$, and
examine in what point~${\sc a}'$ this joining line crosses the
given line~${\sc a} {\sc b}$.  The point~${\sc a}'$ {\it thus\/}
found will be entirely independent of the assumed initial
point~${\sc p}$, and will satisfy the condition required: in such
a matter that if, from any {\it other\/} assumed superficial
point~${\sc p}'$, we draw the chords ${\sc p}' {\sc a} {\sc q}'$,
${\sc q}' {\sc b} {\sc r}'$, and the parallel ${\sc p}' {\sc s}'$
to ${\sc a} {\sc b}$, the chord ${\sc r}' {\sc s}'$ shall pass
through the {\it same\/} point~${\sc a}'$.  All this follows
easily from principles perfectly well known.

\bigbreak

2.
Since then for {\it two\/} given guide-points we may thus
substitute the system of a guide-star and a guide-point, it
follows that for {\it three\/} given guide-points we may
substitute a guide-star and two guide-points; and, therefore, by
a repetition of the same process, may substitute anew a system of
two stars and one point.  And so proceeding, for a system of $n$
given guide-points, through which $n$ successive and rectilinear
chords of the surface are to pass, we may substitute a system of
$n - 1$ guide-stars, and of a single guide-point.  The problem of
inscribing, in a given surface of the second order, a gauche
polygon of $n$ sides, which are required to pass successively
through $n$ given points, is, therefore, in general, reducible,
by operations with straight lines alone, to the problem of
inscribing in the same surface another gauche polygon, of which
the {\it last\/} side shall pass through a new fixed point,
while all its {\it other\/} $(n - 1)$ sides shall be parallel to
so many fixed straight lines.  And if the {\it first\/} $n$ sides
of an inscribed polygon of $n + 1$ sides,
${\sc p} {\sc p}_1 {\sc p}_2 \, \ldots \, {\sc p}_n$,
be obliged to pass, in order, through $n$ given points,
${\sc a}_1 {\sc a}_2 \, \ldots \, {\sc a}_n$,
namely, the side or chord ${\sc p} {\sc p}_1$ through
${\sc a}_1$, \&c., it will then be possible, in general, to
inscribe also {\it another\/} polygon
${\sc p} {\sc q}_1 {\sc q}_2 \, \ldots \, {\sc p}_n$,
having the same first and $n$th points, ${\sc p}$ and
${\sc p}_n$, and therefore the {\it same final\/} or closing side
${\sc p}_n {\sc p}$, but having the other $n$ sides
{\it different}, and such that the $n - 1$ first of these sides,
${\sc p} {\sc q}_1, {\sc q}_1 {\sc q}_2,\ldots \,
   {\sc q}_{n-2} {\sc q}_{n-1}$,
shall be respectively parallel to $n - 1$ given right lines,
while the $n$th side ${\sc q}_{n-1} {\sc p}_n$ shall pass through
a fixed point~${\sc b}_n$.  The analogous reductions for polygons
in conic sections have long been familiar to geometers.

\bigbreak

3.
Let us now consider the inscribed gauche quadrilateral
${\sc p} {\sc q}_1 {\sc q}_2 {\sc q}_3$, of which the four
corners coincide with the four points of the last-mentioned
polygon.  In the plane ${\sc q}_1 {\sc q}_2 {\sc q}_3$ of the
second and third sides of this gauche quadrilateral, draw a new
chord ${\sc q}_1 {\sc r}_2$, which shall have its direction
conjugate to the direction of ${\sc p} {\sc q}_1$, with respect
to the given surface.  This new direction will itself be fixed,
as being parallel to a fixed plane, and conjugate to a fixed
direction, not generally conjugate to that plane; and hence in
the plane inscribed quadrilateral
${\sc r}_2 {\sc q}_1 {\sc q}_2 {\sc q}_3$,
the three first sides having fixed directions, the fourth side
${\sc q}_3 {\sc r}_2$ will also have its direction fixed: which
may be proved, either as a limiting form of the theorem referred
to in (1), respecting four points in one line, or from principles
still more elementary.  And there is no difficulty in seeing that
because ${\sc p} {\sc q}_1$ and ${\sc q}_1 {\sc r}_2$ have fixed
and conjugate directions, the chord ${\sc p} {\sc r}_2$ is
bisected by a fixed diameter of the surface, whose direction is
conjugate to both of their's; or in other words, that if
${\sc o}$ be the centre of the surface, and if we draw the
{\it variable\/} diameter ${\sc p} {\sc o} {\sc n}$, the variable
{\it chord\/} ${\sc n} {\sc r}_2$ will then be parallel to the
{\it fixed\/} diameter just mentioned.  So far, then, as we only
concern ourselves to construct the fourth or closing side
${\sc q}_3 {\sc p}$ of the gauche quadrilateral
${\sc p} {\sc q}_1 {\sc q}_2 {\sc q}_3$, whose three first sides
have given or fixed directions, we may substitute for it another
gauche quadrilateral ${\sc p} {\sc n} {\sc r}_2 {\sc q}_3$,
inscribed in the same surface, and such that while its first
side~${\sc p} {\sc n}$ passes through the centre~${\sc o}$, its
second and third sides, ${\sc n} {\sc r}_2$ and
${\sc r}_2 {\sc q}_3$, are parallel to two fixed right lines.  In
other words, we may substitute, for a system of {\it three
guide-stars}, a system of the {\it centre and two stars}, as
guides for the three first sides; or, if we choose, instead of
drawing successively three chords,
${\sc p} {\sc q}_1$, ${\sc q}_1 {\sc q}_2$, ${\sc q}_2 {\sc q}_3$,
parallel to three given lines, we may draw a first chord
${\sc p} {\sc r}_2$, so as to be bisected by a given diameter,
and then a second chord ${\sc r}_2 {\sc q}_3$, parallel to a
given right line.

