The Complete Pascal Figure Graphically Presented
J. Chris Fisher and Norma Fuller
Abstract. A hexagon inscribed in a conic determines 60 Pascal lines which generate a remarkable configuration of 146 points and 110 lines. Although the complete Pascal figure has been studied since the middle of the 19th century, it is only with today's computer graphics that we are able to actually see it. Our paper is designed to take full advantage of electronic publishing: we explain the notation and present the interactive graphics so that the viewer can explore the dozens of subconfigurations and see how they fit together to form the whole figure. Our paper comes in two pieces -- the first is the published article (which can be read independently) and the second is a computer web page
whose dynamic graphics accompany the published article.
We have designed a web page to accompany this paper. On that page is a copy of the text with links to the appropriate dynamic graphics. We recommend making a copy of Figure 3 (at the end of section 2) as an aid to following the terminology. For those readers who would like to pass conveniently between the original six points and the corresponding combinatorial structure we recommend making a copy also of Figure 1 (in section 2).
AB·DE, BC·EF, and CD·FA,
lie on a line, the Pascal line of the hexagon ABCDEF. The order is obviously important: these same six points in a different order generally determine a different Pascal line. Of course, we could name the hexagon starting with any of its six vertices (without changing the order); also, the hexagon described in a particular order would have the same Pascal line as that described in the reverse order; otherwise, the hexagons generally have distinct Pascal lines. In other words, any six points in general position on a conic determine = 60 distinct Pascal lines. Usually these Pascal lines intersect in pairs. It is clear, however, that they sometimes must meet in fours; for example, there are four Pascal lines through AB·DE (for the hexagons ABCDEF, ABFDEC, ABCEDF, and ABFEDC). Here AB·DE is one of three diagonal points of quadrangle ABDE. Our initial set of six points contains = 15 quadrangles, so there are 45 such diagonal points; together with the 60 Pascal lines, these diagonal points form a 454 603 — a configuration of 45 points on 60 lines, with 4 lines through each point and 3 points on each line. It is considerably more surprising that the Pascal lines sometimes meet in threes, and those new points lie on lines that intersect in threes, and so on. The discovery of these unexpected coincidences came gradually out of the work of several 19th century mathematicians — Steiner, Kirkman, Cayley, Salmon, Veronese, Cremona, among others. The complete figure is called (following Pascal) the hexagrammum mysticum or, more simply, the complete Pascal figure. In all it consists of 6+45+60+20+15 = 146 points and 15+60+20+15 = 110 lines.
There is a vast literature devoted to the complete figure, but most of it is expository, providing little information that was not already known in the 19th century. Following in this expository tradition, we offer no new results. Our purpose is to provide an informative picture of the complete figure. The motivation came from H.S.M. Coxeter, whose career as an eminent geometer began in his teenage years and ended 80 years later in 2003: When he first learned of the power of computer software that produces clear geometric figures more accurately and simply than anything achievable with pen and paper, he commented that he would like to see the complete Pascal figure. So would many of us. Although we delayed our project too long for Professor Coxeter’s benefit, we feel that there are enough curious geometers around to justify our work. It should be pointed out that there exists at least one accurate drawing of the complete figure . The Lintons’s picture is remarkable in many ways — indeed, it is unbelievable that an accurate india-ink drawing could have been accomplished at all, but they produced their picture in 1921, long before the era of interactive computer graphics. Our picture takes full advantage of Cinderella, the remarkable geometry software created and developed by Jürgen Richter-Gebert and Ulrich Kortenkamp. Our secondary purpose here is to bring to a wider audience a diagram of Stanley Payne that simplifies the task of labeling the points and lines.
We follow closely the two-dimensional descriptions of Baker , illustrating his principal ideas. His work is encyclopedic, with complete references to the original sources. We felt no need to provide further references except in those places where we used ideas that are not found in Baker.
