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0 if C = 0 and AB 2 1 −1 A B cot − if C = 0 and AB 2 2 2C 6 π if C < 0 and A = B 4 3π if C > 0 and A = B 4 In [H2], we also derived the following results about ellipses inscribed in parallelograms. Lemma F: Let Z be the rectangle with vertices (0, 0), (l, 0), (0, k),and (l, k), where l,k > 0. (A) The general equation of an ellipse, E, inscribed in Z is given by
k2x2 + l2y2 2l (k 2v) xy 2lkvx 2l2vy + l2v2 =0, 0 < v < k. − − − − (B) The corresponding points of tangency of Ψ are
lv l , 0 , (0, v), (k v) , k , and (l, k v). k k − − (C) If a and b denote the lengths of the semi–major and semi–minor axes, respectively, of E, then
2 2 2l (k v) v a = − and k2 + l2 (k2 + l2)2 16l2 (k v) v − 2 − − 2 2l (k v) v b = p − . k2 + l2 + (k2 + l2)2 16l2 (k v) v − − Proposition G: Let P be thep parallelogram with vertices O = (0, 0), P = (l, 0),Q = (d, k), and R = (d + l, k), where l,k > 0, d 0. The general equation of an ellipse, E, inscribed in P is given by ≥
3 2 2 2 k x + k(d + l) 4dlv y 2k (k(d + l) 2lv) xy 2 − − 2 2 − 2k lvx +2klv (d l) y + kl v =0, 0 < v < k. − − Of course, any two triangles are affine equivalent, while the same is not true of quadrilaterals–thus it is not surprising that not all of the results about ellipse inscribed in triangles extend nicely to quadrilaterals. For example, there is not necessarily an ellipse inscribed in a given quadrilateral, D, which is tangent to
2 the sides of D at their respective midpoints. There is such an ellipse, which we call the midpoint ellipse, when D is a parallelogram(see Proposition 2.2). We are able to prove an inequality(see Theorem 2.1), similar to (trineq), which holds for all convex quadrilaterals. If E is any ellipse inscribed in a quadrilateral, D, Area (E) π then , and equality holds if and only if D is a parallelogram and Area(D) ≤ 4 E is the midpoint ellipse. Not suprisingly, Theorem B also does not extend in general to ellipses in- scribed in convex quadrilaterals. However, such a characterization does hold again when D is a parallelogram. We prove in Theorem 3.1 that the foci of the unique ellipse of maximal area inscribed in a parallelogram, D, lie on the orthogonal least squares line for the vertices of D. It is also well known that if p(z) is a cubic polynomial with roots at the vertices of a triangle, then the roots of p′(z) are the foci of the Steiner inellipse. A proof of this fact goes all the way back to Siebeck in 1864(see [MP]) and is known in the literature as Marden’s Theorem. Now it is easy to show that the orthogonal least squares line, £, from Theorem A is identical to the line through the roots of p′(z). Thus Marden’s Theorem implies Theorem B and hence is a stronger statement than Theorem B. There is an obvious way to try to generalize such a result to convex quadrilaterals, D. If p(z) is a quartic polynomial with roots at the vertices of D, must the foci of the unique ellipse of maximal area inscribed in D equal the roots of p′′(z) ? We give an example in section 3 that shows that such a stronger statement does not hold for parallelograms, or even for rectangles.
