Conics Tangent at the Vertices to Two Sides of a Triangle

Transcription

1 Forum Geometricorum Volume FORUM GEOM ISSN onics Tangent at the Vertices to Two Sides of a Triangle Nikolaos Dergiades bstract. We study conics tangent to two sides of a given triangle at two vertices, and construct two interesting triads of such conics, one consisting of parabolas, the other rectangular hyperbolas. We also construct a new triangle center, the barycentric cube root of X 25, which is the homothetic center of the orthic and tangential triangles. 1. Introduction In the plane of a given triangle, a conic is the locus of points with homogeneous barycentric coordinates x : y : z satisfying a second degree equation of the form fx 2 + gy 2 + hz 2 +2pyz +2qzx +2hxy =0, 1 where f, g, h, p, q, r are real coefficients see [4]. In matrix form, 1 can be rewritten as x y z f r q r g p x y =0. q p h z If we have a circumconic, i.e., a conic that passes through the vertices,,, then from 1, f = g = h =0. Since the equation becomes pyz + qzx + rxy =0, the circumconic is the isogonal or isotomic conjugate of a line. If the conic passes through and, wehaveg = h = 0. The conic has equation fx 2 +2pyz +2qzx +2rxy = onic tangent to two sides of onsider a conic given by 2 which is tangent at, to the sides,. We call this an -conic. Since the line has equation z =0, the conic 2 intersects at x : y :0where xfx +2ry =0. The conic and the line are tangent at only if r =0. Similarly the conic is tangent to at only if q =0. The -conic has equation fx 2 +2pyz =0. 3 Publication Date: May 3, ommunicating Editor: Paul Yiu. The author thanks Paul Yiu for his help in the preparation of this paper.

2 42 N. Dergiades We shall also consider the analogous notions of - and -conics. These are respectively conics with equations gy 2 +2qzx =0, hz 2 +2rxy =0. 4 If triangle is scalene, none of these conics is a circle. Lemma 1. Let be a chord of a conic with center O, and M the midpoint of. If the tangents to the conic at and intersect at, then the points, M, O are collinear. M O Figure 1. Proof. The harmonic conjugate S of the point M relative to, is the point at infinity of the line. The center O is the midpoint of every chord passing through O. Hence, the polar of O is the line at infinity. The polar of is the line see Figure 1. Hence, the polar of S is a line that must pass through O, M and. From Lemma 1 we conclude that the center of an -conic lies on the -median. In the case of a parabola, the axis is parallel to the median of since the center is a point at infinity. In general, the center of the conic 1 is the point 1 r q 1 g p 1 p h : f 1 q r 1 p q 1 h : f r 1 r g 1. 5 q p 1 Hence the centers of the -, -, -conics given in 3 and 4 are the points p : f : f, g : q : g, and h : h : r respectively. We investigate two interesting triads when the three conics i are parabolas, ii rectangular hyperbolas, and iii all pass through a given point P =u : v : w, and shall close with a generalization.

3 onics tangent at the vertices to two sides of a triangle triad of parabolas The -conic 3 is a parabola if the center p : f : f is on the line at infinity. Hence, p + f + f =0, and the -parabola has equation x 2 4yz =0. We label this parabola P a. Similarly, the - and -parabolas are P b : y 2 4zx =0and P c : z 2 4xy =0respectively. These are also known as the rtzt parabolas see Figure onstruction. conic can be constructed with a dynamic software by locating 5 points on it. The -parabola P a clearly contains the vertices,, and the points M a =2:1:1, P b =4:1:4and P c =4:4:1. learly, M a is the midpoint of the median 1, and P b, P c are points dividing the medians 1 and 1 in the ratio P b : P b 1 = P c : P c 1 =8:1. Similarly, if M b, M c are the midpoints of the medians 1 and 1, and P a divides 1 in the ratio P a : P a 1 =8:1, then the -parabola P b is the conic containing,, M b, P c, P a, and the -parabola P c contains,, M c, P a, P b. M a 1 1 P c P b M b G M c P a 1 Figure 2. Remarks. 1 P a, P b, P c are respectively the centroids of triangles G, G, G. 2 Since 1 is the midpoint of and M a 1 is parallel to the axis of P a, by rchimedes celebrated quadrature, the area of the parabola triangle M a is 4 3 M a = The region bounded by the three rtzt parabolas has area 5 27 of triangle Foci and directrices. We identify the focus and directrix of the -parabola P a. Note that the axis of the parabola is parallel to the median 1. Therefore,