2. The Complete Pascal figure.
We turn to the task of labeling the figure, basing our discussion on Baker , Note II, pages 219-236. Baker provides a thorough discussion that includes an outline of the original proofs, which he calls “interesting but intricate.” They depend upon successive applications of Desargues’ theorem; see his Example 6, pages 233-236. Baker preferred to investigate a corresponding configuration in 4-dimensional projective space, then dualize and project into the plane. For those who are happy working in higher dimensions, the approach gives the whole theory plus more in a simple, satisfactory way. We will not concern ourselves with proofs; our interest is in the notation. Baker took his notation from J.J. Sylvester , (1844) volume I p. 92, and (1861) volume 2 p. 265, who devised it in connection with an unrelated problem.
We let A, B, C, D, E, F be six points in general position on a conic. An unordered pair of these letters will be called a duad — the duad represents a secant of the given conic. A syntheme is a partition of our points into a set of three duads — it represents a triple of secants that cover our six points (and because we suppose the points to be in general position, the secants intersect in three noncollinear points off the conic). A hexagon is an ordering of these six points, with the convention that it may be named starting with any one of its letters, and it is equivalent to its reverse order. In this way there is a one-to-one correspondence between hexagons and Pascal lines. We need, however, a more informative labeling of the Pascal lines if we want to conveniently label the rest of the complete figure. To this end we form number duads and synthemes from the integers from 1 to 6. It was Sylvester who noticed that the two systems could be combined so that each number duad is associated with a unique letter syntheme, and each letter duad is associated with a unique number syntheme. Stanley Payne, in his investigation of generalized quadrangles, devised a neat way to display the relationship in a diagram , p. 224 and , pp122-123.
Figure 1. Payne’s diagram displaying the relationship between duads and synthemes.
The diagram (figure 1) consists of 15 points and 15 lines (of which 5 are circular arcs). Each line contains three points; through each point there passes three lines. The picture tells us that the duad AC, for example, corresponds to the syntheme 24.16.35, while a duad such as 16 corresponds to the syntheme AC.DF.BE. More precisely, a line corresponds to a letter duad and, therefore, to the syntheme of number duads attached to the three points on that line. Dually, a number duad corresponds to the syntheme consisting of letter duads attached to the lines through its point. This correspondence has properties that have been used in a variety of different settings. All we require is the property ,
For each number i from 1 to 6, no two of the five duads ij (j ≠ i) belong to the same syntheme.
It is easy to check the property in a straightforward manner, but it is more informative to note that the diagram is combinatorially unchanged by any permutation of the numbers from 1 to 6. It is therefore sufficient to check the property only for i = 1. The property implies that a pair of duads that share a number, such as 15 and 16, determine a pair of disjoint partitions of the original six points:
15 corresponds to AF.BC.DE,
16 corresponds to AC.DF.BE.
Alternating from one syntheme to the next we get the hexagon AFDEBC described by the secants AF, FD, DE, EB, BC, and CA:
Figure 2. How to label a Pascal line.
Since any hexagon corresponds to its Pascal line, we thus have a one-to-one correspondence between the 60 pairs of number duads that share a digit and the 60 Pascal lines determined by six points A, B, C, D, E, F of a conic. We represent Pascal lines by the symbol ij-ik (with the numbers appearing in increasing order). So the Pascal line of our hexagon AFDEBC is the line 15-16. In the other direction, a duad pair 23-34, say (sharing the digit 3), gives us the synthemes AF.CD.BE and AE.DF.BC, and thus corresponds to the hexagon AFDCBE. There seems to be no simpler way to start the labeling process. However, once the Pascal lines have been labeled, the rest is easy.
The most efficient way to present Baker’s labeling convention is by typical example. To check one’s understanding, one could count the number of each item with respect to all permutations of the integers from 1 to 6. So, for example, we write 12-13 for the typical Pascal line. This notation implies that there are six choices for the repeated digit (in place of 1) and = 10 choices for the other two (replacing 2 and 3); thus, the 6·10 = 60 available symbols agrees with the known number of Pascal lines (obtained by counting hexagons). We modify Baker’s notation in just one instance. He sometimes places square brackets around duads, something that has a natural interpretation in his higher-dimensional model. This idea works fine in the plane (that is the subject of our study) except for what he calls I-points. An I point is defined by a 6-tuple of bracketed duads which omit two digits; for example , , , , ,  omit 5 and 6. He chooses to abbreviate this information with a somewhat arbitrary choice of an unambiguous triple of duads such as .. or .. (but not ..). We find this convention awkward to use (and hard to program into a computer). We therefore substitute the symbol , even though it might lack a natural interpretation.
