DUALITY AND INSCRIBED ELLIPSES

DUALITY AND INSCRIBED ELLIPSES MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE Abstract. We give a constructive proof for the existence of inscribed family of ellipses in convex n-gons for 3 n 5 using duality of curves. In the case of a pentagon, we also show that there are simultaneously two ellipses, one inscribed in the pentagon and the other inscribed in its diagonal pentagon. The two ellipses are as intrinsically linked as are the pentagon and its diagonal pentagon. 1. Introduction The goal of this paper is to show how duality of curves lends itself to investigating families of inscribed ellipses in convex n-gons. The foci of such inscribed ellipses (or similar higher class curves) are closely related to location of the critical points of polynomials vis-a-vis their roots. For instance, Marden s Theorem [3] states that the non-trivial critical points of the product of complex power functions p(z) n k1 (z z k) m k for m k R, m k 0 are the foci of a curve of class n 1 that touches the side z j z k of the n-gon formed by z 1, z 2,, z n, in the ratio m j : m k. In this work we prove the following theorem: Theorem 1.1 (Main Theorem). Inscribed ellipses in convex non degenerate n-gons: (1) In triangles there exists a unique two-parameter family of inscribed ellipses - the parameters are the points on two sides of the triangle. That is, by choosing two points on different sides of the triangle there is a unique inscribed ellipse passing through those two points. (2) In quadrilaterals there exists a unique one-parameter family of inscribed ellipses - the one parameter is the points on one side of the quadrilateral. That is, by choosing one point on a side of the quadrilateral there is a unique inscribed ellipse that passes through that point. (3) In pentagons there exists precisely one inscribed ellipse - this gives a zero-parameter family. In fact, our construction gives another ellipse which is inscribed in the diagonal pentagon formed by the diagonals of the original pentagon. 2010 Mathematics Subject Classification. Primary 14H52, 51E10. Key words and phrases. Linfield s Theorem and Ellipses. 1

2 MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE (4) For n 6, there exist n-gons for which there are no inscribed ellipses; whenever there is an inscribed ellipse, it is unique. Our exposition builds on the work of Linfield [2] where he provides a proof of Marden s theorem for n 3. Since, we are only interested in the existence of families of ellipses in n-gons, we generalize only the relevant aspects of Linfield s proof. Besides, the already known connection between foci of ellipses and critical points of polynomials alluded to in Marden s Theorem, these families carry some more interesting information. For instance, in an ongoing project the first author [1] shows that, the loci of the foci of one of the families of ellipses formed by keeping one point fixed on the side AB while varying the other point on the side BC in a triangle ABC is an elliptic curve! In the process of proving our theorem we also give an explicit construction of the ellipses when they exist. Though the result is not new, our goal is to provide an exposition using projective geometry (an area which has long been neglected in the undergraduate curriculum) and some vector calculus. Our results also follow from Brianchon s Theorem. Since this proof is very elementary and quick we present it at the very outset. Theorem 1.2 (Brianchon s Theorem). If a hexagon circumscribes an ellipse, then its three diagonals meet in a point; conversely, if the three diagonals of a hexagon meet in a point, then there exists an inscribed ellipse. r 1-r s 1-s Figure 1. Implications of Brianchon s Theorem for inscribed ellipses For example, to see that there exists a two-parameter family of ellipses inscribed in a triangle, fix 0 < r < 1 and 0 < s < 1 as in the triangle in Figure 1. The five solid dots and one open dot form vertices of a degenerate hexagon; these vertices satisfy the hypothesis of the converse of Brianchon s Theorem, guaranteeing the existence of an inscribed ellipse. For an ellipse to be tangent to the six sides of the hexagon, it must be tangent at the points designated by r, s, and the open dot. The fact that r and s are arbitrary