\bigbreak

4.
Since, for a system of {\it three\/} stars, we may substitute a
system of the centre and {\it two\/} stars, it follows that for a
system of {\it four\/} stars we may substitute a system of the
centre and {\it three\/} stars; or, by a repetition of the same
process, may substitute a system of the centre, the same centre
{\it again}, and two stars; that is, ultimately, a system of
{\it two\/} stars may be substituted for a system of {\it four\/}
stars, the two employments of the centre as a guide having simply
neutralized each other, as amounting merely to a {\it return\/}
from ${\sc n}$ to ${\sc p}$, after having {\it gone\/} from
${\sc p}$ to the diametrically opposite point~${\sc n}$.  For
five stars we may therefore substitute three; and for six stars
we may substitute four, or two.  And so proceeding we perceive
that, for {\it any\/} proposed system of guide-stars, we may
substitute {\it two\/} stars, if the proposed number be even; or
{\it three}, if that number be odd.  And by combining this result
with what was found in (2), we see that for any given system of
$n$ guide-points we may substitute a system of {\it two stars and
a point}, if $n$ be {\it odd\/}; or if $n$ be {\it even}, then in
that case we may substitute a system of {\it three\/} stars and a
point: which may again be changed, by (3), to a system of the
{\it centre, two stars, and one point}.

\bigbreak

5.
Let us now consider more closely the system of two guide-stars,
and one guide-point; and for this purpose let us conceive that
the two first sides ${\sc p} {\sc q}_1$ and ${\sc q}_1 {\sc q}_2$
of an inscribed gauche quadrilateral
${\sc p} {\sc q}_1 {\sc q}_2 {\sc p}_3$
are parallel to two given right lines, while the third side
${\sc q}_2 {\sc p}_3$ is obliged to pass through a fixed
point~${\sc b}_3$; the first point~${\sc p}$, and therefore also
the quadrilateral itself, being in other respects variable.  In
the plane ${\sc p} {\sc q}_1 {\sc q}_2$ of the two first sides,
which is evidently parallel to a fixed plane, inscribe a
chord~${\sc q}_2 {\sc s}$, whose direction shall be conjugate to
that of the fixed line ${\sc o} {\sc b}_3$, and therefore shall
itself also be fixed, ${\sc o}$ being still the centre of the
surface; and draw the chord~${\sc p} {\sc s}$.  Then, in the
plane inscribed quadrilateral
${\sc p} {\sc q}_1 {\sc q}_2 {\sc s}$,
the three first sides have fixed directions, and therefore, by
(3), the direction of the fourth side~${\sc s} {\sc p}$ is also
fixed.  In the plane~${\sc s} {\sc q}_2 {\sc p}_3$, which
contains the given point~${\sc b}_3$, draw through that point an
indefinite right line ${\sc b}_3 {\sc c}_3$, parallel to
${\sc s} {\sc q}_2$; the line so drawn will have a given
position, and will be intersected, at some finite or infinite
distance from ${\sc b}_3$, by the chord ${\sc s} {\sc p}_3$,
which is situated in the same plane with it, namely, in the plane
${\sc s} {\sc q}_2 {\sc p}_3$.  But if we consider the section of
the surface, which is made by this last plane, and observe that
the two first sides of the triangle ${\sc s} {\sc q}_2 {\sc p}_3$
pass, by the construction, through a star or point at infinity
conjugate to ${\sc b}_3$, and through the point~${\sc b}_3$
itself, we shall see that, in virtue of a well-known and
elementary principle respecting triangles in conics, the third
side ${\sc p}_3 {\sc s}$ must pass through the point~${\sc d}_3$,
if ${\sc d}_3$ be the pole of the right line
${\sc b}_3 {\sc c}_3$, which contains upon it the two conjugate
points; this {\it pole\/} being taken with respect to the plane
{\it section\/} lately mentioned.  If then we denote by
${\sc d}_3 {\sc e}_3$ the indefinite right line which is, with
respect to the {\it surface}, the {\it polar\/} of the fixed line
${\sc b}_3 {\sc c}_3$, we see that the chord ${\sc s} {\sc p}_3$
must intersect this reciprocal polar also, besides intersecting
the line ${\sc b}_3 {\sc c}_3$ itself.  Conversely this
condition, of intersecting these two fixed polars, is sufficient
to enable us to draw the chord~${\sc s} {\sc p}_3$ when the
point~${\sc s}$ has been determined, by drawing from the assumed
point~${\sc p}$ the chord~${\sc p} {\sc s}$ parallel to a fixed
right line.  We may then {\it substitute}, for a system of two
guide-stars and one guide-point, the system of {\it one
guide-star\/} and {\it two guide-lines\/}; these {\it lines\/}
being (as has been seen) a pair of {\it reciprocal polars}, with
respect to the given surface.