Here then are the points and lines of a complete Pascal figure. We are interested in three types of lines and three types of points.
Pascal lines are denoted by k.
There are 60 k-lines, such as 12-13.
On each k-line are 3 K-points 12.13.14, 12.13.15, 12.13.16,
and 1 G-point 12.13.23.
Kirkman points are denoted by K.
There are 60 K-points, such as 15.25.35.
Through each K-point pass 3 k-lines 15-25, 15-35, 25-35,
and 1 g-line --.
Cayley-Salmon lines are denoted by g.
There are 20 g-lines, such as --.
On each g-line are 3 K-points 14.24.34, 15.25.35, 16.26.36,
and 1 G-point 12.13.23,
and 3 I-points , , .
Steiner points are denoted by G.
There are 20 G-points, such as 12.13.23.
Through each G point pass 3 k-lines 12-23, 12-13, 13-23,
1 g-line --,
and 3 i-lines 12-, 13-, 23-.
Steiner-Plücker lines denoted by i.
There are 15 i-lines, such as 13-.
On each i-line are 4 G-points 12.13.23, 13.14.34, 13.15.35, 13.16.36.
Salmon points denoted by I.
There are 15 I-points, such as .
Through each I-point pass 4 g-lines --, --,
We summarize the notation in the following schematic diagram (based on ).
60 k-lines (Pascal lines)
60 K-points (Kirkman points)
20 g-lines (Cayley Salmon lines)
20 G-points (Steiner points)
15 i-lines (Steiner-Plücker lines)
15 I-points (Salmon points
Figure 3. Summary of the relationship among the points and lines of the complete Pascal figure.
3. How to determine Kirkman and Steiner points from hexagons.
Before turning to the subconfigurations, let us first address the question of how to recognize pairs of hexagons whose Pascal lines meet in either a Kirkman or a Steiner point. If nothing else, this exercise might lead to an appreciation of Baker’s duad-syntheme notation. Consider the hexagon ABCDEF, whose Pascal line (from figure 1) is 14-15. According to the theory, that line will meet a unique pair of Pascal lines in a Steiner point, and will meet three other pairs of Pascal lines in Kirkman points.
Rule 1, Steiner Points. To find the Steiner mates for the Pascal line of ABCDEF, fix the positions of three alternate letters, say A, C, and E, while cyclically permuting the order of the other three letters.
So for our example, we obtain the hexagons
ADCFEB with Pascal line 14-45
These Pascal lines meet in the Steiner point 14.15.45.
Remark. The other three permutations of B,D,F lead to the Pascal lines 26-36 of ABCFED, 23-26 of ADCBEF, and 23-36 of AFCDEB. One can easily confirm that the Steiner point 23.26.36 where these three lines intersect is conjugate in the conic to the previous Steiner point.
Perhaps a more natural way to obtain our set of three hexagons is to start with the triple of secants AB, CD, EF associated with the duad 14, and the triple AF, ED, CB associated with 15. Duad 14 is also mated with 45 in the hexagon obtained by interchanging the duads CD and EF:
AB CD EF --> AB EF CD.
Duad 15 is matched with 45 in the hexagon obtained by interchanging the duads ED and CB:
AF ED CB --> AF CB ED.
We obtain once again the three hexagons ABCDEF (= AFEDCB), ABEFCD (=ADCFEB), and AFCBED whose Pascal lines meet in the Steiner point 14.15.45. Hence,
Alternative Rule 1. To find the Steiner mates for the Pascal line of ABCDEF,
(a) fix the first duad AB and interchange CD with EF to obtain ABEFCD, then
(b) write the name in the reverse order AFEDCB, fix its first duad AF, and interchange ED with CB to obtain AFCBED.
We now present a rule for finding the Kirkman points.