DUALITY AND INSCRIBED ELLIPSES 3 explains the existence of a two-parameter family. Similar arguments can be made for the quadrilateral and pentagonal cases above. 2. Duality and Dual Curves In this section we fix notations and definitions that will be used for the rest of the paper. For more details we refer the reader to [4]. Let R be the field of real numbers and P 2 the real projective plane. To each point (x, y) in R 2 we associate a homogeneous point [x : y : 1] in P 2, and to each line ax + by + c 0 in R 2 we associate a homogeneous line ax + by + ch 0 in P 2. Just as λax + λby + λch 0 for λ 0 identifies the same line ax + by + ch 0, so it is understood that [λx : λy : λ] [x : y : 1] for λ 0. That is, it is the ratio between components that unambiguously distinguishes among homogeneous points. The homogeneous point [b : a : 0] can be seen to lie on the homogeneous line ax + by + ch 0, and is referred to as a point at infinity. All the points at infinity lie on the homogeneous line at infinity, h 0. To each homogeneous point Q and line L P 0 in the xyh-plane there exists a unique line L Q 0 and point P, respectively, in a dual αβγ-plane. Specifically, Q [x : y : h] L Q : Q (α, β, γ) 0 L P : P (x, y, h) 0 P [α : β : γ], where L Q : Q (α, β, γ) xα + yβ + hγ and similarly, L P : P (x, y, h) αx + βy + γh. Note that L Q is a linear homogeneous polynomial in the variables (α, β, γ), while L P is a linear homogeneous polynomial in the variables (x, y, h). The point Q and line L Q 0 are duals of one another; as are P and L P 0. Note that L P P, so that the dual of a homogeneous line can be seen to be its gradient. In general, to any homogenous curve ϕ ϕ(α, β, γ) P 2 of degree n 1, n 1, we can associate a homogenous dual curve ϕ ϕ(x, y, h) P 2, whose generic points are tangents to ϕ P 2. Symbolically, ϕ {[x : y : h] P (x, y, h) 0, ϕ(p ) 0}. So the duals of the points of ϕ 0 are tangent lines to ϕ 0; analogously, tangent lines to ϕ 0 are duals of the points of ϕ 0. To determine the points of ϕ 0 we require the set of tangent lines to ϕ 0. To this end the homogeneous tangent line to the curve ϕ 0 at any point P 0 on the curve can be written as ϕ(p 0 ) (P P 0 ) 0. According to Euler s homogeneous function theorem, ϕ(p ) P (n 1)ϕ(P ), simplifying the representation of the tangent line: ϕ(p 0 ) P 0. Thus if P 0 lies on ϕ 0, then Q 0 ϕ(p 0 ) lies on ϕ 0.

4 MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE Remark 2.1 (Explicit formula for a dual of a conic). If ϕ is quadratic then (x, y, h) ϕ(α, β, γ) is a linear system. If α α(x, y, h), β β(x, y, h), and γ γ(x, y, h) is a solution to this system then the dual conic is given by ϕ(x, y, h) ϕ(α(x, y, h), β(x, y, h), γ(x, y, h)). This is generally not true for higher degree ϕ. We now prove a key lemma which we need to prove our main results. Lemma 2.2. Let P denote a convex n-gon, with vertices labeled V k, k 1,..., n and let m k, k 1,..., n be positive real numbers. Then there exists a class n 1 curve, which is tangent to the n(n 1)/2 lines joining the vertices of P with the points of tangency given by: (1) m j m j + m k V k + m k m j + m k V j. Proof. Following Linfield [2], we associate to each vertex V k (x k, y k ) of the polygon a homogeneous point and homogeneous linear polynomial Q k [x k : y k : 1] L k Q k [α : β : γ] x k α + y k β + γ. The intersection point P jk of the lines L j 0 and L k 0 is either finite or a point at infinity: { [y j y k : x k x j : x j y k x k y j ], L j 0, L k 0 not parallel; P jk [y k : x k : 0], L j 0, L k 0 parallel. The dual of the point P jk is the line joining Q j to Q k. Let (2) ϕ n m l L 1 L l 1 L l+1 L n. l1 The homogeneous polynomial ϕ is of degree n 1 and interpolates P jk since each summand in its definition contains as a factor either L j or L k. Let ϕ 0 denote the dual to the curve ϕ 0. The fact that ϕ is of degree n 1 implies that ϕ is of class n 1. By definition, ϕ 0 is tangent to the line joining Q j and Q k, since ϕ interpolates the dual of this line, P jk. To determine the point of tangency, Q jk, we compute the tangent line to ϕ 0 at P jk. Since each of the summands of ϕ contain at least one if not both of the factors L j and L k, it can be expressed as ϕ (m j L k + m k L j )A jk + L j L k B jk for appropriate A jk and B jk. Thus ϕ(p jk ) A jk (P jk )(m j Q k + m k Q j ). Normalizing the homogenous component of ϕ(p jk ) we see that the dual of the tangent line to ϕ 0 at P jk corresponds to m j m k Q jk : Q k + Q j, m j + m k m j + m k