\bigbreak

6.
If, then, it be required to inscribe a polygon
${\sc p} {\sc p}_1 {\sc p}_2 \, \ldots \, {\sc p}_{2n}$
with any odd number $2n + 1$ of sides, which shall pass
successively through the same number of given points,
${\sc a}_1 {\sc a}_2 \, \ldots \, {\sc a}_{2n+1}$,
we may begin by {\it assuming\/} a point~${\sc p}$ upon the given
surface, and drawing through the given points $2n + 1$ successive
chords, which will in general conduct to a final
point~${\sc p}_{2n+1}$, {\it distinct\/} from the assumed initial
point~${\sc p}$.  And then, by processes of which the nature has
been already explained, we can find a point~${\sc s}$ such that
the chord ${\sc p} {\sc s}$ shall be parallel to a fixed right
line, or shall have a direction independent of the assumed and
variable position of ${\sc p}$; and that the
chord~${\sc s} {\sc p}_{2n+1}$ shall at the same time cross two
other fixed right lines, which are reciprocal polars of each
other.  In order then to find a {\it new\/} point~${\sc p}$,
which shall satisfy the conditions of the proposed problem, or
shall be such as to {\it coincide\/} with the
point~${\sc p}_{2n+1}$, deduced from it as above, we see that it
is necessary and sufficient to oblige this sought point~${\sc p}$
to be situated at one or other extremity of a certain
chord~${\sc p} {\sc s}$, which shall at once be parallel to a
fixed line, and shall also cross two fixed polars.  It is clear
then that we need only draw two planes, containing respectively
these two polars, and parallel to the fixed direction; for the
right line of intersection of these two planes will be the
{\it chord of solution\/} required; or in other words, it will
cut the surface in the two (real or imaginary) points, ${\sc p}$
and ${\sc s}$, which are adapted, and are alone adapted, to be
positions of the first corner of the polygon to be inscribed.

\bigbreak

7.
But if it be demanded to inscribe in the same surface a polygon
${\sc p} {\sc p}_1 {\sc p}_2 \, \ldots \, {\sc p}_{2n-1}$,
with an {\it even\/} number $2n$ of sides, passing successively
through the same {\it even\/} number of given points,
${\sc a}_1 \, {\sc a}_2 \, \ldots \, {\sc a}_{2n}$,
{\it the problem then acquires a character totally distinct}.
For if, after assuming an initial point~${\sc p}$ upon the
surface, we pass, by $2n$ successive chords, drawn through the
given points ${\sc a}_1$,~\&c., to a final point~${\sc p}_{2n}$
upon the surface, which will thus be in general distinct from
${\sc p}$; it will indeed be possible to assign generally two
fixed polars, across which, as two given guide-lines, a certain
variable chord ${\sc s} {\sc p}_{2n}$ is to be drawn, like the
chord~${\sc s} {\sc p}_{2n+1}$ of (6); but the
chord~${\sc p} {\sc s}$ will {\it not}, in {\it this\/} question,
be {\it parallel to a given line}, or directed to a given star;
it will, on the contrary, by (3) (4) (5), be {\it bisected by a
given diameter}, which we may call ${\sc a} {\sc b}$; or, we
prefer to state the result so, it will be now the
{\it supplementary chord\/}~${\sc n} {\sc s}$ of the same
diametral section of the surface (${\sc n}$ being still the point
of that surface {\it opposite\/} to ${\sc p}$), which will have a
given direction, and {\it not\/} the chord~${\sc p} {\sc s}$
itself. In fact, at the end of (4), we reduced the system of $2n$
guide-points to a system of the centre, two stars, and one point;
and in (5) we reduced the system of two stars and a point to the
system of a star and two polars.  In order then to find a
point~${\sc p}$ which shall {\it coincide\/} with the
point~${\sc p}_{2n}$ deduced from it as above, or which shall be
adapted to be the first corner of an inscribed polygon of $2n$
sides passing respectively through the $2n$ given points,
${\sc a}_1 \, \ldots \, {\sc a}_{2n}$,
we must endeavour to find a chord~${\sc p} {\sc s}$ which shall
be at once bisected by the fixed diameter~${\sc a} {\sc b}$, and
shall {\it also\/} intersect the two fixed polars above
mentioned.  And conversely, if we can find any such
chord~${\sc p} {\sc s}$, it will necessarily be at least
{\it one chord of solution\/} of the problem; understanding
hereby, that if we set out with {\it either\/} extremity,
${\sc p}$ or ${\sc s}$, and draw from it $2n$ successive chords
${\sc p} {\sc p}_1$, \&c., or ${\sc s} {\sc s}_1$, \&c., through
the $2n$ given points ${\sc a}_1$, \&c., we shall be brought
{\it back\/} hereby (as the question requires) to the point with
which we started.  For, in a process which we have proved to
admit of being {\it substituted\/} for the process of drawing the
$2n$ chords, we shall be brought first from ${\sc p}$ to ${\sc
s}$, and then back from ${\sc s}$ to ${\sc p}$; or else first
from ${\sc s}$ to ${\sc p}$, and then back from ${\sc p}$ to
${\sc s}$: provided that the chord of solution~${\sc p} {\sc s}$
has been selected so as to satisfy the conditions above assigned.

