coincides with K, both these excesses and consequently their difference vanish; in every case, therefore, TP-PK = TQ – QK.

COR. Fagnani's theorem, "That an elliptic quadrant can be so divided, that the difference of its parts may be equal to the difference of the semi-axes," follows immediately from this Article, since we have only to draw tangents at the extremities of the axes, and through their intersection to draw a hyperbola confocal with the given ellipse. The coordinates of the points where it meets the ellipse are found to be

401. If a polygon circumscribe a conic, and if all the vertices but one move on confocal conics, the locus of the remaining vertex will be a confocal conic.

In the first place, we assert that if the vertex T of an angle PTQ circumscribing a conic, move on a confocal conic (see fig., Art. 399); and if we denote by a, b, the diameters parallel to TP, TQ; and by a, B, the angles TPT', TQ'T', made by each of the sides of the angle with its consecutive position, then aa=b3. For (Art. 399) TR= T'S; but TR = TP.a;T'S= T'Q'.B, and (Art. 149) TP and TQ are proportional to the diameters to which they are parallel.

Conversely, if aa=bB, T moves on a confocal conic. For by reversing the steps of the proof we prove that TR=T'S; hence that TT' makes equal angles with TP, TQ, and therefore coincides with the tangent to the confocal conic through T; and therefore that T' lies on that conic.

If, then, the diameters parallel to the sides of the polygon be a, b, c, &c., that parallel to the last side being d, we have aa = bß, because the first vertex moves on a confocal conic; in like manner bẞ=cy, and so on until we find aa=d8, which shows that the last vertex moves on a confocal conic.*

This proof is taken from a paper by Dr. Hart; Cambridge and Dublin Mathematical Journal, vol. IV. 193.

NOTES.

PASCAL'S THEOREM, Art. 267.

M. STEINER was the first who (in Gergonne's Annales) directed the attention of geometers to the complete figure obtained by joining in every possible way six points on a conic. M. Steiner's theorems were corrected and extended by M. Plücker (Crelle's Journal, vol. v. p. 274), and the subject has been more recently investigated by Messrs. Cayley and Kirkman, the latter of whom, in particular, has added several new theorems to those already known (see Cambridge and Dublin Mathematical Journal, vol. v. p. 185). We shall in this note give a slight sketch of the more important of these, and of the methods of obtaining them. The greater part are derived by joining the simplest principles of the theory of combinations with the following elementary theorems and their reciprocals: "If two triangles be such that the lines joining corresponding vertices meet in a point (the centre of homology of the two triangles), the intersections of corresponding sides will lie in one right line (their axis)." "If the intersections of opposite sides of three triangles be for each pair the same three points in a right line, the centres of homology of the first and second, second and third, third and first, will lie in a right line."

Now let the six points on a conic be a, b, c, d, e, f, which we shall call the points P. These may be connected by fifteen right lines, ab, ac, &c., which we shall call the lines C. Each of the lines C (for example) ab is intersected by the fourteen others; by four of them in the point a, by four in the point b, and consequently by six in points distinct from the points P (for example the points (ab, cd), &c.). These we shall call the points p. There are forty-five such points; for as there are six on each of the lines C, to find the number of points p, we must multiply the number of lines C by 6, and divide by 2, since two lines C pass through every point p. If we take the sides of the hexagon in the order abcdef, Pascal's theorem is, that the three p points, (ab, de), (ed, fa), (bc, ef), lie in one right line, which we may call either the Pascal abcdef, or else we may denote as the Pascal

Sab .cd.ef de. fa.be a form which we sometimes prefer, as showing more readily the three points through which the Pascal passes. Through each point p four Pascals can be drawn. Thus through (ab, de) can be drawn abcdef, abfdec, abcedf, abfedc. We then find the total number of Pascals by multiplying the number of points p by 4, and dividing by 3, since there are three points p on each Pascal. We thus obtain the number of Pascal's lines 60. We might have derived the same directly by considering the number of different ways of arranging the letters abcdef.