establishing (1). DUALITY AND INSCRIBED ELLIPSES 5 Definition 2.3. The function ϕ defined in (2) plays a critical role in this paper. In what follows we shall refer to the constants m i s as weights and the function ϕ as Linfield s function. 3. Proof of Main Theorem In the case of a triangle (n 3), Linfield s function ϕ is a quadratic, and hence its dual ϕ must also be a quadratic. We exhibit a simple set of constraints on the weights m k, that depend upon two parameters, that compel the de-homogenized quadratic to be an ellipse. In the case of a convex quadrilateral (n 4), Linfield s function ϕ is a cubic. We produce a set of weights, depending upon a single parameter, that enable ϕ to be factored into the product of a linear and quadratic polynomial. Its dehomogenized dual, ϕ, in this instance is a point and an inscribed ellipse. For a convex pentagon (n 5), ϕ is a quartic. There exists a unique set of constraints on the weights that permit ϕ to be factored into the product of two quadratic polynomials. The de-homogenized duals of these factors will be shown to be inscribed in the pentagon and in the diagonal pentagon, respectively. 3.1. In triangles there exists a unique two-parameter family of inscribed ellipses. Proof. Let T denote a nondegenerate triangle with homogeneous vertices Q k, k 1, 2, 3, labeled clockwise. Linfield s function ϕ m 1 L 2 L 3 + m 2 L 1 L 3 + m 3 L 1 L 2 is a quadratic in term of α, β and γ. To establish the existence of the two-parameter family of ellipses inscribed in T, fix two parameters 0 < r, s < 1. Geometrically, these values designate two points Q 12 (1 r)q 1 + rq 2 and Q 23 (1 s)q 2 + sq 3 on adjacent sides of T (see Figure 1). Choose positive m 1, m 2 and m 3 so that m 1 m 2 r 1 r, and m 2 m 3 s 1 s. Then the dual curve ϕ 0 is also quadratic, a conic upon de-homogenization. Since the curve is tangent to the three segments, and all m i s are positive, it follows that ϕ 0 must be a conic and in particular an ellipse. There are no other ellipses inscribed in the triangle other than those of the two-parameter family. If such an ellipse did exist, then it would be tangent to the sides Q 1 Q 2, Q 2 Q 3, and Q 1 Q 3 ; the points of contact on the first two sides define parameters r 0, s 0. The equation of this hypothesized ellipse satisfies five independent homogenous constraints (arising from the 3 tangency conditions and 2 points of contact on the sides Q 1 Q 2, Q 2 Q 3 ) in