\bigbreak

8.
{\it To inscribe then}, for example, {\it a gauche chiliagon in
an ellipsoid},
${\sc p} {\sc p}_1 \, \ldots \, {\sc p}_{999}$, or
${\sc s} {\sc s}_1 \, \ldots \, {\sc s}_{999}$,
under the condition that {\it its thousand successive sides shall
pass successively through a thousand given points\/}
${\sc a}_1 {\sc a}_2 \, \ldots \, {\sc a}_{1000}$,
we are conducted to seek to inscribe, in the same given
ellipsoid, a {\it chord\/}~${\sc p} {\sc s}$, which shall be at
once {\it bisected by a given diameter\/}~${\sc a} {\sc b}$, and
also {\it crossed by a given chord~${\sc c} {\sc d}$ and by the
polar of that given chord}.  Now in general when any two proposed
right lines intersect each other, their respective polars also
intersect, namely, in the pole of the plane of the two lines
proposed.  Since then the sought chord~${\sc p} {\sc s}$
intersects the polar of the given chord~${\sc c} {\sc d}$, it
follows that the polar of the same sought chord~${\sc p} {\sc s}$
must intersect the given chord~${\sc c} {\sc d}$ itself.  We may
therefore reduce the problem to this form: to find a
chord~${\sc p} {\sc s}$ of the ellipsoid which shall be bisected
by a given diameter~${\sc a} {\sc b}$, and shall also be such
that while it intersects a given chord~${\sc c} {\sc d}$ in some
point~${\sc e}$, its polar intersects the prolongation of that
given chord, in some other point~${\sc f}$.

\bigbreak

9.
The two sought points ${\sc e}$,~${\sc f}$, as being situated
upon two polars, are of course {\it conjugate\/} relatively to
the {\it surface\/}; they are therefore also conjugate relatively
to the {\it chord\/}~${\sc c} {\sc d}$, or, in other words, they
cut that given chord {\it harmonically}.  The four diametral
planes ${\sc a} {\sc b} {\sc c}$, ${\sc a} {\sc b} {\sc e}$,
${\sc a} {\sc b} {\sc d}$, ${\sc a} {\sc b} {\sc f}$, compose
therefore an harmonic pencil; the second being, {\it in this
pencil}, harmonically conjugate to the fourth; and being at the
same time, on account of the polars, conjugate to it also with
respect to the {\it surface}, as one diametral plane to another.
When the ellipsoid becomes a {\it sphere}, the conjugate planes
${\sc a} {\sc b} {\sc e}$, ${\sc a} {\sc b} {\sc f}$ become
{\it rectangular\/}; and consequently the sought
plane~${\sc a} {\sc b} {\sc e}$ {\it bisects the angle\/} between
the two given planes ${\sc a} {\sc b} {\sc c}$ and
${\sc a} {\sc b} {\sc d}$.  {\it This solves at once the problem
for the sphere\/}; for if, conversely, we thus bisect the given
dihedral angle ${\sc c} {\sc a} {\sc b} {\sc d}$ by a plane
${\sc a} {\sc b} {\sc e}$, cutting the chord~${\sc c} {\sc d}$ in
${\sc e}$, and if we take the harmonic conjugate~${\sc f}$ on the
same given chord prolonged, and draw from ${\sc e}$ to ${\sc f}$
lines meeting ordinately the given diameter~${\sc a} {\sc b}$,
these two right lines will be situated in two rectangular or
conjugate diametral planes, and will satisfy all the other
conditions requisite for their being polars of each other; but
each intersects the given chord~${\sc c} {\sc d}$, or that chord
prolonged, and therefore each intersects also, by (8), the polar
of that chord; each therefore satisfies all the transformed
conditions of the problem, and gives a chord of solution, real or
imaginary.  More fully, the ordinate~${\sc e} {\sc e}'$ to the
diameter~${\sc a} {\sc b}$, drawn from the {\it internal\/} point
of harmonic section~${\sc e}$ of the chord~${\sc c} {\sc d}$,
gives, when prolonged both ways to meet the surface, the
{\it chord of real solution},~${\sc p} {\sc s}$; and the other
ordinate~${\sc f} {\sc f}'$ to the same
diameter~${\sc a} {\sc b}$, which is drawn from the
{\it external\/} point of section~${\sc f}$ of the same
chord~${\sc c} {\sc d}$, and which is itself wholly external to
the surface, is the {\it chord of imaginary solution}.  But
because when we return from the sphere to the {\it ellipsoid}, or
other surface of the second order, the condition of
{\it bisection\/} of the given dihedral angle
${\sc c} {\sc a} {\sc b} {\sc d}$
is no longer fulfilled by the sought
plane~${\sc a} {\sc b} {\sc e}$, a slight generalization of the
foregoing process becomes necessary, and can easily be
accomplished as follows.