2 An Area Inequality
Theorem 2.1: Let E be any ellipse inscribed in a convex quadrilateral, D. Area (E) π Then , and equality holds if and only if D is a parallelogram and Area(D) ≤ 4 E is tangent to the sides of D at the midpoints. Before proving Theorem 2.1, we need the following lemma. Lemma 2.2: Suppose that s and t are positive real numbers with s + t> 1 and s =1 = t. Let 6 16 1 h = st + t 2s 1+ (t 1)2 + s2(t2 t + 1) s(t2 3t + 2) . a 6 t 1 − − − − − − − 1 1 Then h I = the open interval withp endpoints and s. a ∈ 2 2 Proof of Lemma 1:
1 st 2 (s + t 1)+ (t 1)2 + s2(t2 t + 1) s(t2 3t + 2) ha = − − − − − − , and − 2 p 6 (t 1) − (1)
1 1 h s = (t 1)2 + s2(t2 t + 1) s(t2 3t + 2) (2st (s + t 1)) . a−2 6 (t 1) − − − − − − − − p (2)
3 There are four cases to consider: s,t > 1, s< 1
s + t> 1 and s =1 = t. (3) 6 6 1 It also follows easily that Area(D) = (s + t). Let E denote any ellipse 2 inscribed in D and AE = Area(E). The midpoints of the diagonals of D are 1 1 1 1 M1 = , and M2 = s, t . Let I denote the open interval with 2 2 2 2 1 1 endpoints and s. By Theorem C the locus of centers inscribed in D is 2 2
4 precisely the set (h,L(h),h I , where { ∈ } 1 s t +2x(t 1) L(x)= − − (4) 2 s 1 − is the line thru M1 and M2. In proving Theorem D[H1, Theorem 3.3], we also showed that 2 2 π AE = 2 A(h), A(h)=(2h 1) (s 2h) (s +2h(t 1)) ,h I. (5) 4 (s 1) − − − ∈ − 1 Setting A′(h) = 0 yields h = h = st + t 2s 1 (t 1)2 + s2(t2 t + 1) s(t2 3t + 2) . a 6 (t 1) − − ± − − − − − By Lemma 2.2, the positive square root always yields ha I.p Thus ∈ 1 h = st + t 2s 1+ (t 1)2 + s2(t2 t + 1) s(t2 3t + 2) a 6 (t 1) − − − − − − − p is the unique point in I such that A(h) A (h ) ,h I. Let ≤ a ∈ 2 b(s,t) = (st (s + t 1)) + st(s + t 1). − − − A simple computation using (5) shows that 1 A (ha)= 2 2ts s t +1 b(s,t) ts 2t 2s +2+ b(s,t) s + ts + t 1+ b(s,t) . 27 (t 1) − − − − − − − p p p 1 Thus by (5) and the fact that Area(D) = (s + t), we have 2
2 2 2ts s t +1 b(s,t) ts 2t 2s +2+ b(s,t) s + ts + t 1+ b(s,t) AE π − − − − − − 2 = 2 2 . (Area(D)) 27 p (s 1) (t 1) (sp+ t)2 p − − (6) A2 Clearly E is symmetric in s and t. Thus we may assume, without (Area(D))2 loss of generality, that s t. (7) ≤ It is convenient to make the change of variable
u = s + t 1, v = st. (8) − 1 Solving (8) for s and t yields s = u +1 u +1 (u + 1)2 4v ,t = − 2 ± − 1 2 u +1 (u + 1) 4v . By (7), p 2 ± − p 1 1 s = u +1 (u + 1)2 4v ,t = u +1+ (u + 1)2 4v (9) 2 − − 2 − p p 5 Substituting for s and t using (9) gives b(s,t) = c(u, v), where c(u, v) = u2 uv + v2. Then − 2 2 A π 1 E = (2v u c(u, v))(v 2u + c(u, v))(v + u + c(u, v)) / (u + 1)2 , (Area(D))2 2 − − − 4 108(v u) ! which simplifies to − p p p
A2 π2 E = d(u, v), where (10) (Area(D))2 27
3 2 (v 2u) (2v u) (u + v)+2(c(u, v)) / d(u, v)= − − 2 . (11) (u + 1)2 (v u) − We consider two cases: s> 1 and 0 6 v Now s< 1 implies that v 0 since s,t > 0. Now we let w = , u which again implies that 0 u 27 Area (E) π By (14) and (16), z(w) . < then follows immedi- (u + 1)2 ≤ 16 Area(D) 4 ately by (10) and (15). Area (E) π We have proven that < when D is not a trapezoid. Using a Area(D) 4 Area (E) π limiting argument, one can then show immediately that when D Area(D) ≤ 4 is a trapezoid. However, that still does not give the strict inequality when D is not a parallelogram. We shall omit the details here, but