4 44 N. Dergiades the parallel through to the median 1 is parallel to the axis, and its reflection in the tangent passes through the focus F a. Similarly the reflection in of the parallel through to the median 1 also passes through F a. Hence the focus F a is constructible in the Euclidean sense. Let D, E be the orthogonal projections of the focus F a on the sides,, and D, E the reflections of F a with respect to,. It is known that the line DE is the tangent to the -parabola P a at its vertex and the line D E is the directrix of the parabola see Figure 3. E D D E F a 1 Figure 3. Theorem 2. The foci F a, F b, F c of the three parabolas are the vertices of the second rocard triangle of and hence are lying on the rocard circle. The triangles and F a F b F c are perspective at the Lemoine point K. Proof. With reference to Figure 3, in triangle DE, the median 1 is parallel to the axis of the -parabola P a. Hence, 1 is an altitude of triangle DE. The line F a is a diameter of the circumcircle of DE and is the isogonal conjugate to 1. Hence the line F a is a symmedian of, and it passes through the Lemoine point K of triangle. Similarly, if F b and F c are the foci of the - and -parabolas respectively, then the lines F b and F c also pass through K, and the triangles and F a F b F c are perspective at the Lemoine point K. Since the reflection of F a in is parallel to the median 1,wehave F a = D = 1 = F a, 6 and the circle F a is tangent to at. Similarly, F a = E = 1 = F a, 7

5 onics tangent at the vertices to two sides of a triangle 45 and the circle F a is tangent to at. From 6 and 7, we conclude that the triangles F a and F a are similar, so that F a = F a = π and F a =2. 8 K F a O 1 Figure 4. If O is the circumcenter of and the lines F a, F a meet the circumcircle again at the points, respectively then from the equality of the arcs and, the chords =. Since the triangles F a and F a are similar to triangle F a, they are congruent. Hence, F a is the midpoint of, and is the orthogonal projection of the circumcenter O on the -symmedian. s such, it is on the rocard circle with diameter OK. Likewise, the foci of the - and -parabolas are the orthogonal projections of the point O on the - and -symmedians respectively. The three foci form the second rocard triangle of. Remarks. 1 Here is an alternative, analytic proof. In homogeneous barycentric coordinates, the equation of a circle is of the form a 2 yz + b 2 zx + c 2 xy x + y + zpx+ Qy + Rz =0, where a, b, c are the lengths of the sides of, and P, Q, R are the powers of,, relative to the circle. The equation of the circle F a tangent to at is a 2 yz + b 2 zx + c 2 xy b 2 x + y + zz =0, 9

6 46 N. Dergiades and that of the circle F a tangent to at is a 2 yz + b 2 zx + c 2 xy c 2 x + y + zy =0. 10 Solving these equations we find F a =b 2 + c 2 a 2 : b 2 : c 2. It is easy to verify that this lies on the rocard circle a 2 yz + b 2 zx + c 2 xy a2 b 2 c 2 x a 2 + b 2 + c 2 x + y + z a 2 + y b 2 + z c 2 =0. 2 This computation would be more difficult if the above circles were not tangent at to the sides and. So it is interesting to show another method. We can get the same result, as we know directed angles defined modulo π from θ =F a,f a =2, ϕ =F a, F a =, ψ =F a, F a =. Making use of the formula given in [1], we obtain 1 F a = cot cot θ : 1 cot cot ϕ : 1 cot cot ψ 1 = cot cot 2 : 1 cot +cot : 1 cot +cot sin sin 2 sin sin sin sin = : : sin sin + sin + = 2cos sin sin :sin 2 :sin 2 = b 2 + c 2 a 2 : b 2 : c Rectangular -, -, -hyperbolas The -conic fx 2 +2pyz = 0 is a rectangular hyperbola if it contains two orthogonal points at infinity x 1 : y 1 : z 1 and x 2 : y 2 : z 2 where x 1 + y 1 + z 1 =0 and x 2 + y 2 + z 2 =0. and. Putting x = sy and z = x + y we have fs 2 2ps 2p =0with roots s 1 + s 2 = 2p f and s 1s 2 = 2p f. The two points are orthogonal if S x 1 x 2 + S y 1 y 2 + S z 1 z 2 =0. From this, S x 1 x 2 +S y 1 y 2 +S x 1 +y 1 x 2 +y 2 =0, S s 1 s 2 +S +S s 1 + 1s 2 +1=0, and p f = S + S 2S = a 2 b 2 + c 2 a 2. This gives the rectangular -hyperbola H a : b 2 + c 2 a 2 x 2 +2a 2 yz =0, with center O a =a 2 : b 2 + c 2 a 2 : b 2 + c 2 a 2.