\bigbreak

10.
Conceive, as before, that on the diameter~${\sc a} {\sc b}$ the
ordinate~${\sc e} {\sc e}'$ is let fall from the internal point
of section~${\sc e}$, and likewise the
ordinates~${\sc c} {\sc c}'$ and ${\sc d} {\sc d}'$ from
${\sc c}$ and ${\sc d}$; and draw also, parallel to that
diameter, the right lines
${\sc c} {\sc c}''$, ${\sc d} {\sc d}''$, ${\sc e} {\sc e}''$,
from the same three points ${\sc c}$,~${\sc d}$,~${\sc e}$, so as
to terminate on the diametral plane through ${\sc o}$ which is
conjugate to the same diameter; in such a manner that
${\sc o} {\sc c}''$, ${\sc o} {\sc d}''$, ${\sc o} {\sc e}''$
shall be parallel and equal to the ordinates
${\sc c}' {\sc c}$, ${\sc d}' {\sc d}$, ${\sc e}' {\sc e}$;
and that the segments ${\sc c} {\sc e}$, ${\sc e} {\sc d}$ of the
chord~${\sc c} {\sc d}$ shall be proportional to the segments
${\sc c}'' {\sc e}''$, ${\sc e}'' {\sc d}''$ of the base
${\sc c}'' {\sc d}''$ of the triangle
${\sc c}'' {\sc o} {\sc d}''$, which is situated in the diametral
plane, and has the centre~${\sc o}$ for its vertex.  For the case
of the {\it sphere}, the vertical angle
${\sc c}'' {\sc o} {\sc d}''$ of this triangle is, by (9),
bisected by the line ${\sc o} {\sc e}''$; wherefore the sides
${\sc o} {\sc c}''$, ${\sc o} {\sc d}''$, or their equals, the
ordinates ${\sc c}' {\sc c}$, ${\sc d}' {\sc d}$, are, in this
case, proportional to the segments
${\sc c}'' {\sc e}''$, ${\sc e}'' {\sc d}''$
of the base, or to the segments ${\sc c} {\sc e}$,
${\sc e} {\sc d}$ of the chord: while the squares of the
ordinates are, for the same case of the sphere, equal to the
rectangles ${\sc a} {\sc c}' {\sc b}$,
${\sc a} {\sc d}' {\sc b}$,
under the segments of the diameter~${\sc a} {\sc b}$.  Hence,
{\it for the sphere, the squares of the segments of the given
chord are proportional to the rectangles under the segments of
the given diameter}, these latter segments being found by letting
fall ordinates from the ends of the chord; or, in symbols, we
have the proportion,
$${\sc c} {\sc f}^2 : {\sc d} {\sc f}^2 ::
  {\sc c} {\sc e}^2 : {\sc e} {\sc d}^2 ::
  {\sc a} {\sc c}' {\sc b} : {\sc a} {\sc d}' {\sc b}.$$
But, by the general principles of {\it geometrical deformation,
the property, thus stated, cannot be peculiar to the sphere.  It
must extend}, without any further modification, {\it to the
ellipsoid\/}; and it gives at once, for that surface, the two
points of harmonic section, ${\sc e}$ and ${\sc f}$, of the given
chord~${\sc c} {\sc d}$, through which points the two sought
{\it chords of real and imaginary solution\/} are to pass;
{\it these chords of solution are therefore completely
determined}, since they are to be also ordinates, as before, to
the given diameter~${\sc a} {\sc b}$.  {\it The problem of
inscription for the ellipsoid is therefore fully resolved; not
only\/} when, as in (6), the number of sides of the polygon is
{\it odd}, but {\it also\/} in the more difficult case~(7), when
the number of sides is {\it even}.