the author will provide them upon request. To finish the proof of Theorem 2.1, we need to consider the case when D is a parallelogram. That case will follow from Theorem 2.3 below. First we prove that, just like any triangle in the plane, there is an ellipse tangent to a parallelogram at the midpoints of its sides. Proposition 2.2: The ellipse of maximal area inscribed in a parallelogram, P, is tangent to P at the midpoints of the four sides. Proof: Proposition 2.2 was proven in [H2, Theorem 3] when P is a rectangle. The reader can prove it directly themselves using Lemma F. Ratios of areas of ellipses, points of tangency, and midpoints of line segments are preserved under one–one affine transformations. Since any given parallelogram is affinely equiv- alent to a rectangle, that proves Proposition 2.2 for parallelograms in general. Theorem 2.3: Let E be any ellipse inscribed in a parallelogram, P. Then Area (E) π , with equality if and only if E is the midpoint ellipse. Area(P) ≤ 4 Proof: As noted earlier, ratios of areas of ellipses and four–sided convex polygons, are preserved under one–one affine transformations. In addition, points of tangency and midpoints of line segments are also preserved under one–one affine transformations. Thus we may assume that P is a rectangle(or even a square, though we don’t need that much simplification here), Z, with ver- tices (0, 0), (l, 0), (0, k),and (l, k), where l,k > 0. Let E be any ellipse inscribed in Z, and let a and b denote the lengths of the semi–major and semi–minor axes, respectively, of E. 7 By Lemma F(C), 2 2 2 2l (k v) v a b = − k2 + l2 (k2 + l2)2 16l2 (k v) v × − − − 2l2 (k v) v p − = k2 + l2 + (k2 + l2)2 16l2 (k v) v − − p 2 4l4 (k v) v2 2 − = (k2 + l2) ((k2 + l2)2 16l2 (k v) v) 4 − 2 2 − − 4l (k v) v 1 2 − = l (k v) v. 16l2 (k v) v 4 − − π Hence Area(E) = πa2b2 = l v (k v). Since the function v (k v) at- 2 − − k tains its unique maximum on (0, vp) when v = , the unique ellipse of maximal 2 π k k π Area (E ) area, E , satisfies Area(E )= l k = lk, and thus A = A A 2 2 − 2 4 Area (Z) s π Area (E) π . Furthermore, < when E = E . By Lemma F(B), letting 4 Area (Z) 4 6 A k l k l k v = gives the points of tangency , 0 , 0, , , k ,and l, , which 2 2 2 2 2 are the midpoints of Z. That completes the proof of Theorem 2.3 as well as the equality part of Theorem 2.1. 3 Orthogonal least squares & Inscribed Ellipses If l is a line in the plane, we let d (zk,l) denote the distance from zk to l. Theorem 3.1: Let P be the parallelogram with vertices z1,z2,z3,z4 (A) If P is not a square, then there is a unique line, £, which minimizes 4 2 d (zk,l) among all lines, l. Furthermore, the foci of the midpoint ellipse for k=1 PP lie on £. (B) If P is a square, then the midpoint ellipse is a circle, and every line through the center of P is a line of best fit for the vertices of P. Remarks: (1) For part (A), an equivalent statement is that the foci of the midpoint ellipse for P lie on the line through the second derivative of the polynomial with roots at the vertices of D. (2) Note that £ is not the standard regression line in general. Proof: The line through the foci of an ellipse is not preserved in general under nonsingular affine transformations of the plane, but it is preserved under translations and rotations. Thus we can assume that