7 onics tangent at the vertices to two sides of a triangle 47 This point is the orthogonal projection of the orthocenter H of on the - median and lies on the orthocentroidal circle, i.e., the circle with diameter GH. Similarly the centers O b, O c lie on the orthocentroidal circle as projections of H on the -, -medians. Triangle O a O b O c is similar to the triangle of the medians of see Figure 5. The construction of the -hyperbola can be done since we know five points of it: the points,, their reflections, in O a, and the othocenter of triangle. H c H b O c H a G O a H O b Figure Triad of conics passing through a given point Let P =u : v : w be a given point. We denote the -conic through P by P ; similarly for P and P. These conics have equations vwx 2 u 2 yz =0, wuy 2 v 2 zx =0, uvz 2 w 2 xy =0. Let XY Z be the cevian triangle of P and X Y Z be the trillinear polar of P see Figure 6. It is known that the point X is the intersection of YZ with, and is the harmonic conjugate of X relative to the points, ; similarly for Y and Z are the harmonic conjugates of Y and Z relative to, and,. These points have coordinates X =0: v : w, Y =u :0: w, Z = u : v :0,

8 48 N. Dergiades and they lie on the trilinear polar x u + y v + z w =0. Now, the polar of relative to the conic P is the line. Hence, the polar of X passes through. Since X is harmonic conjugate of X relative to,, the line X is the polar of X and the line X P is tangent to the conic P at P. Y Y E X P Z F M 2 M 3 X O a Z Figure 6. Remark. The polar of an arbitrary point or the tangent of a conic at the point P = u : v : w is the line given by x y z f r q r g p u v =0. q p h w Hence the tangent of P at P is the line x y z 2vw u 2 u v =0, 0 u 2 0 w or 2vwx wuy uvz =0. This meets at the point X =0: v : w as shown before.

9 onics tangent at the vertices to two sides of a triangle onstruction of the conic P. We draw the line YZ to meet at X. Then X P is tangent to the conic at P. Let this meet and at E and F respectively. Let M 2 and M 3 be the midpoints of P and P. Since, and EF are tangents to P at the points,, P respectively, the lines EM 3 and FM 2 meet at the center O a of the conic P.If, are the symmetric points of, relative to O a, then P is the conic passing through the five points, P,,,. The conics P and P can be constructed in a similar way. G Figure 7. In the special case when P is the centroid G of triangle, X is a point at infinity and the line YZis parallel to. The center O a is the symmetric of relative to G. It is obvious that P is the translation of the Steiner circumellipse with center G, by the vector G; similarly for P, P see Figure 7. Theorem 3. For an arbitrary point Q the line X Q intersects the line P at the point R, and we define the mapping hq =S, where S is the harmonic conjugate of Q with respect to X and R. The mapping h swaps the conics P and P. Proof. The mapping h is involutive because hq =S if and only if hs =Q. Since the line P is the polar of X relative to the pair of lines, we have h =, hp =P, h =, so that h =, and h =. Hence, the conic h P is the one passing through, P, and tangent to the sides, at, respectively. This is clearly the conic P. Therefore, a line passing through X tangent to P is also tangent to P. This means that the common tangents of the conics P and P intersect at X. Similarly we can define mappings with pivot points Y, Z swapping P, P and P, P.

10 50 N. Dergiades onsider the line X P tangent to the conic P at P. It has equation 2vwx + wuy + uvz =0. This line meets the conic P at the point X b =u : 2v :4w, and the conic P at the point X c =u :4v : 2w. Similarly, the tangent Y P of P intersects P again at Y c =4u : v : 2w and P again at Y a = 2u : v :4w. The tangent Z P of P intersects P again at Z a = 2u :4v : w and P again at Z b =4u : 2v : w respectively. Theorem 4. The line Z b Y c is a common tangent of P and P,soisX c Z a of P and P, and Y a X b of P and P. Y Z b X X c Y c P X b Z a Y a Z Figure 8. Proof. We need only prove the case Y c Z b. The line has equation vwx +4wuy +4uvz =0. This is tangent to P at Z b and to P at Y c as the following calculation confirms. 0 0 v2 0 2wu 0 4u 2v = v vw 4wu, v w 4uv 0 w2 0 w u v = w vw 4wu uv 2w 4uv Theorem 5. The six points X b, X c, Y c, Y a, Z a, Z b lie on a conic see Figure 7. Proof. It is easy to verify that the six points satisfy the equation of the conic 2v 2 w 2 x 2 +2w 2 u 2 y 2 +2u 2 v 2 z 2 +7u 2 vwyz +7uv 2 wzx +7uvw 2 xy =0, 11

