Aspects of the Butterfly Theorem

Transcription

1 Aspects of the Butterfly Theorem Rudolf Fritsch Ludwig-Maximilians-Universität München 1 Basics The object of the following study is the complete quadrangle consisting of four points (the vertices) in the real affine plane, no three on a line, and the six lines (the sides) connecting each pair of vertices Given a side connecting any two vertices there is an opposite side connecting the remaining two We refer to a given side and the corresponding opposite side as a pair of opposite sides Even though there is now no distinction between the sides and diagonals of the quadrangle, the term diagonal point will designate the intersection of a pair of opposite sides According to the Fano property, a complete quadrangle in the real affine plane has at least one proper diagonal point since there are three pairs of opposite sides, one has three diagonal points from which two might be improper In the main sections of this paper, the classification of the complete quadrangles by properties of the diagonal points is needed and will be used in the following form: 1 The class of convex quadrangles consists of all quadrangles with a (unique) diagonal point which is an interior point of both sides intersecting in that point The terminology reflects the fact that in this case the convex hull of the vertices is a convex quadrangle; the distinguished diagonal point is an interior point of this convex set and will be referred to as the interior diagonal point of the quadrangle The class of convex quadrangles has three subclasses; they are parallelogram trapezium general convex affine kite (a) parallelograms containing all quadrangles with two improper diagonal points, in particular, rectangles and squares; 0 The main part of this paper was written during a stay at the Kaliningrad State University which I thank for hospitality Thanks also to Roland Eddy and Chris Fisher for careful reading and improvement of the English text 1

2 (b) proper trapezia containing all quadrangles with one improper diagonal point; and (c) general convex quadrangles consisting of all convex quadrangles with no improper diagonal point Within this subclass, the subsubclass of (convex) affine kites where one side (a bisector) bisects its opposite side, deserves special attention Note that a trapezium which is an affine kite is a parallelogram 2 The class of nonconvex quadrangles consists of all quadrangles with three proper diagonal points such that each of them is an interior point of one of the sides intersecting there and an exterior point of the other In this case, the convex hull of the vertices is a triangle and one vertex is an interior point of this triangle; this vertex will be called the interior vertex of the nonconvex quadrangle Furthermore, the subclass of nonconvex affine kites quadrangles has the notable property that such a quadrangle may have one or three side bisectors In the latter case, the interior vertex is the centroid of the triangle formed by the other three vertices 1 general nonconvex one bisector three bisectors This general (affine) point of view will be essential in Sections 4, 5 and 6 of this paper The following two sections deal with some known results surrounding the Butterfly Theorem and are based on the usual Euclidean plane 2 The Butterfly Theorems of the 19th Century Coxeter and Greitzer [6] trace the Butterfly Theorem back to a paper published in 1815 by William Horner ( ), the namesake of the Horner scheme, for evaluating polynomials [34] Because of the lack of access to the original version, we state the theorem in a form similar to that given in [6] Theorem 1 (Butterfly Theorem) Through the midpoint M of a chord [P Q] of a circle K, two chords [AB] and [C] are drawn with A and C on the same side of the line P Q 2 If chords [A] and [BC] meet [P Q] at points X and Y, then M is the midpoint of the segment [XY ] 1 When one considers a quadrangle to be an ordered quadruple of points (with the four sides defined as the segments joining consecutive points) one has three main classes: convex, concave and crossed quadrangles, [27]; from our point of view there is no difference between convex and crossed quadrangles 2 The assumption A and C on the same side of the line P Q is not explicitly formulated in the statement of the Butterfly Theorem in [6], but implicitly used in the proof The case A and C on different sides of the line P Q is given there as a problem for the reader 2

3 P A C X M Y Q K B This theorem has gained a lot of interest in the past, in particular, in the last quarter of the 20th century, see the list of references, in particular [1], and for an Einstein-like treatment [16], [22] It is interesting to look at it from the point of view of real projective geometry, but this will not be done here A variety of proofs can even be found on the internet [31] where one can also see the nice demonstration on page [32] Here is a proof using complex numbers that I have not found in any reference Such a proof is appropriate for theorems in plane Euclidean geometry whenever a circle plays an essential role This circle will be taken as the unit circle in the Gaussian plane Proof of the Butterfly Theorem Without loss of generality one can assume the circle K to be the unit circle of a Gaussian plane and the line P Q parallel to the real axis Thus, the coordinates a, b, c, d of the points A, B, C and satisfy the relations ā = 1 a, 1 b = b, c = 1 c, 1 d = d while the coordinate m of M is purely imaginary, ie, m = m with the line P Q described by the equation z z = 2m In general, the line connecting two points R and S with the coordinates r and s has the equation [18] ( r s) z + (s r) z = r s r s This can be seen by taking Z as an arbitrary point on the line RS and noting that the vector RZ is a scalar multiple of the vector RS Thus the quotient (z r) : (s r) must be real, and so, z r z r = s r s r, which is equivalent to the former equation If R and S belong to the unit circle, then the righthand side of the latter equation reduces to r s z s, r s 3

4 which yields the simple form z + r s z = r + s as the equation of the connecting line Since the point M belongs to the lines AB and C, one has m a b m = a + b and m c d m = c + d, which gives b = m a 1 + a m, d = m c 1 + c m Note that the above denominators do not vanish since 1 + a m = 0 m = m = 1 = ā m = a, a which is impossible because A K and M K This allows one to compute the coordinates x and y of the points X A P Q and Y BC P Q to be x = m + 2am2 acm c + a 1 + (a + c)m ac, y = m + 2cm2 acm a + c 1 + (a + c)m ac in view of the fact that x = x 2m, ȳ = y 2m The denominators 1 + ad and 1 + bc cannot be zero: multiplying 1 + ad = 0 by ā yields ā = d implies A P Q, which is impossible, since the points A and lie on different sides of the line P Q by assumption One can now compute x + y 2 = m, which is the desired result This computational proof has the following advantages over the geometric proof in [6] 1 It is really easy to remember which, in our opinion fails to be true for the latter (although the authors claim that their proof is easily remembered 2 It points out the role of the condition that the points A and C lie on the same side of the line P Q: the lines A and P Q are not parallel Therefore, one can interchange the labelling of the points C and, as long as the lines AC and P Q are not parallel, without any change of the argument That means that M is not only the midpoint of the segment [XY ] but also of the segment [V W ], where V and W denote the intersections of the line P Q with the lines AC and B respectively 4