\bigbreak

11.
If the given surface be a hyperboloid of {\it two sheets, one\/}
of the two fixed polars will still intersect that surface, and
the fixed chord~${\sc c} {\sc d}$ may still be considered as
{\it real}.  If the given diameter~${\sc a} {\sc b}$ be also
real, the proportion in (10) still holds good, without any
modification from imaginaries, and determines still a real
point~${\sc e}$, with its harmonic conjugate~${\sc f}$, through
one of other of which two points still passes a {\it chord of
real solution}, while through the other point of section still is
drawn a {\it chord of imaginary solution}, reciprocally polar to
the former.  But if the {\it diameter\/}~${\sc a} {\sc b}$ be
{\it imaginary}, or in other words, if it fail to meet the
proposed hyperboloid at all, we are led to consider, instead of
it, an {\it ideal diameter\/}~${\sc a}' {\sc b}'$, having the
same {\it real direction}, but terminating, in a well-known way,
on a certain {\it supplementary surface\/}; in such a manner that
while ${\sc a}$ and ${\sc b}$ are now {\it imaginary points}, the
points ${\sc a}'$ and ${\sc b}'$ are {\it real}, although
{\it not really situated on the given surface\/}; and that
$${\sc o} {\sc a}^2
   =  {\sc o} {\sc b}^2
   =  - {\sc o} {\sc a}'^2
   =  - {\sc o} {\sc b}'^2.$$
The points ${\sc c}'$ and ${\sc d}'$ are still real, and so are
the rectangles ${\sc a} {\sc c}' {\sc b}$ and
${\sc a} {\sc d}' {\sc b}$, although ${\sc a}$ and ${\sc b}$ are
imaginary; for we may write,
$${\sc a} {\sc c}' {\sc b}
   =  {\sc o} {\sc a}^2 - {\sc o} {\sc c}'^2,\quad
  {\sc a} {\sc d}' {\sc b}
   =  {\sc o} {\sc a}^2 - {\sc o} {\sc d}'^2,$$
and the proportion in (10) becomes now,
$${\sc c} {\sc f}^2 : {\sc d} {\sc f}^2 ::
  {\sc c} {\sc e}^2 : {\sc e} {\sc d}^2 ::
  {\sc o} {\sc c}'^2 + {\sc o} {\sc a}'^2 :
  {\sc o} {\sc d}'^2 + {\sc o} {\sc a}'^2.$$
It gives therefore still a {\it real point of
section\/}~${\sc e}$, and a {\it real conjugate
point\/}~${\sc f}$; and through these two points of section of
${\sc c} {\sc d}$ we can still draw {\it two real right lines},
which shall still ordinately cross the real direction of
${\sc a} {\sc b}$, and shall still be two reciprocal polars,
satisfying all the transformed conditions of the question, and
coinciding with two chords of real and imaginary solution.
{\it For the double-sheeted hyperboloid}, therefore, {\it as
well\/} as for the ellipsoid, the problem of inscribing a
{\it gauche chiliagon}, or other {\it even-sided polygon}, whose
sides shall pass successively, and in order, through the same
given number of points, is solved by a system of {\it two polar
chords}, which we have assigned geometrical processes to
determine; and the solutions are {\it still}, in general,
{\it four\/} in number; {\it two\/} of them being still
{\it real}, and {\it two imaginary}.

\bigbreak

12.
If the given surface be a hyperboloid of {\it one sheet}, then
not only may the diameter~${\sc a} {\sc b}$ be real or imaginary,
but also the chord~${\sc c} {\sc d}$ may or may not cease to be
real; for the two fixed polars will {\it now\/} either {\it both
meet\/} the surface, or else {\it both fail\/} to meet it in any
two real points.  When ${\sc a} {\sc b}$ and ${\sc c} {\sc d}$
are both real, the proportion in (10), being put under the form
$${\sc c} {\sc f}^2 : {\sc d} {\sc f}^2 ::
  {\sc c} {\sc e}^2 : {\sc e} {\sc d}^2 ::
  {\sc o} {\sc a}^2 - {\sc o} {\sc c}'^2 :
  {\sc o} {\sc a}^2 - {\sc o} {\sc d}'^2,$$
shews that the point of section~${\sc e}$ and its
conjugate~${\sc f}$ will be real, if the points ${\sc c}'$ and
${\sc d}'$ fall {\it both\/} on the diameter ${\sc a} {\sc b}$
{\it itself}, or {\it both\/} on that diameter {\it prolonged\/};
that is, if the extremities ${\sc c}$and ${\sc d}$ lie {\it both
within\/} or {\it both without\/} the interval between the two
parallel tangent planes to the surface which are drawn at the
points~${\sc a}$ and ${\sc b}$: under these conditions therefore
there will still be {\it two real right lines}, which may still
be called the {\it two chords of solution\/}; but because these
lines will still be two reciprocal polars, they will now (like
the two fixed polars above mentioned) either {\it both meet\/}
the hyperboloid, or else {\it both fail\/} to meet it; and
consequently there will now be either {\it four real}, or else
{\it four imaginary\/} solutions.  If ${\sc a} {\sc b}$ and
${\sc c} {\sc d}$ be still both real, but if the
chord~${\sc c} {\sc d}$ have {\it one\/} extremity {\it within\/}
and the {\it other\/} extremity {\it without\/} the interval
between the two parallel tangent planes, the proportion above
written will assign {\it a negative ratio\/} for the squares of
the segments of ${\sc c} {\sc d}$; the points of section
${\sc e}$ and ${\sc f}$, and the {\it two polar chords\/} of
solution, become therefore, in {\it this\/} case, {\it themselves
imaginary\/}; and of course, by still stronger reason, the four
solutions of the problem become then imaginary likewise.  If
${\sc c} {\sc d}$ be real, but ${\sc a} {\sc b}$ imaginary, the
proportion in (11) conducts to two real points of section, and
consequently to two real chords, which may, however, correspond,
as above, either to four real or to four imaginary solutions of
the problem.  And, finally, it will be found that the same
conclusion holds good also in the remaining case, namely, when
the chord~${\sc c} {\sc d}$ becomes imaginary, whether the
diameter ${\sc a} {\sc b}$ be real or not; that is, when the two
fixed polars do not meet, in any real points, the single-sheeted
hyperboloid.