6 MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE ΑΒΓ Q 1 r P 12 x,y,h 0 L 3 0 Q 12 1 r P 23 Φ 0 Q 2 Φ 0 P 13 L 1 0 Q 3 P 12 L 2 0 Q 12 Α,Β,Γ 0 Figure 2. Illustration of Linfield s theorem for n 3 common with the de-homogenized dual ϕ for this choice of parameters r 0, s 0. Hence the ellipse is among the two-parameter family. Example 3.1. The computation of the dual of a polynomial curve is in general a nontrivial problem. Since the coefficients of the tangent lines to ϕ 0 constitute homogenous points of the dual ϕ 0, we must solve the generally nonlinear system (x, y, h) ϕ(α, β, γ). In the case of a quadratic polynomial, however, the system is linear and consequently more accessible. We illustrate the construction, by using Linfield s function corresponding to triangle with vertices (0, 0), (1, 0), and (0, 1). In this instance Q 1 [0 : 0 : 1], Q 2 [1 : 0 : 1], Q 3 [0 : 1 : 1], and Linfield s function is ϕ(α, β, γ) m 3 γ(α + γ) + m 2 γ(β + γ) + m 1 (α + γ)(β + γ). In this example, the system (x, y, h) ϕ(α, β, γ) is x m 3 γ + m 1 (β + γ), y m 2 γ + m 1 (α + γ), h m 1 (α + β + 2γ) + m 2 (β + 2γ) + m 3 (α + 2γ). and it always has a solution provided m 1 m 2 m 3 0. Here, α hm 1 (m 1 + m 2 ) (m 1 + m 2 ) 2 x (m 2 1 m 2 m 3 + m 1 (m 2 + m 3 ))y, β hm 1 (m 1 + m 3 ) (m 2 1 m 2 m 3 + m 1 (m 2 + m 3 ))x (m 1 + m 3 ) 2 y, γ m 1 ( hm 1 + (m 1 + m 2 )x + (m 1 + m 3 )y). Substituting these values for α, β, γ into ϕ(α, β, γ) yields the dual of ϕ as a function of x, y, h: ϕ(x, y, h) m 2 1( 1+x+y) 2 (m 2 x m 3 y) 2 2m 1 ( 1+x+y)(m 2 x+m 3 y).

DUALITY AND INSCRIBED ELLIPSES 7 3.2. In convex quadrilaterals there exists a one-parameter family of inscribed ellipses. Proof. Let Q denote a convex nondegenerate quadrilateral with homogeneous vertices Q k, k 0,..., 3, labeled clockwise. In this instance Linfield sfunction (3) ϕ m 0 L 1 L 2 L 3 + m 1 L 0 L 2 L 2 + m 2 L 0 L 1 L 3 + m 3 L 0 L 1 L 2 is cubic. Without loss of generalization we assume that the intersection of the diagonals of Q lies at the origin and that the vertices Q 1 and Q 3 lie on the vertical axis. Thus there exist constants 0 < θ, φ < 1 for which (4)(1 θ)q 1 + θq 3 [0 : 0 : 1] and (1 φ)q 2 + φq 0 [0 : 0 : 1]. For a parameter 0 < r < 1. Choose positive real numbers m 0, m 1, m 2 and m 3 such that (5) m 1 m 2 r 1 r, m 1 m 3 θ 1 θ, and m 2 m 0 φ 1 φ. Conditions (4) imply (1 θ)l 1 + θl 3 γ (1 φ)l 2 + φl 0. So, upon writing (3) as ϕ ( m 2 L 0 + m 0 L 2 ) L1 L 3 + ( m 1 L 3 + m 3 L 1 ) L2 L 0, and taking into account the constraints (5), it can be verified that Linfield sfunction can be factored into a linear term times a quadratic one: (6) γ (1 r)θφ ( (1 r)θl1 L 3 + rφl 2 L 0 ). The dual of the linear factor γ/((1 r)θφ) is the origin; the dual of the quadratic is a conic. Consequently, the dual of ϕ is a conic, tangent to the interior of the four segments defining Q, and hence must be an ellipse. As in the case of the triangle, any ellipse inscribed in the quadrilateral would necessarily share five independent conditions with ϕ for a particular choice of r. Hence this family is unique. Remark 3.2. The linear factor in the function ϕ in (6) corresponds to the line at infinity because we constrained the intersection of the diagonals of Q to lie at the origin. If this were not the case, then the linear factor of ϕ would correspond to the dual of that point of intersection. Remark 3.3. Since we can translate and rotate a quadrilateral Q so that it fulfills the conditions in the above proof, we can apply the rotation and translation to the computed two-parameter family in reverse to obtain the desired family inscribed in Q.