Rule 2, Kirkman Points. To find the three pairs of Kirkman mates for the Pascal line of ABCDEF, first write its name in reverse order,
ABCDEF = AFEDCB,
(a) Fix the first duad of each order. For the name on the left, fix the positions of the duads CD and EF and permute their names to produce AB DC FE; for the name on the right, interchange the duads ED and CB and permute the name of the duad that ends up on the right to produce AF CB DE. The resulting three hexagons ABCDEF, ABDCFE, AFCBDE have respective Pascal lines 14-15, 12-14, 12-15 that meet in the Kirkman point 12.14.15.
(b) Reverse the roles of the right and left names from part (a): Fix the first duad of each. For the name on the left interchange the duads CD and EF, and permute the name that ends up on the right to produce AB EF DC; for the name on the right fix the positions of the duads ED and CB and permute their names to produce AF DE BC. The three hexagons ABCDEF, ABEFDC, AFDEBC have respective Pascal lines 14-15, 14-16, 15-16 that meet in the Kirkman point 14.15.16.
(c) For right and left names fix the first duad (AB on left, AF on right), interchange the other two and permute the name of the duad ending up on the left to produce AB FE CD on left and AF BC ED on right. The three hexagons ABCDEF, ABFECD, AFBCED have respective Pascal lines 14-15, 13-14, 13-15 that meet in the Kirkman point 13.14.15.
We summarize the rules in the following diagram.
Figure 4. The Pascal lines meeting 14-15 in Kirkman and Steiner points.
We look at the subconfigurations that Baker discusses under the heading, “examples of the theory.”
A. The partition of the complete figure into six Desargues configurations.
The set of all K-points and k-lines form a 603 603 (meaning, a configuration of 60 points with three lines through each, and 60 lines with three points on each). This subconfiguration itself comes in 6 disconnected pieces, each of which is a Desargues configuration 103 103: for each digit from 1 to 6, the Desargues configurations consist of all those Kirkman points and Pascal lines whose labels are composed of duads that contain that digit. Below is the picture for the digit 1.
Comment. One can obtain the complete set of ten hexagons (whose Pascal lines form a Desargues configuration) from one of its members by the following operation: (i) Pick a pair of adjacent vertices (say, for example, BC in ABCDEF), then (ii) interchange the other two adjacent pairs (BC FA DE), and (iii) switch the elements of just one of the three pairs (to get CBFADE, BCAFDE, and BCFAED). For a systematic way to list the ten hexagons associated with any given hexagon, start with any hexagon (say, ABCDEF) and reverse its order (AF ED CB); for the second hexagon and for all hexagons in the even positions on the list, fix the first pair of letters, interchange the other two, and switch the order of the new middle pair (AF ED CB ® AF CB ED ® AFBCED). For the third hexagon and for all hexagons in the odd positions on the list, proceed as for the even case, but switch the order of the new right-hand pair (AFBCED ® AD EC BF ® AD BF EC ® ADBFCE). Starting with the hexagon ABCDEF that corresponds to the Pascal line 14-15, we get the list of ten hexagons that correspond to the Pascal lines of Figure 5.
Pascal Line Hexagons in both orders
14-15 AB CD EF = AF ED CB
15-13 AF BC ED = AD EC BF
13-12 AD BF CE = AE CF BD
12-15 AE DB CF = AF CB DE
15-16 AF DE BC = AC BE DF
16-12 AC FD BE = AE BD FC
12-14 AE FC DB = AB DC FE
14-16 AB EF DC = AC DF EB
16-13 AC EB FD = AD FB EC
13-14 AD CE FB = AB FE CD
14-15 AB CD EF
Figure 5. The Desargues configuration of K-points and k-lines whose labels consist of duads that contain the digit 1.