5 A C P Q V X M Y W B K This confirms the following remark by H Lenz in [15, p 46]: the arguments of synthetic geometry require, quite often, the careful discussion of many exceptions Thus one can understand why some mathematicians refuse synthetic methods on principle and work only with the algebraic representation Sometimes one can find synthetic proofs shorter and more elegant than the algebraic but this needs luck 3 The coordinates of the points P and Q don t appear in the proof It suffices to know that the point M is the foot of the line perpendicular to the line P Q drawn from the center of the circle So one can translate the point M and the line P Q almost arbitrarily along this perpendicular line without changing the statement of the theorem The only problems which might occur are caused again by vanishing denominators like 1 + a m = 0 implying M K In this case, the two chords would have a common endpoint and ABC would not be a quadrangle, while the hypotheses of the Butterfly Theorem in the given form imply that the points A, B, C, are the vertices of a complete quadrangle Thus, our proof yields more than the original Butterfly Theorem by establishing the following Theorem 2 Given a complete cyclic quadrangle and a line g not parallel to any side of the quadrangle If one diagonal point of the quadrangle is the foot M of the line perpendicular to g through the center of the circumcircle, then the two other pairs of opposite sides cut the line g in equal distances from the point M V X M Y W A C B K 5

6 Just for fun, note the special case of this theorem if M is the center of the circle Then, by Thales theorem, the quadrangle under consideration is a rectangle and, the other way around, any diameter of the the circle may serve as the line g A M Y X V W B C The following variation of the Butterfly Theorem is taken from [12, p 78] which gives credit to an 1884 publication [17] by John Sturgeon Mackay ( ), the first president of the Edinburgh Mathematical Society This society, in contrast to other learned societies, was founded mainly by school teachers [33] Theorem 3 Given a complete cyclic quadrangle If any line not parallel to a side of the quadrangle cuts two opposite sides in two different points 3 equidistant from the center of the circumcircle, it cuts every pair of opposite sides at equal distances from the center of the circumcircle W T X Y U V C A K B Proof Again we assume the circumcircle K to be the unit circle in the Gaussian plane with the line g parallel to the real axis described by the equation z z = 2m Here m, a purely imaginary number, is the coordinate of the foot of the line perpendicular to g through the center of the circle The vertices of the complete quadrangle under consideration are labelled as in the original Butterfly Theorem, as are the intersection points V, W, X, Y In addition, we 3 Again, the assumption in two different intersection points is not formulated in [12] But without this assumption the statement of the theorem is evidently wrong That would mean that any line through a diagonal point of the complete quadrangle under consideration would cut the two remaining pairs of opposite sides at equal distances from the circle 6

7 denote by T and U, with coordinates t, u, the intersections of the line g with the lines AB and C respectively, and we assume T U Now, let the points T and U be equidistant from the center of circle (the origin); this gives t t = u ū Since the points T and U belong to the line g, it follows that which is equivalent to t (t 2m) = u (u 2m), t 2 u 2 = 2m (t u) Since T U one can cancel t u and obtain t + u = 2m From T AB and U C one derives and t = a + b + 2mab 1 + ab a + b + 2mab 1 + ab, u = c + d + 2mcd 1 + cd + c + d + 2mcd 1 + cd The latter equation is equivalent to the equation = 2m a + b + c + d + abcd(ā + b + c + d + 2m) = 2m which is invariant under permutations of a, b, c, d and, thus, it is also equivalent to x + y = 2m as well as to v + w = 2m From these one obtains x x = x (x 2m) = x ( y) = y ( x) = y (y 2m) = y ȳ and, analogously, v v = w w, which proves the theorem Remarks 1 The unmotivated computation in the ultimate step of the preceding proof could be replaced by noting that x + y = 2m means that the points X and Y are equidistant from the point M and thus, by Pythagoras Theorem, from the origin 2 The six points T, U, V, W, X, Y form a quadrangular set in the sense of [26, p 49] and [5, p 20], denoted by Q(T V X, UW Y ) and (T U)(V W )(XY ) respectively Note that also the five points M, V, W, X, Y in Theorem 2 constitute a quadrangular set Q(MV X, MW Y ) or (MM)(V W )(XY ) A quadrangular set of four points cannot appear in this situation; it would require a further diagonal point on the line g But, according to Jungius Polar Theorem [12, Nr 139], the diagonal points different from the diagonal point M belong to the polar of M which is parallel to the line g but different from g; so none of them can lie on the line g 7

8 3 Generalizations of the 21st Century In this section, the results of [28], [3], [29], [4], and [25] are recalled 4 These papers inspired the considerations presented in the following sections of this paper In [28] a version of Mackay s theorem (Theorem 3) is presented which contains Theorem 2 as a special case Theorem 4 Given a complete cyclic quadrangle and a line g not parallel to a side of the quadrangle If the line g cuts a pair of opposite sides in two points equidistant from the foot M of the line perpendicular to it through the center of the circumcircle, then it cuts the other pairs of opposite sides at equal distances from the point M W T XM Y U V A C K B The proof of this theorem can be easily extracted from our proof of Theorem 3 A first generalization to conics can be found in [3] which is reproduced without proof Theorem 5 Given a complete quadrangle inscribed in a conic and a line g not parallel to a side of the quadrangle but perpendicular at a point M to an axis of the conic If g cuts one pair of opposite sides in two points equidistant from M then it cuts the other pairs of opposite sides at equal distances from the point M 4 It seems that the Zagreb mathematicians Vladimir Volenec and Zvonko Čerin tried to surpass each other with generalizations of the Butterfly Theorem 8