8 MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE 3.3. In pentagons there exists a unique inscribed ellipse. Proof. In this case Linfield s function takes the form ϕ m 1 L 2 L 3 L 4 L 5 + m 2 L 1 L 3 L 4 L 5 + m 3 L 1 L 2 L 4 L 5 + m 4 L 1 L 2 L 3 L 5 + m 5 L 1 L 2 L 3 L 4. and we shall show that ϕ can be expressed as the product of two quadratic polynomials for appropriate choices of the weights m 1,..., m 5. To see this let P denote a convex nondegenerate pentagon. Extend two of the sides of P to form a convex quadrilateral Q from which the pentagon can be thought to have been cut (see Figure 3). Label the vertices of Q in a clockwise fashion Q k, for k 0, 1, 2, 3, beginning with the vertex not among those of P. Label the remaining vertices of P consecutively Q 4 and Q 5. Without loss of generalization we assume that, as in the n 4 case, the diagonals of Q intersect at the origin, and additionally that the vertices Q 1 and Q 3 lie on the vertical axis (see Figure 3). With this structure in place there exist constants 0 < θ, φ < 1 for which (1 θ)q 1 + θq 3 0 and (1 φ)q 0 + φq 2 0. The first factor of ϕ can be modeled on the quadratic factor in (6), (7) ϕ P (1 r)θl 1 L 3 + rφl 0 L 2. and for the second factor of ϕ, we let Q A, Q B, and Q C denote the clockwise labeling of the homogeneous vertices of the diagonal pentagon that do not lie on the vertical axis, and define ϕ DP (1 s)l A L C + sl B L 2. Q 1 Q 5 P C24 Φ DP 0 Q A Φ P P 0 23 P 034 P BC35 Q 0 Q B 0 Φ DP 0 Q 2 0:0:1 P 45 Q 4 Q C Φ P 0 P 12 P 015 P AB14 P A25 Q 3 Figure 3. Illustration of Linfield s theorem for n 5, here P 23 denotes the intersection of lines L 2 0, L 3 0.

DUALITY AND INSCRIBED ELLIPSES 9 We start with ϕ P. Since each L k vanishes at any point with k among its subscript, ϕ 0 can be seen to interpolate the four points P 12, P 23, P 034, and P 015 for any choice of the free parameter r. We now show r in (7) can be chosen so that ϕ P also interpolates a fifth point, P 45. Once we have shown that ϕ P 0 interpolates the duals of the five sides of P, it follows immediately that ϕ P 0 must be tangent to the sides of P. Because Q 0 and Q 1 are multiples of Q 2 and Q 3, respectively, it follows that L 0 0 and L 2 0 are parallel, as are L 1 0 and L 3 0. Furthermore, since Q 1 and Q 3 are assumed to lie on the vertical axis, L 1 0 and L 3 0 must be parallel to the α-axis, and intersect at P 13 [1 : 0 : 0]. Because the dual image of a strictly convex polygon is also strictly convex, the point P 45 must lie between L 1 0 and L 3 0, and to one side of L 0 0 and L 2 0 (see Figure 3). Since each of the homogeneous linear polynomials L 0, L 1, L 2, and L 3 is positive at the origin [0 : 0 : 1], it follows that the product L 1 L 3 at P 45 is positive, and the product L 0 L 2 at P 45 is negative. Thus there exists a unique parameter value 0 < r 0 < 1 for which (1 r 0 )θl 1 L 3 (P 45 ) + r 0 φl 0 L 2 (P 45 ) 0, hence ϕ P (P 45 ) 0. Likewise, for the second factor of ϕ, we let Q A, Q B, and Q C denote the clockwise labeling of the homogeneous vertices of the diagonal pentagon that do not lie on the vertical axis, and define ϕ DP (1 s)l A L C + sl B L 2. By analogy with ϕ P above, ϕ DP interpolates the points P A25, P AB14, P BC35, and P C24, duals of four of the sides of the diagonal pentagon DP. Let x A, x B, x C denote the first components of Q A, Q B, and Q C respectively and recall that P 13 [1 : 0 : 0] is the dual of the side of DP containing the origin. Since L A L C (P 13 ) x A x C > 0 and L B L 2 (P 13 ) x B x 2 < 0, there exists a unique parameter 0 < s 0 < 1 for which ϕ DP (P 13 ) 0. With this parameter choice, ϕ DP interpolates the duals of all five sides of the diagonal pentagon of P. By Lemma 2.2 and the arguments above, ϕ and the product ϕ P ϕ DP both interpolate the duals of the ten segments joining the vertices Q j to Q k, for distinct j, k 1, 2,..., 5. With the constraints (8) (9) m 1 r 0 m 1 θ m 2 1 r 0 m 3 1 θ m 1 Q 5 + m 5 Q 1 ϕ P (P 015 ) m 3 Q 4 + m 4 Q 3 ϕ P (P 034 ) we shall show that ϕ and the product ϕ P ϕ DP additionally have common gradients at four specific points. It is this observation that forces the two quartic polynomials ϕ and ϕ P ϕ DP to be proportional to one another. The gradient of Linfield s function ϕ at the points P 12, P 23, P 015, and P 034 corresponds to points of interpolation for ϕ 0. Each of these interpolated points subdivides a line segment into fixed proportions according the the values of the weights. With the weights defined by (8) and (9), it follows

