A Quadrilateral and resulting Conics Part 2 A Special Case showing the role of the New Points X and Y

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1 A Quadrilateral and resulting Conics Part 2 A Special Case showing the role of the New Points X and Y Christopher Bradley Article: CJB/231/2012 To 2 11 T G U To F To 7 X P B A S D 5 6 E C R Y 12 8 To 10 To 12 Q Figure 1 To 31 Abstract: In Article 230 we investigated properties of a configuration of a quadrilateral ABCD inscribed in an ellipse and the common points of tangents to the ellipse at the vertices A, B, C, D. It turns out that there are two remarkable points X and Y through which many lines may be drawn and also through which many conics pass. The analysis of Part 1 was too complicated to illustrate the results, so in this Article we give a numerical presentation that illustrates the properties of the point X and Y, for which the results all hold in the general case. It seems unlikely the general case can be treated analytically, but possibly a pure approach would succeed as the points are members of a Desargues involution. Figure 1 above illustrates the line properties of X and Y and Figure 2 below illustrates their Conical properties. 1

2 To V T Z G U X 41 P A 13 S E 34 D 42 R W Y 24 B L C 23 Q To 31 To Introduction, the initial points and lines We use areal co-ordinates throughout with ABC as triangle of reference and D the point with coordinates D(3, 2, 6). We take the circumscribing ellipse to be the Steiner circum-ellipse with equation yz + zx + xy = 0. (1.1) 2

3 The equations of BC, CA, AB are, of course x = 0, y = 0, z = 0. It is easy to show that those of AD, BD, CD are respectively 3y + z = 0, 2x z = 0, 3y + 2x = 0. (1.2) We can now calculate the co-ordinates of the diagonal points of ABCD and these are E = BD^AC: x = 1, y = 0, z = 2, (1.3) F = AD^BC: x = 0, y = 1, z = 3, (1.4) G = AB^CD: x = 3, y = 2, z = 0. (1.5) The tangent at (f, g, h) to the ellipse with equation (1.1) is (hy + gz) + (hx + fz) + (gx + fy) = 0. (1.6) Therefore the tangent at A has equation y + z = 0, (1.7) the tangent at B has equation z + x = 0, (1.8) the tangent at C has equation x + y = 0, (1.9) and the tangent at D has equation 4x + 9y + z = 0. (1.10) We can now compute the co-ordinates of the second quadrilateral PQRS. These are P = t A^t B : x = 1, y = 1. z = 1, (1.11) Q = t B^t C : x = 1, y = 1, z = 1, (1.12) R = t C^t D : x = 1, y = 1, z = 5, (1.13) S = t D^t A : x = 2, y = 1, z = 1. (1.14) The two external diagonal points are U = t A^t C: : x = 1, y = 1, z = 1, (1.15) T = t B^t D : x = 3, y = 1, z = 3. (1.16) It transpires that the external diagonal lines of the quadrilaterals ABCD and PQRS are identical. That is F, G, T, U are collinear on the line with equation 2x + 3y + z = 0. (1.17) The sides of the quadrilateral PQRS have equations as follows: PQ: z + x = 0, (1.18) QR: x + y = 0, (1.19) RS: 4x + 9y + z = 0, (1.20) SP: y + z = 0. (1.21) This is because they are identical to the tangents at A, B, C, D. 2. The eight intersections of the sides of the two quadrilaterals 3