9 A C B The next step for a generalization needs the notion of the conjugate diameter of a line with respect to a conic Recall the definitions and the basic properties of a conic in the real affine plane: A line is called a diameter in the case of a central conic if it passes through the center and in the case of a parabola if it is parallel to the axis X U V M g W The affine hull of the locus of the midpoints of all chords parallel a line g is a diameter, the conjugate diameter of the line g; An asymptote of a hyperbola is considered to be selfconjugate, ie, conjugate to itself If the conic is an ellipse, every line g has a conjugate diameter d and is parallel to the tangents at the intersection points of the diameter d and the ellipse; d T Y a parabola, a line g has the conjugate diameter d if and only if it is parallel to the tangent at the intersection point of d and the parabola; g 9

10 d g a hyperbola, a line g different from the asymptoptes has a conjugate diameter d if it is not parallel to an asymptote; if the diameter parallel to such a line g does not meet the hyperbola then g is parallel to the tangents at the intersection points of the diameter d and the hyperbola d g g d The version of a generalized butterfly theorem stated and proved in [29] is Theorem 6 Given a complete quadrangle inscribed in a conic and a line g not parallel to a side of the quadrangle but conjugate to a diameter of the conic If g cuts a pair of opposite sides in two points equidistant from the point of intersection M of the line g with the conjugate diameter, then it cuts the other pairs of opposite sides at equal distances from the point M V T Y X A U C B W An interesting degenerate case has been overlooked in the proof of the preceding theorem in [29] If the conic is a hyperbola and the line g an asymptote then the line g itself is the diameter appearing in the statement of the theorem and every point on the line g may serve as point M It is worthwhile to state and prove this special case separately 10

11 Theorem 7 If a complete quadrangle is inscribed in a hyperbola, the three pairs of opposite sides cut any asymptote h in segments with the same midpoint B h C A Proof In [29], the following pair of functions ϕ(λ) = 1 λ 2 1, ψ(λ) = λ λ 2 1 is given as a useful parametric representation of a hyperbola A quadrangle inscribed to the hyperbola is given by four different values λ A, λ B, λ C, and λ for the parameter λ while the equation x y = 0 describes an asymptote g Furthermore, the midpoint of the segment cut from g by the lines AB and C has the coordinates ( 1 4 σ 4 3 σ σ 2 σ 1 4 (σ 4 σ 3 + σ 2 σ 1 + σ 0 ), σ ) 4 3 σ σ 2 σ 1, 4 (σ 4 σ 3 + σ 2 σ 1 + σ 0 ) where σ i denotes the elementary symmetric functions of the variables λ A, λ B, λ C, and λ of degree i From the symmetry of these coordinates the claim follows In [25], the point of view is slightly different and indicates the direction of the following developments in this paper The main result of [25] is here reproduced without proof Theorem 8 Given a complete cyclic quadrangle there is an infinite number of points which obey the conditions of Theorem 4 All these points are located on the equilateral hyperbola containing the diagonal points of the quadrangle and the center of the circumcircle 11

12 Indeed, the hyperbola is the locus of the points M in Theorem 4 which will be clear later on 4 Butterfly points and lines Consider a complete quadrangle Q = ABC and an arbitrary line g This line is parallel to at most two sides of the quadrangle and, if parallel to two of them, then those two form a pair of opposite sides Thus, the line g is cut at least by two pairs of opposite sides in line segments The line g is called a butterfly line (of the quadrangle Q) if two of these segments have the same midpoint which, in turn, is called a butterfly point of the quadrangle Q associated with the line g Recall Theorem 1 in [4]: Theorem 9 If a butterfly line is cut by all three pairs of opposite sides in honest line segments, then the midpoints of all three segments coincide Proof The proof in [4] uses complex numbers as coordinates But the claim is an easy consequence of basic properties of quadrangular sets The three pairs of endpoints of the line segments under consideration are three pairs of a unique involution [5, Theorems 532 and 533] If two pairs have the same midpoint, this involution is in affine terms the reflection in this midpoint Thus, the points of the third pair are interchanged by this reflection and so they have the same midpoint Indeed, another proof of this fact is implicitly contained here in the subsequent developments Note that all the sides of the quadrangle Q are butterfly lines and the midpoints of the sides are the associated butterfly points In the following discussion, butterfly points and lines will be studied by the means of analytic geometry Since, the butterfly property of a point or a line is invariant under affine transformations, one can restrict the considerations to a special affine coordinate system Given a quadrangle Q, choose a proper diagonal point O and a labelling A, B, C, and of the vertices such that O is the intersection point of the lines AC and B It should be noted that, later on, a more specific choice of the diagonal point and the labelling of the vertices will be useful If one now chooses the point O as origin of the coordinate system 12