the vertices of P are O = (0, 0), P = (l, 0),Q = (d, k), and R = (l + d, k), where l,k > 0, d 0. ≥ 8 Using complex notation for the vertices, z1 = 0, z2 = l, z3 = d + ki, and 1 4 1 z4 = l + d + ki, the centroid, g, of P is given by g = zk = (l + d + ki), 4 k=1 2 4 4 2 2 1 P and Z = (zj g) = zj (l + d + ki) , which simplifies to =1 − =1 − 2 j j P P 2 2 2 Z = d + l k +2idk. (17) − It is well known, and easy to show, that the center of any ellipse inscribed in 1 1 P equals the point of intersection of the diagonals of P, which is (d + l) , k . 2 2 k Also, as noted in the proof of Theorem 2.3, v = yields the unique ellipse of 2 maximal area, EA, inscribed in P. Let the major axis line of EA refer to the line through the foci of EA. Thus the major axis line of EA passes through g. We take care of some special cases first. d = 0: Then P is a rectangle. If l = k, then P is a square and letting • k v = shows that E is a circle. Also, Z = 0. Thus every line through the 2 A center of P is a line of best fit for the vertices of P. We may assume now that 1 l = k, which implies that Z = 0. Since g = (l + ki), the major axis line of 6 6 2 2 2 2 2 EA passes through g; Z = l k ; If l > k, then √Z = √l k is real, and thus the line, £, thru g parallel− to the vector from (0, 0) to √Z−has slope 0. By Lemma F(A), the major axis line of any ellipse inscribed in P is parallel to the x axis. Since the major axis line passes through g, it must be identical to £. If l < k, then √Z = √l2 k2 is imaginary, and thus £ is vertical. By Lemma F(A) again, the major axis− line of any ellipse inscribed in P is parallel to the y axis. Again, the the major axis line must be identical to £. Assume now that d = 0, which implies that Z = 0 since dk = 0. • k 6 6 6 Letting v = in Proposition G gives the coefficients for the equation of E . 2 A In particular, 3 2 2 2 A = k ,B = k d + l , C = k d. (18) − Note that C < 0 for all d, k, and l, and A 0 since k> 0. ⇐⇒ ⇐⇒ Let −φ denote the counterclockwise angle of rotation from the line thru the center parallel to the x axis to the major axis of EA. We use the formula Im √Z Z Re Z = | |− , Im Z =0. (19) Re √Z Im Z 6 Im √Z 2dk If k2 d2 l2 = 0, then by (17) and (19), Z =2idk = = 1. − − ⇒ Re √Z 2dk π By (18), A = B and C < 0, which implies, by Lemma E, that φ = . Thus the 4 9 π major axis line has slope equal to tan φ = tan = 1, which is the same slope 4 as that of £. Since both the major axis line and £ pass through g, they must be identical. Assume now that k2 d2 l2 = 0, which implies that A = B. 2 2 − 2− 6 6 Z = d2 + l2 k2 + (2dk) = (k + l)2 + d2 (k l)2 + d2 , which im- plies| that| − − Z = ((k + l)2 + d2) ((k l)2 + d2). | | − Hence p 2 2 2 Z Re Z ((k + l)2 + d2) ((k l)2 + d2) d + l k | |− = − − − . (20) Im Z p 2dk A B A B By Lemma E, cot2φ = − or cot(2φ π)= − , which implies that 2C − 2C 2C 2C tan2φ = or tan(2φ π) = . Since tan(x π) = tan x, in either A B − A B − case, − − 2 tan φ 2C = , 1 tan2 φ A B − − which implies that tan φ C = = 1 tan2 φ A B − − k2d dk − = k3 k (d2 + l2) d2 + l2 k2 − − by (18). tan φ The equation = r has solution 1 tan2 φ − 1 tan φ = 1 1+4r2 . (21) 2r − ± p 2 2 2 2 dk 2 d + (k l) d + (k + l) Letting r = 2 2 2 yields 1 + 4r = − 2 . d + l k (k2 d2 l2) − By (21), − − k2 d2 l2 ((k l)2 + d2) ((k + l)2 + d2) tan φ = − − 1 − . 