B. Quadrangles of Steiner points each in fourfold perspective with a quadrangle of Kirkman points.
Consider the four G-points and four K-points whose labels do not contain a specific pair of integers, say for example, they omit the integers 5 and 6. The unique I-point that omits 5 and 6 has through it four g-lines, each containing one G- and one K-point that omit 5 and 6. In other words the quadrangle of G-points and the quadrangle of K-points are perspective from that I-point. Moreover, the join of any of these G-points to any of the other three K-points is a Pascal line, and these Pascal lines are easily seen to meet in sets of four in three diagonal points. Consequently, the two quadrangles are perspective in four ways, and the four centers of perspectivity form a quadrangle. Altogether, these four G-points, four K-points, three diagonal points, and the I point form with the four g-lines and twelve k-lines a 124 163. Each pair of the three quadrangles is perspective in four ways, with the four centers of perspectivity being the vertices of the third quadrangle. We say they form a desmic system of quadrangles. The name was coined by Greek mathematician Cyparissos Stephanos (Darboux Bull. II ser 3 (1879), 424-456) to refer to a particular system of three tetrahedra. The original Greek word means bound together: If two tetrahedra in 3-dimensions are perspective from three centers, then necessarily they are perspective from a fourth center; the four centers form a third tetrahedron with the property that any two of the three desmic tetrahedra are in perspective in four ways with the centers of perspectivity being the vertices of the third tetrahedron. Our desmic system of quadrangles is a projection into the plane of a desmic system of tetrahedra.
There is another way to view our 124 163: it consists of three triangles that are perspective in pairs from three collinear centers of perspectivity. The three pairs are not arbitrary: if two of the triangles are and , then the vertices of the third are . This observation comes from , section 4. Although the triangles do not arise in our figure in a natural way, they simplify its drawing.
Figure 6. A desmic system of quadrangles: a 124 163 consisting of 4 G-points in perspective from an I-point with 4 K-points. (There is one such desmic system for each I-point).
C. The configuration of Steiner points and Steiner-Plücker lines.
The collection of all G-points and i-lines forms a 203 154. For an instructive picture, consider six planes in general position in three dimensions; these planes intersect in = 20 points and = 15 lines. Those points and lines project into a planar 203 154 of the type shown in figure 7. Imagine three of the planes to be the sides of a triangular pyramid with vertex 12.13.23, and the other three planes as cross sections. In the projection this arrangement appears as three triangles in perspective from the same center (so that the three axes guaranteed by Desargues’ theorem necessarily pass through a point 45.46.56). Note that the triangles play no role in the interpretation here, but were used only to simplify drawing and analyzing the configuration. Indeed, the intersection point of the three axes can be considered the center of perspectivity of a second triple of perspective triangles.
Figure 7. The G-points and i-lines form a 203 154.
D. The configuration of Salmon points and Cayley-Salmon lines.
The set of all I-points and g-lines forms a 154 203 that can be thought of as three triangles in perspective from a line (in which case the centers necessarily lie on a line). In figure 8 we have, somewhat arbitrarily, emphasized three triangles: (i) whose I-points contain 5,6 and omit 4; (ii) they contain 4,6 and omit 5; (iii) they contain 4,5 and omit 6. An alternative way of viewing the figure would be to consider any point of the configuration to be the center of perspectivity of two quadrangles whose six pairs of corresponding sides meet in the vertices of a complete quadrilateral.
Figure 8. The I-points and g-lines form a 154 203.
 H. Baker, Principles of Geometry, volume II, 2nd edition. Cambridge University Press (1954). Note II, pages 219-236.
 Keith Dean and Floor van Lamoen, Geometric construction of reciprocal conjugations. Forum Geom. 1 (2001), 115-120.
 Anne and Elizabeth Linton, Pascal’s Mystic Hexagram, Its History and Graphical Representation. PhD. Thesis, Univ. of Pennsylvania, Philadelphia, PA 1921.
 Stanley Payne, Finite generalized quadrangles: a survey. Proceedings of the International Conference on Projective Planes, Washington State Univ. 1973, edited by M.J. Kallaher and T.G. Ostrom. Washington State Univ. Press (1973), 219-261.
 Stanley Payne and J.A. Thas, Finite Generalized Quadrangles. Research Notes in Mathematics, 110. Boston - London - Melbourne: Pitman Advanced Publishing Program (1984).
 J.J. Sylvester, Collected Mathematical Papers, Cambridge University Press. Volume 1 (1904) and Volume 2 (1908).
J. Chris Fisher, Department of Mathematics
Norma Fuller, First Nations University of Canada
University of Regina
Canada S4S 0A2