13 and the points A and B as the unit points on the x- and y-axis respectively, the vertices have the coordinates of the form A = (1, 0), B = (0, 1), C = (c, 0), = (d, 0) with c, d 0, B C A Given a line g the intersection points with the sides of the quadrangle are labelled as in Section 2, but now affine coordinates are used For future computations the following observation is helpful Theorem 10 Given a quadrangle, a line parallel to a side but different from it can be a butterfly line only if the side is parallel to its opposite side Proof In our coordinate system, it will be shown that no line x = k with k 0 is a butterfly line Such a line g is cut only by the pairs (AB, C) and (BC, A) in segments, thus being a butterfly line requires that the midpoints of these two segments coincide The coordinates of T AB g, Y BC g, U C g, and X A g are computed as T = (k, 1 k), Y = (k, 1 k c ), U = (k, d k d ), X = (k, d (1 k)) c while the midpoints of the segments [T, U] and [X, Y ] are ( ) ( ) c(1 + d) k(c + d) c(1 + d) k(1 + cd) k,, k, 2c 2c They agree only for k(c + d) = k(1 + cd), ie, k(1 c)(1 d) = 0; this, according to the assumptions on the numbers c and d, implies k = 0, in which case the line is the y-axis As a consequence of the preceding theorem, the search for butterfly lines using the chosen coordinate system can be restricted to lines which can be represented in the form y = m x + t, (1) where m and t are arbitrary real numbers For these lines the y-coordinates of the midpoints of the segments under consideration are easily computed Furthermore, if the labelling of the endpoints of the segments is taken as in Theorem 13

14 3, then one obtains [T U] : c(1 + d) m 2 + ((c + d) t + d (1 + c)) m + 2d t 2(1 + m) (d + c m), [V W ] : t 2, [XY ] : c(1 + d) m 2 + ((1 + c d) t + d (1 + c)) m + 2d t 2(1 + c m) (d + m) where the midpoints exist if the denominators don t vanish From this, one easily derives again Theorem 9 (ie, Theorem 1 of [4]) and the necessary and sufficient condition c (t 1 d) m 2 d (1 + c) m d t = 0 (2) for a line (1) to be a butterfly line different from the y-axis Next we turn our attention to the butterfly points In addition to the midpoints of the sides, there are other remarkable points of the quadrangle which are butterfly points Theorem 11 A proper diagonal point O of a complete quadrangle is a butterfly point of the quadrangle If the quadrangle is not a parallelogram, then there is exactly one butterfly line whose associated butterfly point is the point O Proof We again use our coordinate system with the diagonal point O as origin The quadrangle is a parallelogram if and only if c = d = 1 Setting t = 0 in equation (2), one obtains c (1 + d) m 2 + d (1 + c) m = 0 In the case of a parallelogram, the equation does not pose a condition on the slope m of a butterfly line and, thus, every line through origin is a butterfly line with the origin as associated butterfly point If d = 1 c then m = 0 and thus, the x-axis is a butterfly line However, the associated butterfly point has the coordinates ( ) 1 + c 2, 0, and is different from the origin, so the y-axis is the only butterfly line with the origin as associated butterfly point If both parameters c and d are different from 1, then the equation for the slope m has the roots 0 and d (1 + c) c (1 + d), giving rise to two butterfly lines, the x-axis and another one which has the origin as associated butterfly point The second part of the previous statement holds in a more general form, 14

15 Theorem 12 If the point P is a butterfly point of the quadrangle Q then it is contained in exactly one butterfly line of the quadrangle passing through P except when the quadrangle Q is a parallelogram and P is its only proper diagonal point Proof If the point P = (x o, y o ) does not lie on the axes of the coordinate system, the line meeting the axes in the points (2x o, 0) and (0, 2y o ) is the only candidate for a butterfly line passing through this point By symmetry, one can restrict the argument to butterfly points on the x- axis With respect to the previous lemma, one can also assume the point under consideration to be different from the origin, ie, P = (x o, 0) with x o 0 The corresponding butterfly line can t be parallel to the y-axis so it must have an equation of the form y = mx + t with t = m x o If the slope of such a butterfly line does not vanish, then the associated butterfly point can t lie on a coordinate axis It follows that the slope must be zero, and the only possible butterfly line is the x-axis Note also that this implies x o = (1 + c)/2 In the following, a special choice of the origin and a more careful labelling of the vertices will be used In particular, the origin is taken as the interior diagonal point in the case of a convex quadrangle, and as an interior point of the side [B] in the nonconvex case Moreover, the following affinely invariant rules should be satisfied O AC [B], in the case of a convex quadrangle in the case nonconvex quadrangle OB O OA OC 1, OB O 1, OA OC 1 The parameters c and d then obey the following restrictions: c = d = 1 for a parallelogram, c = d < 1 for a proper trapezium, d < c 1 for a general convex quadrangle, in particular, d < c = 1 for a convex bisecting quadrangle, c > 1 and d 1 for a nonconvex quadrangle, in particular, c > 1 and d = 1 for a nonconvex affine kite quadrangle and c = 3 and d = 1 for the nonconvex quadrangle whose interior vertex is the centroid of the triangle formed by the other vertices 15

16 Remark From the Euclidean point of view the picture for the proper trapezium (c = d < 1) shows an isosceles trapezium Sometimes it is necessary to get rid of the additional symmetry due to the isosceles property Then in contrast to the conventions before a proper trapezium will be characterized by 1 < c = 1 d < 0; note that one still has d < c, 1 by the lack of c < 1 5 The locus of butterfly points The section starts out by recalling two well-known theorems from elementary geometry The first, by Pierre Varignon ( ), states that the midpoints of two pairs of opposite sides of a quadrangle form a parallelogram whose sides are parallel to the remaining pair of opposite sides The second, formulated by Blaise Pascal ( ) in a more general form, states that the vertices of a hexagon with parallel opposite sides belong to a conic It follows from these theorems that the midpoints of the six sides of a complete quadrangle belong to a conic that also contains the diagonal points; moreover, the center of this conic is the vertex centroid of the quadrangle This midpoint conic is the locus of the centers of the conics passing through all vertices of the quadrangle It is a hyperbola in the case of a general convex quadrangle; one branch contains the centers of the circumscribed ellipses, the other the centers of the circumscribed hyperbolas, while the points at infinity belong to the axes of the circumscribed parabolas It degenerates to a pair of intersecting lines for trapezia If the quadrangle is a proper trapezium one line, g say, is the line midway between the parallel sides, the other, h, connects the midpoints of the parallel sides The lines supporting the parallel sides form a degenerate circumscribed conic with all points of the midline g as centers The line h is divided by g into two rays The ray which meets the smaller of the parallel sides consists of the centers of the circumscribing hyperbolas, the other of the centers of the circumscribing ellipses If the quadrangle is a parallelogram, then the midpoint conic consists of the midlines; their intersection point, the proper diagonal point, is the center of many circumscribed ellipses and hyperbolas, the other points on the midlines are centers of the degenerate conics consisting of the pairs of parallel lines which support the pairs of parallel sides In the nonconvex case the midpoint conic is an ellipse; all circumscribed conics are hyperbolas 16