10 MAHESH AGARWAL, JOHN CLIFFORD, AND MICHAEL LACHANCE that ϕ(p 12 ) m 1 Q 2 + m 2 Q 1, ϕ(p 23 ) m 2 Q 3 + m 3 Q 2, ϕ(p 015 ) m 1 Q 5 + m 5 Q 1, ϕ(p 034 ) m 3 Q 4 + m 4 Q 3 We next compute the gradient of the product ϕ P ϕ DP. In general (ϕ P ϕ DP ) ϕ P ϕ DP + ϕ P ϕ DP, but, since ϕ P is zero at the points P 12, P 23, P 015, and P 034, satisfies (ϕ P ϕ DP )(P ) ϕ P (P ) at those points. The proof of the theorem is complete if we show that ϕ ϕ P at the four points P 12, P 23, P 015, and P 034. The proportionality is obvious at the latter two points from the constraints on m 1, m 3, m 4, and m 5 in (9). To see that ϕ and ϕ P are proportional at P 12 we note that ϕ(p 12 ) and ϕ P (P 12 ) each correspond to a point between Q 1 and Q 2. That is, comparison of the properties of the weights in (8) with those in (5), with r r 0, imply that ϕ(p 12 ) ϕ P (P 12 ). A similar argument holds at the point P 23. Thus Linfield s function ϕ is proportional to the product of ϕ P ϕ DP. The dual of each factor is an ellipse: that of ϕ P is inscribed in the pentagon, and that of ϕ DP is inscribed in its diagonal pentagon. The uniqueness of each inscribed ellipse follows because another ellipse would need to share five independent tangent conditions, forcing them to be the same. 3.4. If there exists an inscribed ellipse in a convex n-gon for n 6, then it is unique. Proof. If a convex hexagon contains an inscribed ellipse, extend two suitable sides of the hexagon to form a pentagon from which the hexagon can be thought to have been cut out. The ellipse that is inscribed in the hexagon, is also inscribed in the pentagon. Since any pentagon can only inscribe a unique ellipse, this ellipse must be unique. A similar argument can be applied to n-gons for n > 6. 3.5. For n 6 there exist convex n-gons with no inscribed ellipses. Proof. By Brianchon s theorem, a convex hexagon contains an inscribed ellipse if and only if the diagonals intersect at a point. It is unlikely, that a generic hexagon has an inscribed ellipse. This follows from the observation that three generic lines (these can be extended to form the diagonals of a hexagon) are unlikely to be concurrent (see Figure 1).

DUALITY AND INSCRIBED ELLIPSES 11 References [1] M. Agarwal and N. Natarajan. Elliptic curves from ellipses. In preparation. [2] Ben-Zion Linfield. On the relation of the roots and poles of a rational function to the roots of its derivative. Bull. Amer. Math. Soc., 27(1):17 21, 1920. [3] Morris Marden. Geometry of polynomials. Second edition. Mathematical Surveys, No. 3. American Mathematical Society, Providence, R.I., 1966. [4] Joseph H. Silverman and John Tate. Rational points on elliptic curves. Undergraduate Texts in Mathematics. Springer-Verlag, New York, 1992. Department of Mathematics and Statistics, University of Michigan, Dearborn, Dearborn, MI 48128 E-mail address: mkagarwa@umich.edu Department of Mathematics and Statistics, University of Michigan-Dearborn, Dearborn, MI 48128 E-mail address: jcliff@umich.edu Department of Mathematics and Statistics, University of Michigan-Dearborn, Dearborn, MI 48128 E-mail address: malach@umich.edu