\bigbreak

13.
Although the case last mentioned may still be treated by a
modification of the proportion assigned in (10), which was
deduced from considerations relative to the sphere, yet in order
to put the subject in a clearer (or at least in another) point of
view, we may now resume the problem for the ellipsoid as follows,
without making any use of the spherical deformation.  It was
required to find two lines, reciprocally polar to each other, and
ordinately crossing a given diameter~${\sc a} {\sc b}$ of the
ellipsoid, which should also cut a given chord~${\sc c} {\sc d}$
of the same surface, internally in some point~${\sc e}$, and
externally in some other point~${\sc f}$.  Bisect
${\sc c} {\sc d}$ in ${\sc g}$, and conceive ${\sc e} {\sc f}$ to
be bisected in ${\sc h}$; and besides the four old ordinates to
the diameter~${\sc a} {\sc b}$, namely
${\sc c} {\sc c}'$, ${\sc d} {\sc d}'$, ${\sc e} {\sc e}'$,
and ${\sc f} {\sc f}'$, let there be now supposed to be drawn, as
two new ordinates to the same diameter, the lines
${\sc g} {\sc g}'$ and ${\sc h} {\sc h}'$.  Then ${\sc g}'$ will
bisect ${\sc c}' {\sc d}'$, and ${\sc h}'$ will bisect
${\sc e}' {\sc f}'$; while the centre~${\sc o}$ of the ellipsoid
will still bisect ${\sc a} {\sc b}$.  And because the points
${\sc e}'$ and ${\sc f}'$ are harmonic conjugates, not only with
respect to the points ${\sc a}$ and ${\sc b}$, but also with
respect to the points ${\sc c}'$ and ${\sc d}'$, we shall have
the following equalities:
$$\eqalign{
{\sc h}' {\sc f}'^2
   &= {\sc h}' {\sc e}'^2
    = {\sc h}' {\sc a}' \mathbin{.} {\sc h}' {\sc b}'
    = {\sc h}' {\sc c}' \mathbin{.} {\sc h}' {\sc d}' \cr
   &= {\sc h}' {\sc o}^2 - {\sc o} {\sc a}^2
    = {\sc h}' {\sc g}'^2 - {\sc g}' {\sc c}'^2.\cr}$$
Hence,
$${\sc o} {\sc h}'^2 - {\sc g}' {\sc h}'^2
   =  {\sc o} {\sc a}^2 - {\sc c}' {\sc g}'^2,$$
that is,
$${\sc o} {\sc h}'
   =  {{\sc o} {\sc a}^2 + {\sc o} {\sc g}'^2 - {\sc c}' {\sc g}'^2
         \over 2 {\sc o} {\sc g}'}
   =  {{\rm o} {\sc a}^2 + {\sc o} {\sc c}' \mathbin{.} {\sc o} {\sc d}'
         \over {\sc o} {\sc c}' + {\sc o} {\sc d}'}.$$

Now each of these two last expressions for ${\sc o} {\sc h}'$
remains real, and assigns a real and determinate position for the
point~${\sc h}'$, even when the points ${\sc c}'$,~${\sc d}'$, or
the points ${\sc a}$,~${\sc b}$, or when both these pairs of
points at once become imaginary; for the points ${\sc o}$ and
${\sc g}'$ are still in all cases real, and so are the squares of
${\sc o} {\sc a}$ and ${\sc c}' {\sc g}'$, the rectangle under
${\sc o} {\sc c}'$ and ${\sc o} {\sc d}'$, and the sum
${\sc o} {\sc c}' + {\sc o} {\sc d}'$.  Thus ${\sc h}'$ can
always be found, as a real point, and hence we have a real
value for the square of ${\sc h}' {\sc e}'$, or
${\sc h}' {\sc f}'$, which will enable us to assign the
points~${\sc e}'$ and ${\sc f}'$ themselves, or else to pronounce
that they are imaginary.