17 B B C A A C The center of the midpoint conic belongs to a side if and only if this side is a bisector If there are three bisectors (which is only possible for a nonconvex quadrangle) then the interior vertex is the center and the midpoint conic is the Steiner ellipse inscribed in the triangle formed by the three other vertices Returning for a moment to the Euclidean plane, we note that the equation of the locus describes a circle if and only if the quadrangle is orthocentric In this case, the circle is the familiar 9-point (or Feuerbach) circle of the convex hull and the diagonal points are the feet of the altitudes Remark From the point of view of dynamical geometry if one has a dynamical process with transforms a hyperbola into an ellipse, one might expect a parabola at the transition point Here moving the point C along the line AC from outside the triangle AB C and A on different sides of the line B inside one gets a pair of parallel lines if C is located on the line B Theorem 13 The locus of the butterfly points of a complete quadrangle is its midpoint conic Proof The equation for the desired locus is obtained by eliminating the slope m and the axis segment t from the equations (2) and This yields y = m x + t = t 2 ( d x 1 + c ) 2 ( c y 1 + d ) 2 = 1 (c d 1) (c d) Thus, the butterfly points not on the y-axis satisfy this equation as do the butterfly points on the y-axis, the origin and the point (0, (d + 1)/2) Clearly, the set of butterfly points is contained in the conic described by this equation It remains to show that every point of this conic is a butterfly point This follows immediately for the two points of intersection of the conic with the y- axis On the other hand, for a point (x o, y o ) on the conic but not on the y-axis 17

18 set m = y o x o, t = 2 y o in order to obtain a butterfly line corresponding to (x o, y o ) Note that it does not make sense to ask for the main axes of this conic because these are not invariant under affine transformations But the equation shows that the pair of opposite sides meeting in the origin is parallel to a pair of conjugate diameters Since we are interested in affine properties, each proper diagonal point can be taken in turn as the origin, and therefor each pair of non-parallel opposite sides is parallel to a pair of conjugate diameters If the quadrangle is a proper trapezium, then the vertex centroid belongs to the middle parallel and the conic is a degenerated hyperbola composed of middle parallel and the line connecting the two proper diagonal points In the case of a parallelogram, one has the degenerate conic consisting of the two middle parallels which could be seen immediately without any computation Moreover, Theorem 4 of [25] states that, in the case of a cyclic quadrangle, the locus of butterfly points coincides with the locus of the centers of all central conics passing through the vertices of the quadrangle This pencil of conics is, in the coordinate system considered here, given by the equation d x 2 + λ d x y + c y 2 d (c + 1) x c (d + 1) y + c d = 0 (3) with λ R, while the conic ( d x c + 1 ) 2 ( c y d + 1 ) 2 = 1 (c + d + 2) (c d) is locus of the centers From the Euclidean point of view, one has a cyclic quadrangle if c = d which is an isosceles trapezium In this case the locus of the butterfly points and the locus of the centers of the circumscribed central conics agree, in fact, they form the degenerate conic consisting of the angular bisector of the first and third quadrant and the middle parallel line of the trapezium 6 Sets of butterfly lines We now describe the set of butterfly lines for arbitrary quadrangles, which is the principal aim of this paper According to equation (2), the set of butterfly lines envelops an algebraic curve C of class 3 Our main result is the following Theorem 14 The algebraic curve C enveloped by the set of butterfly lines of a quadrangle is 1 for a parallelogram, the union of three pencils of lines determined by the diagonal points, one proper and two improper; 18

19 2 for a proper trapezium, the union of the parabola touching all four nonparallel sides with the pencil of lines determined by the improper pencil point B C A (The picture shows the case with the vertices given in Section 4, specialized to c = 05, d = 2 see the remark at the end of Section 4; the parabola is given by the equation 16x xy + y 2 16x + 8y + 4); 3 for the case of a general convex quadrangle, an irreducible rational quartic with one cusp and the line at infinity as a double tangent; C B A 19

20 4 for the case of a nonconvex quadrangle, an irreducible rational quartic with three cusps not meeting the line at infinity in real points Proof The coordinate system described previously is used Case 1: The quadrangle is a parallelogram (c = d = 1) in which case the defining equation (2) reduces to t (1 m) (1 + m) = 0 The curve thus decomposes into three pencils of lines determined by the three diagonal points (one proper and two improper) These comprise the pencil of all lines through the center of symmetry of parallelogram (the origin, where t = 0), the pencil of all lines parallel to one pair of parallel opposite sides (lines with slope 1 parallel to the angle bisector of the first and third quadrants), and the pencil of lines parallel to the other pair of parallel opposite sides (lines with slope 1 parallel to the angle bisector of the second and fourth quadrants) Case 2: The quadrangle is a proper trapezium Assume see the remark at the end of section 4 cd = 1, c 1, which gives more insight in the geometric situation Now the defining equation reduces to (1 + c m) ((c (t 1) 1) m t) = 0 In this case, the curve decomposes into the pencil of lines through the improper diagonal point (with slope 1/c), and the parabola with the point form (x y) 2 (1 + c) (2 x + 2 y 1 c) = 0 Among the tangents to this parabola, there are the midline and the sides through the proper diagonal points of the trapezium The tangency point of a side is obtained by reflecting the diagonal point on this side in the side s midpoint Since the latter is the butterfly point associated with the side, one finds that, in general for a butterfly line, the associated butterfly point and the point where it touches the enveloping curve are different In particular, the chord of the parabola determined by the tangency points on a pair of opposite sides is 20