11 onics tangent at the vertices to two sides of a triangle 51 which has center u 11u +7v +7w :v7u 11v +7w :w7u +7v 11w. Remark. The equation of the conic 11 can be rewritten as u 2 2v 2 7vw +2w 2 yz + v 2 2w 2 7wu +2u 2 zx + w 2 2u 2 7uv +2v 2 xy 2x + y + zv 2 w 2 x + w 2 u 2 y + u 2 v 2 z=0. This is a circle if and only if u 2 2v 2 7vw +2w 2 a 2 = v2 2w 2 7wu +2u 2 b 2 = w2 2u 2 7uv +2v 2 c 2. Equivalently, the isotomic conjugate of P, namely, 1 u : 1 v : 1 w is the intersection of the three conics defined by 2y 2 7yz +2z 2 a 2 = 2x2 7zx +2x 2 b 2 = 2x2 7xy +2y 2 c The type of the three conics P, P, P. The type of a conic with given equation 1 can be determined by the quantity d = 1 f r q 1 r g p 1 q p h For the conic P : vwx 2 u 2 yz =0, this is i a parabola if u 2 4vw =0and P = P a, ii an ellipse if u 2 vw < 0, i.e., P lying inside P a, iii a hyperbola if u 2 vw > 0, i.e., P lying outside P a. Thus the conics in the triad P, P, P are all ellipses if and only if P lies in the interior of the curvilinear triangle P a P b P c bounded by arcs of the rtzt parabolas see Figure 2. Remarks. 1 It is impossible for all three conics P, P, P to be parabolas. 2 The various possibilities of conics of different types are given in the table below. Ellipses Parabolas Hyperbolas Note that the conics in the triad P, P, P cannot be all rectangular hyperbolas. This is because, as we have seen in 4 that the only rectangular -hyperbola is H a. The conic P is a rectangular hyperbola if and only if P lies on H a, and in this case, P = H a.

12 52 N. Dergiades 6. Generalization Slightly modifying 3 and 4, we rewrite the equations of a triad of -, -, -conics as x 2 +2Pyz =0, y 2 +2Qzx =0, z 2 +2Rxy =0. part from the vertices, these conics intersect at Q a = :2Q 3 :2R 3, QR 1 Q b = 2P : :2R 3, RP 1 Q c = 2P :2Q 3 :, PQ where, for a real number x, x 1 3 stands for the real cube root of x. learly, the triangles Q a Q b Q c and are perspective at Q = P : Q 3 : R 3. Let XY Z be the cevian triangle of Q relative to and X, Y, Z the intersection of the sidelines with the trilinear polar of Q. Proposition 6. For an arbitrary point S, the line X S meets the line Q at the point T, and we define the mapping hs =U, where U is the harmonic conjugate of S with respect to X and T. The mapping h swaps the conics - and -conics. Similarly we can define mappings with pivot points Y, Z swapping the - and -conics, and the - and -conics. The tangent to at intersects at the point a Q b c X c = RP :2:2R 2 P 1 3 a Q c b X b = P 2 Q 1 3 :2PQ :2 b Q c a Y a = 2P 2 Q 1 3 : PQ :2 b Q a c Y c = 2: Q 2 R 1 3 :2QR c Q a b Z b = 2:2Q 2 R 1 3 : QR c Q b a Z a = 2RP :2: R 2 P 1 3 From these data we deduce the following theorem. Theorem 7. The line Z b Y c is a common tangent of the - and -conics; so is X c Z a of the - and -conics, and Y a X b of the - and -conics.

13 onics tangent at the vertices to two sides of a triangle 53 Figure 9 shows the case of the triad of rectangular hyperbolas H a, H b, H c. The perspector Q is the barycentric cube root of X 25 the homothetic center of the orthic and tangential triangles. Q does not appear in [3]. Z b Q b Z a H c H b X b Q c O c H a X c Q O a O b Y c Q a Y a Figure 9. References [1] N. Dergiades, simple barycentric coordinates formula, Forum Geom., [2] G. Kapetis, Triangle Geometry, Thessaloniki [3]. Kimberling, Encyclopedia of Triangle enters, available at [4] P. Yiu, Introduction to the Geometry of the Triangle, Nikolaos Dergiades: I. Zanna 27, Thessaloniki 54643, Greece address: ndergiades@yahoo.gr