2dk ± k2 d2 l2 p − − ! There are two cases to consider. Case 1: k2 d2 l2 > 0 Let S = (d,k,l− :−d> 0,k> 0,l> 0 , which is a connected set. { π } k2 d2 l2 = 0 φ = tan φ > 0. Also, tan φ = 0 r = 0. But − − ⇒ 4 ⇒ ⇐⇒ r = 0 since dk = 0. Thus tan φ> 0 on S. 6 6 10 Also, 2 2 4k d > 0 implies that 2 (k l)2 + d2 (k + l)2 + d2 k2 d2 l2 > 0, − − − − which implies that ((k l)2 + d2) ((k + l)2 + d2) > k2 d2 l2. − − − Hence p k2 d2 l2 ((k l)2 + d2) ((k + l)2 + d2) tan φ = − − 1+ − (23) 2dk k2 d2 l2 p − − ! k2 d2 l2 ((k l)2 + d2) ((k + l)2 + d2) since − − 1 − < 0. 2dk − k2 d2 l2 p − − ! Case 2: k2 d2 l2 < 0 2 −2 2− k d l ((k l)2 + d2) ((k + l)2 + d2) Then − − 1 − < 0, so again (23) 2dk − k2 d2 l2 p − − ! holds. By (20), Z Re Z k2 d2 l2 | |− = − − Im Z 2dk × ((k l)2 + d2) ((k + l)2 + d2) 1+ − . k2 d2 l2 p − − Thus the major axis line and £ each have slope given by the right hand side of (23). Since both the major axis line and £ pass through g, they must be identical. That completes the proof of (A). (B) follows immediately from the fact that the ellipse of maximal area inscribed in a square is a circle. We show in the following example that a version of Marden’s Theorem does not hold for ellipses inscribed in parallelograms. Example: Let D be the rectangle with vertices z1 = 0, z2 = 1, z3 =1+2i, k z4 = 2i. Letting l = 1, k = 2, and v = = 1 in Lemma F(A) yields the 2 2 2 equation 4x + y 4x 2y + 1 = 0 for the maximal area ellipse, EA, inscribed − − 2 1 2 in D. Rewriting the equation as 4 x + (y 1) = 1 shows that the foci − 2 − 1 √3 1 √3 of EA are , 1+ and , 1 . If 2 2 ! 2 − 2 ! Q(z)= z(z 1)(z (1+2i))(z 2i)), − − − 11 1 1 1 3 then the roots of Q′′(z) are + i and + i, which would give the points 2 2 2 2 1 1 1 3 , and , . Hence the roots of Q′′(z) do not give the foci of the 2 2 2 2 maximal area ellipse. Remark: Some heuristic reasoning shows that a version of Marden’s The- orem does not hold for ellipses inscribed in convex quadrilaterals. For if it did, then when the quadrilateral is tangential(contains an inscribed circle), Q′′ would have to have a double root. But a square is tangential, while Q′′ has distinct roots. For ellipses circumscribed about a convex quadrilateral, D(that is, passing through the vertices of D), we make a conjecture similar to Theorem 2.1. Conjecture: Let E be any ellipse circumscribed about a convex quadrilat- A π eral, D. Then E , with equality if and only if D is a parallelogram. Area(D) ≥ 2 There is strong evidence that this conjecture is true, but we have not yet been able to prove this inequality. 4 References [CH] G. D. Chakerian, A Distorted View of Geometry, MAA, Mathematical Plums, Washington, DC, 1979, 130-150. [H1] Alan Horwitz, “Ellipses of maximal area and of minimal eccentricity in- scribed in a convex quadrilateral”, Australian Journal of Mathematical Analysis and Applications, 2(2005), 1-12. [H2] Alan Horwitz, Ellipses inscribed in parallelograms, submitted to the Australian Journal of Mathematical Analysis and Applications. [K] D. Kalman, An elementary proof of Marden’s theorem, American Math- ematical Monthly 115(2008), 330–338. [MP] D. Minda and S. Phelps, Triangles, Ellipses, and Cubic Polynomials, American Mathematical Monthly 115(2008), 679–689. [W] Weisstein, Eric W. ”Ellipse.” From MathWorld–A Wolfram Web Re- source. http://mathworld.wolfram.com/Ellipse.html 12