21 parallel to the parallel sides of the trapezium; thus the corresponding conjugate diameter intersects the midline in the point of tangency 5 Case 3: In the case of a general convex or a nonconvex quadrangle, the given line form of the curve C is irreducible Its point form, obtained as described in [2], is the equation f(x, y) = 4 (dx 2 cy 2 ) 2 4 px (3 x 2 3 px + p 2 ) d 2 4 qy (3 y 2 3 qy + q 2 ) c 2 cd (qx + py pq) (20 xy qx py + pq) = 0, where p = c + 1, q = d + 1 Thus, C is an irreducible quartic From this representation of the curve C the claims concerning its relation to the line at infinity in parts 3 and 4 of the theorem are immediately clear In the case of a general convex quadrangle the curve C touches the line at infinity at the points (± c/d, 1, 0) while in the case of a nonconvex quadrangle the equation dx 2 cy 2 = 0 fails to have a nontrivial real root A possible rational parametrization of this curve is given by the pair of functions ϕ(λ) = dλ2 (p(dλ 2 + c(d 1) 2 ) + 2 cq(d 1)λ) (dλ 2 c(d 1) 2 ) 2, ψ(λ) = c(d 1)2 (q(dλ 2 + c(d 1) 2 ) + 2 dp(d 1)λ) (dλ 2 c(d 1) 2 ) 2, although there may be a simpler one Next, one has to look for the singular points of the quartic C According to the general theory of algebraic curves, a quartic has at most three singular points The singular points are obtained as commons zeros of the function f and its partial derivatives f x, f y A simple special case is that of the bisecting quadrangles where, in the convex case (d < 1 and p = 0), the singular points can be easily computed as (0, q), ( ±i 3q 3d, q ) 8d 8 For the nonconvex case (c > 1, q = 0), there are three real singular points of the form ( ) (p, 0), p 8, ±3p 3c 8c obtained by a symmetry argument In the convex case, the case of an affine kite, there is only one bisector which contains the single cusp of the curve C 5 Note as an aside that all five tangency points are easily to obtain for any trapezium; thus any dynamical geometry software producing a conic from five points allows one to visualize this parabola 21

22 C B A Here is a picture of the complete curve after a projective transformation transforming the line at infinity u into a proper line C A u B For nonconvex, one may have one or three bisectors furthermore, a side contains a cusp if and only if it is a bisector B A C Now, general convex and nonconvex quadrangles without bisectors will be 22

23 considered with the main role in the discussion being played by the polynomial g(y) = 64 cy 3 48 cqy ( 9 dp 2 5 cq 2) y cq 3 This polynomial and its derivative with respect to y have greatest common divisor 1, so it has three different roots, y 1, y 2, y 3 say Consider, next, the three points (x i, y i ), where x i = p (q y i) 8 y i + q for i = 1, 2, 3 Since, these points represent all singular points of the curve C, the existence of three singular points is verified Since the determinant of the Hessian matrix of the function f vanishes at these singular points, they are cusps It remains to decide if these singularities are real or complex Since the polynomial g has order 3, it has at least one real root, and the corresponding singular point is real Observe under the given assumptions that the function g has two extrema such that the product of their ordinates is positive in the convex case and negative in the nonconvex case This proves the claim In the picture showing the general convex case in the statement of the preceding theorem, Case 2, it looks as if the cusp might belong to a side of the quadrangle; but the following enlargement of the neighborhood of the cusp shows that this is not the case In the discussion of Case 2 of the previous proof, it was mentioned how, when a butterfly line is a tangent to the parabola, the associated butterfly point is related to the point where the line touches the parabola This turns out to be true in a more general version Theorem 15 If Q is a general convex or a nonconvex complete quadrangle and g a butterfly line of it, then there is exactly one hyperbola circumscribing the quadrangle for which g is an asymptote The point where the line g touches the curve C of the preceding theorem is obtained by reflecting the center of this hyperbola in the butterfly point associated to the butterfly line g 23

24 Proof The previously described coordinates will be used for the vertices of the quadrangle under consideration Since the claim is trivial if the line g is the y-axis, we can restrict our attention to lines g given by an equation of the form y = mx + t The hyperbola in question then has the equation mdx 2 xyd m 2 xyc + mcy 2 mdpx mcqy + dmc, while its center is the point ( cm(cqm2 + 2 dpm + dq) (cm 2 d) 2, dm(cpm2 + 2 cqm + dp) (cm 2 d) 2 and the point where g touches the curve C has the coordinates ( d(cpm cqm + dp) (cm 2 d) 2, cm2 (cqm 2 ) + 2 dpm + dq) (cm 2 d) 2 Adding the second coordinates one obtains using equation (2) m(cqm + dp) cm 2 d which proves the claim, since the second coordinate of the butterfly point on the line g is t/2 The section will close with a curious observation The results presented in the Section 3 ask for some reversion of the argument, but, surprisingly, the following holds true Theorem 16 Let Q be a quadrangle and g a butterfly line for this quadrangle Then the butterfly point associated with g is the intersection of g with its conjugate diameter, where the conjugate is taken with respect to any conic K circumscribing the quadrangle Q Proof We use equation (3) for the pencil of conics circumscribing the quadrangle Q Consider a butterfly line g with the equation y = mx + t Since the associated butterfly point is the midpoint of the segment cut from the line g by such a conic, it belongs to the diameter conjugate to the line g 7 The set of butterfly lines in the space of lines The aim of this section is to describe the set of butterfly lines of a quadrangle Q = ABC considered as the set of lines given by equation (2) plus the y-axis and the line at infinity in the space of lines which is defined as usual as the Grassmann manifold G 2 (R 3 ) It is the real projective plane for whose points homogeneous coordinates (g, h, k) are chosen such that the line at infinity is represented by (0,0,1), the line AC, (ie, the x-axis), by (0,1,0), the line B, (ie, the y-axis), by (1,0,0) and the line AB by (-1,-1,1) The pencil of the lines = t, ), 24