\bigbreak

14.
We see at the same time, from the values
${\sc h}' {\sc o}^2 - {\sc o} {\sc a}^2$ and
${\sc h}' {\sc g}'^2 - {\sc c}' {\sc g}'^2$
above assigned for ${\sc h}' {\sc e}'^2$ or
${\sc h}' {\sc f}'^2$, that these two sought points ${\sc e}'$
and ${\sc f}'$ must both be real, unless the two fixed points
${\sc a}$ and ${\sc c}'$ are themselves both real, since
${\sc o}$,~${\sc g}'$,~${\sc h}'$, are, all three, real points.
But for the ellipsoid, and for the double sheeted hyperboloid, we
can in general {\it oblige\/} the points ${\sc c}$,~${\sc d}$,
and their projections ${\sc c}'$,~${\sc d}'$, to become
imaginary, by selecting {\it that one\/} of the two fixed polars
which does {\it not\/} actually meet the surface; for
{\it these\/} two sorts of surfaces, the two polar chords of
solution of the problem of inscription of a gauche polygon with
an even number of sides passing through the same number of given
points, and therefore found anew to be two {\it real lines},
although only one of them will actually intersect the surface,
and only two of the four polygons will (as before) be real.  And
even for the single-sheeted hyperboloid, in order to render the
two chords of solution {\it imaginary lines}, it is necessary
that the two given polars should actually meet the surface; for
otherwise the polar lines deduced will still be real.  It is
necessary also, for the imaginariness of the two lines deduced,
that the given diameter~${\sc a} {\sc b}$ should be itself a real
diameter, or in other words that it should actually intersect the
hyperboloid.  But even when the given chord~${\sc c} {\sc d}$ and
the given diameter ${\sc a} {\sc b}$ are thus {\it both\/} real,
and when the surface is a single-{\it sheeted\/} hyperboloid, it
does not {\it follow\/} that the two chords of solution
{\it may not\/} be real lines.  We shall only have {\it failed to
prove\/} their reality by the expressions recently referred to.
We must {\it resume}, for this case, the reasonings of (12), or
some others equivalent to them, and we find, as in that section
of this Abstract, for the imaginariness of the two sought polar
lines, the condition that {\it one\/} of the two extremities of
the given and real chord~${\sc c} {\sc d}$ shall fall
{\it within}, and that the {\it other\/} extremity of that chord
shall fall {\it without\/} the interval between the two real and
parallel tangent planes to the single-sheeted hyperboloid, which
are drawn at the extremities of the real
diameter~${\sc a} {\sc b}$.  Sir W.~R. Hamilton confesses that
the case where all these particular conditions are combined, so
as to render {\it imaginary\/} the two polar lines of solution,
had not occurred to him when he made to the Royal Irish Academy
his communication of June, 1849.

\bigbreak

15.
It seems to him worth while to notice here that instead of the
foregoing {\it metric\/} processes for finding (when they exist)
the two lines of solution of the problem, the following {\it
graphic\/} process of construction of those lines may always, at
least in theory, be substituted, although in practice it will
sometimes require modification for imaginaries.  In the diametral
plane~${\sc a} {\sc b} {\sc c}$ draw a chord
${\sc k} {\sc d}' {\sc l}$, which shall be bisected at the known
point~${\sc d}'$ by the given diameter~${\sc a} {\sc b}$; and
join ${\sc c} {\sc k}$, ${\sc c} {\sc l}$.  These joining lines
will cut that diameter in the two sought points
${\sc e}'$,~${\sc f}'$; which being in this manner found, the
two sought lines of solution ${\sc e} {\sc e}'$,
${\sc f} {\sc f}'$, are constructed without any difficulty.  For
the sphere, the ellipsoid, and the hyperboloid of two sheets,
although not always for the single-sheeted hyperboloid, this
simple and graphic process can actually be applied, without any
such modification from imaginaries as was above alluded to.  The
consideration of non-central surfaces does not enter into the
object of the present communication; nor has it been thought
necessary to consider in it any limiting or exceptional cases,
such as those where certain positions or directions become
indeterminate, by some {\it peculiar\/} combinations of the data,
while yet they are {\it in general\/} definitely assignable, by
the processes already explained.

\bigbreak

16.
Sir William Rowan Hamilton is unwilling to add to the length of
this communication by any historical references; in regard to
which, indeed, he does not consider himself prepared to furnish
anything important, as supplementary to what seems to be pretty
generally known, by those who feel an interest in such matters.
He has however taken some pains to inquire, from a few
geometrical friends, whether it is {\it likely\/} that he has
been anticipated in his results respecting the inscription of
{\it gauche\/} polygons in {\it surfaces\/} of the second order;
and he has not hitherto  been able to learn that any such
anticipation is thought to exist.  Of course he knows that he
must, consciously and unconsciously, be in many ways indebted to
his scientific contemporaries, for their instructions and
suggestions on these and on other subjects; and also to his
acquaintance, imperfect as it may be, with what has been done in
earlier times.  But he conceives that he only does justice to the
yet infant Method of Quaternions (communicated to the Royal Irish
Academy for the first time in 1843), when he states that he
considers himself to owe, to that new method of geometrical
research, not merely the {\it results\/} stated to the Academy in
the summer of 1849, respecting these inscriptions of gauche
polygons, and several other connected although hitherto
unpublished results, which to him appear remarkable, but also the
{\it suggestion\/} of the mode of {\it geometrical\/}
investigation which has been employed in the present Abstract.
No doubt the principles used in it have all been very elementary,
and perhaps their combination would have cost no serious trouble
to any experienced geometer who had chosen to attack the problem.
But to his {\it own\/} mind the whole foregoing investigation
presents itself as being (what in fact in his his case it
{\it was\/}) a mere {\it translation of the quaternion analysis
into ordinary geometrical language}, on this particular subject.
And he will not complicate the present Abstract by giving, on
{\it this\/} occasion, any account of those {\it other\/}
theorems respecting polygons in surfaces, to which the Calculus
of Quaternions has conducted him, but of which he has not seen
how to {\it translate the proofs\/} (for it is easy to translate
the {\it results\/}) into the usual language of {\it geometry}.

\bye