4 These intersections are key points in the configuration we have just defined in Section 1. In no particular order their co-ordinates are: 31 = PQ^CD: x = 3, y = 2, z = 3, (2.1) 24 = SP^BC: x = 0, y = 1, z = 1, (2.2) 34 = SP^CD: x = 3, y = 2, z = 2, (2.3) 12 = QR^AB: x = 1, y = 1, z = 0, (2.4) 42 = QR^DA: x = 1, y = 1, z = 3, (2.5) 13 = RS^AB: x = 9, y = 4, z = 0, (2.6) 23 = RS^BC: x = 0, y = 1, z = 9, (2.7) 41 = PQ^DA: x = 3, y = 1, z = 3. (2.8) The notation is such that PQ, QR, RS, SP carry numbers 1, 2, 3, 4 as do AB, BC, CD, DA. It will be observed that as a consequence of the choice of initial ellipse and the co-ordinates of D, the two points 24 and 12 lie on the line at infinity. This in no way invalidates any of the results that follow. Conics containing these points are such that they pass through the line at infinity in one or other (or both) of these points. In the figures, however, we have chosen a case in which 24 and 12 are finite points. The remarkable property of these 8 points is that they all lie on the same conic. This conic has equation ux 2 + vy 2 + wz 2 + 2fyz + 2gzx + 2hxy = 0, (2.9) where u = 4, v = 9, w = 1, f = 5, g = 7/2, h = 13/2. (2.10) 3. Lines through E It is no surprise that six further lines in the figure pass through E = AC^BD with co-ordinates E(1, 0, 2). These are as follows: PR: 2x 3y z = 0, (3.1) QS: 2x + 3y z = 0, (3.2) 41 23: 2x 9y z = 0, (3.3) 24 42: 2x y z = 0, (3.4) 12 34: 2x + 2y z = 0, (3.5) 31 13: 4x + 9y 2z = 0. (3.6) 4. Four 6 point conics and the definition of the remarkable points X and Y 4

5 If one adds the points T, U on the external diagonal line to the eight points defined in Section 2 one can construct four 6 point conics. These are T U: with equation 4x 2 + 9y 2 + z 2 + 4yz + 3zx + 13xy = 0, (4.1) T U with equation 4x 2 + 9y 2 + z yz + 6zx + 10xy = 0, (4.2) T U with equation 10x 2 + 9y 2 + z yz + 9zx + 19xy = 0, (4.3) T U with equation 8x y 2 + 5z yz + 18zx + 38xy = 0. (4.4) The equation of PR is 2x 3y z = 0, (4.5) It may now be checked that conics (4.3) and (4.4) and PR all meet at points X, Y (and the two conics of course also intersect at T and U so that the four intersections of the two conics are T, U, X, Y. We shall not pursue in this article the involution T U, X Y). The co-ordinates of X, Y are X: x = (k + 1)/10, y = (k + 6)/15, z = 1, (4.6) Y: x = (k 1)/10, y = (k 6)/15, z = 1. (4.7) In Equations (4.6) and (4.7) k = 6 and in what follows we shall continue to write k for this numerical value. 5. Lines through X and Y intersecting on the diagonal QS Points X and Y have some remarkable properties. In this section we shall show that whenever X is joined by a line through another key point, then there is a line through Y and another key point such that the two lines intersect on the diagonal QS or the line FGTU. Figure 1 shows such lines and the intersection points 1, 2, 3,..., 14. QS has equation and FGTU has equation 2x + 3y z = 0, (5.1) 2x + 3y + z = 0. (5.2) We now give details of these pairs of lines and their points of intersection. Point 1: XU^YT: x = 3k + 3/2, y = 2k, z = 3. (5.3) 5