25 through the origin is the set {(g, h, 0)} which will serve as the line at infinity of the projective plane G 2 (R 3 ) Then the set of butterfly lines of the quadrangle Q is the the set of points of the cubic B given by the equation c(1 + d)g 2 h + cg 2 k d(1 + c)gh 2 dkh 2 = 0 (4) Theorem 17 1 The cubic B decomposes into three different lines if the quadrangle Q is a parallelogram Taking c = d = 1 and the affine picture with k = 0 as line at infinity, one obtains the angular bisectors of the quadrants and the line at infinity (as in the figure) g The cubic B decomposes into the product of a quadric and a line if the quadrangle Q is a proper trapezium Taking the affine picture as before, one obtains a hyperbola passing through the origin and the diameter through the origin c = 3 8, d = The cubic B has the line at infinity (0,0,1) as a crunode (ordinary double point) if the quadrangle Q is general convex c = 2, d = 3 with the unique real inflection point marked If, moreover, the quadrangle Q is an affine kite, the affine picture is an axial cubic with the only real inflection point being improper 25

26 c = 1, d = The cubic B has the line at infinity (0,0,1) as an acnode (isolated point) if the quadrangle Q is nonconvex c = 2, d = 10, marked acnode, three inflection points The affine picture is the axial mixed cubic 6 with one improper real inflection point if the quadrangle Q is moreover an affine kite; c = 2, d = 1, marked acnode there is no further specialization if there are three bisectors Proof 1 For d = c = 1 equation (4) reduces to (h g)(h + g)k = 0 2 For d = c 1 (and c 0) equation (4) reduces to (g h)((1 + c)gh + (g + h)k) = 0; ie, the curve decomposes into the line B 1 given by g = h and the quadric B 2 given (1 + c)gh + (g + h)k = 0 Taking k = 0 as the line at infinity and 6 In German this cubic has the flowery name Kinderhemd (children s shirt) [9, page 164] 26

27 substituting k = 1 the quadric becomes a hyperbola with the center ( c 1 + c, c ) 1 + c belonging to the line B 1 Here it is worthwhile to consider the alternative possibility, mentioned at the end of Section 4, for assuring the quadrangle Q to be a proper trapezium: In this case equation (4) reduces to d = 1 c 1 (cg h)((1 + c)gh + (cg + h)k) = 0; ie, the cubic decomposes into the line B 3 given by x = y and the quadric B 4 given (1 + c)gh + (g + h)k = 0 In the affine derivation the quadric becomes again a hyperbola with the center ( belonging to the line B c, c 1 + c 3 In the remaining cases the cubic B is irreducible, and it is immediate that (0,0,1) is a double point The classification of singular cubics is given in [9, page 112] The tangents at the double point obey the equations c h = ± d g; so they are real if the parameters c and d have the same sign, ie, if the quadrangle Q is convex In this case there is one and only one real inflection point Now assume that in addition the quadrangle Q is an affine kite, that means according to the conventions in Section 4 d < c = 1 d < 0 Equation (4) reduces to ) g 2 ((1 + d)h + k) + dh 2 k = 0 from which one obtains the claimed symmetry The Hessian of the cubic is computed as (k 2(1 + d)h)g 2 + d(1 + d)h 3 + dkh 2 Thus the real inflection point is the point (1,0,0), the improper point of the y-axis 27

28 4 If the quadrangle Q is nonconvex the parameters c and d have different signs and the tangents are conjugate imaginary There are three real inflection points The same computations as before show the symmetry and one improper real inflection point in the case of a bisecting nonconvex quadrangle Three bisectors are obtained for c = 3 and d = 1 so that the equation of the cubic reduces to 3g 2 + 4gh 2 + h 2 which does not exhibit further special properties Remarks 1 If one looks at part 2 of the preceding theorem from the Euclidean point of view, the first case discussed in the proof deals as mentioned earlier with an isosceles trapezium, and the cubic under consideration decomposes into an isosceles (rectangular) hyperbola and its main axis c = d = 2 2 Note again as in Section 5 an interesting observation from the point of view of dynamical geometry At the stage of a transition from the convex to the nonconvex case one does not obtain a cubic with a cusp Consider for example the case c = 0 Then the cubic decomposes into the x-axis counted twice and the line x = 1 parallel to the y-axis We now investigate the location of the curve of butterfly lines in the partition of the dual plane induced by the fundamental quadrangle Q = ABC will be studied The four lines corresponding to the pencils of lines through the vertices of Q form a complete quadrilateral in the dual plane whose vertices correspond to the sides of Q The complement of this quadrilateral consists of seven domains, four triangular and three quadrangular; these will be called Q-triangles and Q-quadrangles respectively 28