Consider now the three triangles whose sides are

The intersections of corresponding sides of 1 and 2 lie on the same Pascal, therefore the lines joining corresponding vertices meet in a point, but these are the three Pascals,

Lab.de.cf
lcd.fa.be'

Lcd fa.ber
ef.bc.ads'

Jef.be .ad
lab.de.cf

This is Steiner's theorem (Art. 268); we shall call this the g point,

The notation shows plainly that on each Pascal's line there is only one g point; for

given the Pascal

fab.de.cf

{a} the g point on it is found by writing under each term

the two letters not already found in that vertical line. intersect in every point g, the number of points g 20.

Since then three Pascals
If we take the triangles

2, 3; and 1, 3; the lines joining corresponding vertices are the same in all cases therefore, by the reciprocal of the second preliminary theorem, the three axes of the

three triangles meet in a point. This is also a g point de fa.be

(ab.cd.ef)
cf.be. ad

and Steiner

has stated that the two g points just written are harmonic conjugates with regard to the conic, so that the 20 g points may be distributed into ten pairs.* The Pascals which pass through these two g points correspond to hexagons taken in the order respectively, abcfed, afcdeb, adcbef; abcdef, afcbed, adefeb; three alternate vertices holding in all the same position.

The intersections of corresponding sides of 1 and 4 are three points which lie on the same Pascal; therefore the lines joining corresponding vertices meet in a point. But these are the three Pascals,

We may denote the point of meeting as the h point, cd.bf. ae

ef.ac.bd)

The notation differs from that of the g points in that only one of the vertical columns contains the six letters without omission or repetition. On every Pascal there are three h points, viz. there are on

where the bar denotes the complete vertical column. We obtain then Mr. Kirkman's extension of Steiner's theorem :-The Pascals intersect three by three, not only in Steiner's twenty points g, but also in sixty other points h. The demonstration of Art. 268 applies alike to Mr. Kirkman's and to Steiner's theorem.

In like manner if we consider the triangles 1 and 5, the lines joining corresponding **ices are the same as for 1 and 4; therefore the corresponding sides intersect on

*For a proof of this see Staudt (Crelle, LXII. 142).

a right line, as they manifestly do on a Pascal. In the same manner the corresponding sides of 4 and 5 must intersect on a right line, but these intersections are the three k points,

Moreover, the axis of 4 and 5 must pass through the intersection of the axes of

In this notation the g point is found by combining the complete vertical columns of the three h points. Hence we have the theorem, "There are twenty lines G, each of which passes through one g and three h points." The existence of these lines was observed independently by Prof. Cayley and myself. The proof here given is Prof. Cayley's.

It can be proved similarly that "The twenty lines G pass four by four through fifteen points i." The four lines G whose g points in the preceding notation have a common vertical column will pass through the same point.

Again, let us take three Pascals meeting in a point h. For instance,

ab.ce. df de.bf.ac}'

de.bf.ac
cf.ae.bd)'

cf.ae.bd
ab.df.ce

We may, by taking on each of these a point p, form a triangle whose vertices are (df, ac), (bf, ae), (bd, ce) and whose sides are, therefore,

Again, we may take on each a point h, by writing under each of the above Pascals af.cd.be, and so form a triangle whose sides are

But the intersections of corresponding sides of these triangles, which must therefore be on a right line, are the three g points,

I have added a fourth g point, which the symmetry of the notation shows must lie on the same right line; these being all the g points into the notation of which be.cd. af can enter. Now there can be formed, as may readily be seen, fifteen different products of the form be.cd.af; we have then Steiner's theorem, The g points lie four by four on fifteen right lines I. Hesse has noticed that there is a certain reciprocity between the theorems we have obtained. There are 60 Kirkman points h, and 60 Pascal lines H corresponding each to each in a definite order to be explained presently. There are 20 Steiner points g, through each of which passes three Pascals H and one line G; and there are 20 lines G, on each of which lie three Kirkman points and one Steiner g. And as the twenty lines G pass four by four through fifteen points i, so the twenty points g lie four by four on fifteen lines I. The following investigation gives a new proof of some of the preceding theorems and also shews what h point corresponds to the Pascal got by taking the vertices in the order abcdef. Consider the two inscribed triangles ace, bdf; their sides touch a conic (see Ex. 4, Art. 355); therefore we may apply Brianchon's theorem the hexagon whose sides are ce, df, ae, bf, ac, bd. Taking them in this orde