6 Point 2: XT^YU: x = 3k + 3/2, y = 2k, z = 3. (5.4) Point 3: XB^YC: x = (k + 1)/10, y = (k + 6)/15, z = 1. (5.5) Point 4: XC^YB: x = (k 1)/10, y = (6 k)/15, z = 1. (5.6) Point 5: XA^YD: x = (k + 11)/10, y = (k + 6)/15, z = 1. (5.7) Point 6: XD^YA: x = (11 k)/10, y = (k 6)/15, z = 1. (5.8) Point 7: XC^YA: x = (1 k)/10, y = (k 6)/15, z = 1. (5.9) Point 8: XA^YC: x = (k + 1)/10, y = (k + 6)/15, z = 1. (5.10) Point 9: XB^YD: x = (k + 1)/10, y = (k 4)/15, z = 1. (5.11) Point 10: XD^YB: x = (k 1)/10, y = (k + 4)/15, z = 1. (5.12) Point 11: XQ^YS: x = k/4, y = (k 2)/6, z = 1. (5.13) Point 12: XS^YQ: x = k/4, y = (k + 2)/6, z = 1. (5.14) Point 13: X 23^ Y 24: x = (k 2)/2, y = (3 k)/6, z = 1. (5.15) Point 14: X 34^Y 13: x = (7 k)/2, y = (k 6)/3, z = 1. (5.16) It will be observed that the first 12 points are conjugate pairs in which k is replaced by k in the partner. Points 7, 8, 9, 10, 11, 12 lie on FGTU the others on the diagonal QS. The line XY itself has equation 2x 3y z = 0, (5.17) and XY passes through F(0, 1, 3). See Figure 1 to observe the results of this Section. 6. The conics passing through X and Y We have already defined X and Y as both lying on the two conics T U with equation 10x 2 + 9y 2 + z yz + 9zx + 19xy = 0, (6.1) and T U with equation 8x y 2 + 5z yz + 18zx + 38xy = 0. (6.2) We now investigate other such conics having an equation of the form ux 2 + vy 2 + wz 2 + 2fyz + 2gzx + 2hxy = 0. (6.3) First we find the conditions that key points lie on the conic (6.3). A: u = 0, (6.4) B: v = 0, (6.5) C: w = 0, (6.6) D: 24f + 36g 12h + 9u + 4v + 36w = 0, (6.7) X: (2k/15 + 4/5)f (k/5 +1/5)g + (7k/75 + 4/25)h + (k/50 + 7/100)u + 6

7 + (4k/ /75)v + w = 0. (6.8) Y: (2k/15 4/5)f + (k/5 1/5)g + (4/25 7k/75)h + (7/100 k/50)u + + (14/75 4k/75)v + w = 0. (6.9) T: 6f 18g 6h + 9u + v + 9w = 0. (6.10) U: 2f + 2g 2h + u + v + w = 0. (6.11) Q: 2f 2g 2h + u + v + w = 0. (6.12) S: 2f + 4g 4h + 4u + v + w = 0. (6.13) 24: 2f + v + w = 0. (6.14) 23: 18f + v + 81w = 0. (6.15) 13: 72h + 81u + 16v = 0. (6.16) 34: 8f + 12g 12h + 9u + 4v + 4w = 0. (6.17) 41: 6f 18g + 6h + 9u + v + 9w = 0. (6.18) 42: 6f + 6g 2h + u + v + 9w = 0. (6.19) G: 12h + 9u + 4v = 0. (6.20) We now enumerate and give the equations of numerous six or eight point conics, as shown in Figure 2. Conic XYABCD: u = v = w = 0, f = 1, g = 4, h = 10. (6.21) Conic XYBCTU: u = 28, v = w = 0, f = 4, g = 9, h = 19. (6.22) Conic XYADTU: u = 0, v = 63, w = 7, f = 25. g = 9, h = 19. (6.23) Conic XYQSBD: u = 104, v = 0, w = 26, f = 57, g = 20, h = 102. (6.24) Conic XYACQS: u = 0, v = 78, w = 0, f = 19, g = 24, h = 34. (6.25) Conic XYTU : u = 8, v = 45, w = 5, f = 19, g = 9, h = 19. (6.26) Conic XYAD 24 23: u = 0, v = 9, f = 5, g = 7, h = 17. (6.27) Conic XYBC 13 34: u = 16, v = 0, w = 0, f = 3, g = 8, h = 18. (6.28) Conic XYBC 41 42: u = 4, v = 0, w = 0, f = 2, g = 7, h = 17. (6.29) Conic TU : u = 4, v = 9, w = 1, f = 5, g = 3, h = 5. (6.30) Conic XYAD 13 34: u = 0, v = 18, w = 2, f = 7, g = 2, h = 4. (6.31) Cabri II + indicates two further interesting points, which we have called Z and W. These are the intersections of conic XYAD (Equation (6.31)) and XYTU (Equation (4.3)) With Z also lying on L 42 and W also lying on L41, where L = BC^QS. ZW passes through F as do XY and TU and the pair Z and W also have many conics passing through them. We do not investigate their properties further, but it seems likely that other pairs of points lying on a line through F might be found having systems of conics passing through them. It is one of those cases that as the figure gets larger it then gets even larger. Flat 4, Terrill Court, 12-14, Apsley Road, BRISTOL BS8 2SP. 7

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