29 The three Q-quadrangles shaded in three types The geometric meaning of these domains is given in the next theorem, which needs an introductory explanation: The three pairs of opposite sides of a quadrangle meet any line (not through a vertex) in an induced quadrangular set consisting of three pairs of an involution, assigned to the pair (Q, l) [5, Theorem 533], [21, page 138] This involution (ie projectivity on l of period 2) [5, Section 54] and is called elliptic and hyperbolic according as there are 0 or 2 fixed points Theorem 18 Let a quadrangle Q = ABC and a proper line l not passing through a vertex of the quadrangle be given The involution assigned to the pair (Q, l) is elliptic if the l line belongs to Q-triangle and hyperbolic if the l line belongs to a Q-quadrangle In particular, the cubic B is contained in the union of the Q-quadrangles and their vertices Proof A coordinate system as in Section 4 is chosen Then the Q-triangles have one vertex the x-axis or the y-axis on the line at infinity whose other points belong the Q-quadrangles Since the origin in the plane of the quadrangle Q is a diagonal point of Q and thus a fixed point of the involutions induced on the lines containing it, these involutions are hyperbolic This takes care of the cases where the line l passes through the origin, ie, belongs to the line at infinity of the projective plane G 2 (R 3 ) It remains to consider lines l with coordinates (g, h, 1), ie, with coordinates (g, h) (0, 0) in the affine derivation of the projective plane obtained by taking k = 1 The condition that the line l does not pass through a vertex of the quadrangle Q yields the following restrictions on the coordinates (g, h): A l g B l h C l cg l dh Here is a picture of the decomposition of the affine plane assuming c < 0, d < 0 assuring the quadrangle Q to be convex with the Q-quadrangles shaded 29

30 0 A C B 1 + g < 0 0 < 1 + g 0 < 1 + cg 1 + cg < h < 0 0 < 1 + h 0 < 1 + dh 1 + dh < 0 The lines are labelled by the centers of the corresponding pencils A similar picture can be drawn for c > 0, d < 0, when the quadrangle Q is nonconvex In both cases it is easy to check that the line l with the affine coordinates (g, h) belongs to a Q-quadrangle if and only if cd(1 + g)(1 + cg)(1 + h)(1 + dh) > 0 (5) The intersection points of the line l with the sides of the quadrangle Q can be easily computed: T = ( h + 1 h g, g + 1 g h ), U = ( c(dh + 1) dh cg, d(cg + 1) cg hd ), V = ( 1g, 0 ), W = ( 0, 1 h ), X = ( dh + 1 dh g, d(g + 1) g dh ), Y = ( c(h + 1) h cg, cg + 1 cg h ) Note that in these expressions at most two of the denominators may vanish, namely if the line l is parallel to one side or two opposite sides of the quadrangle; if there are two vanishing denominators, they belong to the same row of this table In this case the corresponding intersection point(s) belong to the line at infinity As said in the introduction to this theorem, the pairs (T, U), (V, W ), (X, Y ) are pairs of one involution The task is to determine the type of this involution Assume first the line l not being parallel to the y-axis, ie, h 0 Then the 30

31 projection onto the x-axis yield three pairs of an involution of the same type, which can be considered as an involution on the real line Now recall that an involution on the real line R is given by an assignment [21, page 281]: t a 1 + bt (6) a 2 t b with a 1, a 2, b R, a 1 a 2 +b 2 0 It is elliptical for a 1 a 2 +b 2 < 0 and hyperbolic with the fixed points b ± a 1 a 2 + b 2 for a 1 a 2 + b 2 > 0; in particular, it is a reflection for a 2 = 0 In the case under consideration one may take a 1 = c(1 + h)(1 + dh), c a 2 = d(1 + c)gh 2 + dh 2 c(1 + d)g 2 h cg 2, b = a 1 g According to equation (4) parameter a 2 vanishes if and only if the line l is a butterfly line of the quadrangle Q Now compute a 1 a 2 + b 2 = cd(1 + g)(1 + cg)(1 + h)(1 + dh)h 2 The comparison with the inequality (5) proves the claim for h 0 Finally, if h = 0 projection on the y-axis yields the pairs ( g, d + d ) (, (0, ), d + d cg g, ) cg Now a 1 = d(1 + g)(1 + cg), a 2 = cg 2, b = 0 do the job The decision between elliptic and hyperbolic involution depends on the sign of a 1 a 2 + b 2 = cd(1 + g)(1 + cg)g 2 which agrees with the sign of the left-hand side of the inequality (5) after the substitution h = 0 In order to get a better visualization of the curve B of butterfly lines in the Grassmann manifold G 2 (R 3 ), which is the real projective plane, a finite model of the Grassmann manifold is chosen: the unit disk in R 2 with identification of 31

32 the pairs of diametral points on the boundary circle To a point (g, h, k) in the coordinatisation before the point ( g g2 + h 2 + k 2, ) h g2 + h 2 + k 2 of the unit disk is assigned Then the points represented by points on the unit circle are associated to the pencil of lines through the origin; in particular, the points [(1, 0), ( 1, 0)] and [(0, 1), (0, 1)] correspond to the y-axis and the x-axis respectively The centers of all other pencils of lines produce half ellipses with a diameter of the unit circle as main axes; this diameter joins the points on the unit circle which represent the line joining the center of the pencil with the origin Parallel pencils correspond to diameters Translating equation (4) into this framework yields (cg 2 dh 2 ) 1 g 2 h 2 + gh(c(1 + d)g d(1 + c)h) (7) The corresponding curve B p is depicted together with the decomposition of the unit disk into (curved) Q-triangles and Q-quadrangles: Q convex, c = 1, d = 2 Q nonconvex, c = 2, d = 3 A C A C B B The half ellipses labelled A, B, C, depict the pencils of lines with the respective vertices A, B, C, of the quadrangle Q as centers The points of intersection of these half ellipses correspond as already said to the sides of the quadrangle The curve B p consists of those points (g, h) of the sextic for which c 2 g 6 + c(2(cd + c d) + cd 2 )g 4 h 2 2cd(1 + c)(1 + d)g 3 h 3 + +d(2(cd c + d)d + c 2 d)g 2 h 4 + d 2 h 6 = (cg 2 dh 2 ) 2 gh(d(c + 1)h c(1 + d)g)(cg 2 dh 2 ) > 0 This leads to the version of the pictures explaining Theorem 17 presented in the following table 32