ce.bf.ad

gonals of the hexagon are the three Pascals intersecting in the h point, df.ac.be

ae. bd.cf

And since, if retaining the alternate sides ce, ae, ac, we permutate cyclically the other three, then by the reciprocal of Steiner's theorem, the three resulting Brianchon points lie on a right line, it is thus proved that three h points lie in a right line G. From the same circumscribing hexagon it can be inferred that the lines joining the point a to {bc, df} and d to {ac, ef} intersect on the Pascal abcdef, and that there are six such intersections on every Pascal.

More recently Prof. Cayley has deduced the properties of this figure by considering it as the projection of the lines of intersection of six planes. See Quarterly Journal, vol. IX. p. 348.

Still more recently the whole figure has been discussed and several new properties obtained by Veronese (Nuovi Teoremi sull' Hexagrammum Mysticum in the Memoirs of the Reale Accademia dei Lincei, 1877). He states with some extension the geometrical principles which we have employed in the investigation, as follows: I. Consider three lines passing through a point, and three points in each line; these points form 27 triangles which may be divided into 36 sets of three triangles in perspective in pairs, the axes of homology passing three by three through 36 points which lie four by four on 27 right lines. II. If 4 triangles a,b,c1, a,b,c, &c. are in perspective, the first with the second, the second with the third, the third with the fourth, and the fourth with the first, the vertices marked with the same letters corresponding to each other, and if the four centres of homology lie in a right line, the four axes will pass through a point. III. If we have four quadrangles a,b,c,d1, &c. related in like manner, the four points of the last theorem answering to the triangles bed, cda, dab, abc lie on a right line. Considering the case when all four quadrangles have the same centre of homology, we obtain the corollary: If on four lines passing through a point we take 3 homologous quadrangles a,b,c,d1, ab2łądy, a,b,c,d ̧; then we have four sets of three homologous triangles, a,b,c,, &c. the axes of homology of each three passing through a point and the four points lying on a right line. IV. If we have two triangles in perspective ab11, a,b,c, and if we take the intersections of bc bacca cɑ1; a,b, ab, we form a new triangle in perspective with the other two, the three centres of homology lying on a right line. It would be too long to enumerate all the theorems which Veronese derives from these principles. Suffice it to say that a leading feature of his investigation is the breaking up of the system of Pascals into six groups, each of ten Pascals, the ten corresponding Kirkman points lying three by three on these lines which also pass in threes through these points. It may be added that Veronese states the correspondence between a Pascal line and a Kirkman point as follows: Take out of the 15 lines C the six sides of any hexagon, there remain 9 lines C; out of these can be formed three hexagons whose Pascals meet in the Kirkman point corresponding to the Pascal of the hexagon with which we started.

After the publication of Veronese's paper Cremona obtained very elegant demonstrations of his theorems by studying the subject from quite a different point of view. From the theory of cubical surfaces we know (Geometry of Three Dimensions, Art. 536), that if such a surface have a nodal point, there lie on the surface six right lines passing through the node, which also lie on a cone of the second order, and fifteen other lines, one in the plane of each pair of the foregoing; by projecting this figure Cremona obtains the whole theory of the hexagon.

It may be well to add some formulæ useful in the analytic discussion of the hexagon inscribed in the conic LM-R2. Let the values of the parameter μ (Art. 270) for the six vertices be a, b, c, d, e, f, and let us denote by (ab) the quantity abL (a + b) R+ M, which, equated to zero, represents the chord joining