aa r X i v : . [ m a t h . M G ] F e b Midpoint Diagonal Quadrilaterals
Alan Horwitz2/17/21
Abstract
A convex quadrilateral, $ Q $ , is called a midpoint diagonal quadri-lateral if the intersection point of the diagonals of $ Q $ coincides withthe midpoint of at least one of the diagonals of $ Q $ . A parallelo-gram, P, is a special case of a midpoint diagonal quadrilateral sincethe diagonals of P bisect one another. We prove two results aboutellipses inscribed in midpoint diagonal quadrilaterals, which general-ize properties of ellipses inscribed in parallelograms involving convexquadrilaterals. First, $ Q $ is a midpoint diagonal quadrilateral if andonly if each ellipse inscribed in $ Q $ has tangency chords which areparallel to one of the diagonals of $ Q $ . Second, $ Q $ is a midpointdiagonal quadrilateral if and only if each ellipse inscribed in $ Q $ hasa unique pair of conjugate diameters parallel to the diagonals of $ Q $ .Finally, we show that there is a unique ellipse, $ E I $ , of minimal eccen-tricity inscribed in a midpoint diagonal quadrilateral, $ Q $ , and alsothat the unique pair of conjugate diameters parallel to the diagonalsof $ Q $ are the equal conjugate diameters of $ E I $ . Introduction
Given a diameter, l , of an ellipse, E , there is a unique diameter, m , of E such that the midpoints of all chords parallel to l lie on m . In this case wesay that l and m are conjugate diameters of E , or that m is a diameter of E conjugate to l . l and m are called equal conjugate diameters if | l | = | m | .We say that E is inscribed in a convex quadrilateral, Q , if E lies inside Q and is tangent to each side of Q . A tangency chord is any chord connect-ing two points where E is tangent to two different sides of Q . There aretwo interesting properties(probably mostly known) of ellipses inscribed inparallelograms, P, which involve tangency chords and conjugate diameters:1P1) Each ellipse inscribed in P has tangency chords which are parallelto one of the diagonals of P.(P2) Each ellipse inscribed in P has a pair of conjugate diameters whichare parallel to the diagonals of P.Note that when we say that two lines are parallel, we include the possibiltythat they are equal, which does, in fact, occur for (P2).This author is not sure if P1 is known at all, while P2 appears to beknown([6]) only if E is the ellipse of maximal area inscribed in P. Oneof the purposes of this paper is to examine P1 and P2 for a larger class ofconvex quadrilaterals, which we call midpoint diagonal quadrilaterals(definedbelow). Definition:
A convex quadrilateral, Q , is called a midpoint diagonalquadrilateral if the intersection point of the diagonals of Q coincides withthe midpoint of at least one of the diagonals of Q .We show that not only do P1 and P2 each hold for midpoint diagonalquadrilaterals, but that if Q is not a midpoint diagonal quadrilateral, thenno ellipse inscribed in Q satisfies P1 or P2. Hence each of these propertiescompletely characterizes the class of midpoint diagonal quadrilaterals(seeTheorems 3 and 4 below), and thus they are a generalization of parallelogramsin this sense. A parallelogram, P, is a special case of a midpoint diagonalquadrilateral since the diagonals of P bisect one another. Equivalently, if Q is not a parallelogram, then Q is a midpoint diagonal quadrilateral if andonly if the line, L Q , thru the midpoints of the diagonals of Q contains one ofthe diagonals of Q . The line L Q plays an important role for ellipses inscribedin quadrilaterals due to the following well–known result(see [1] for a proof). Theorem 1 (Newton): Let M and M be the midpoints of the diagonals ofa quadrilateral, Q . If E is an ellipse inscribed in Q , then the center of E must lie on the open line segment, M M , connecting M and M . Remark 1 If Q is a parallelogram, then the diagonals of Q intersect at themidpoints of the diagonals of Q , and thus M M is really just one point. Remark 2
By Theorem 1, Q is a midpoint diagonal quadrilateral if and onlyif the center of any ellipse inscribed in Q lies on one of the diagonals of Q . If E is an ellipse which is not a circle, then E has a unique set of con-jugate diameters, l and m , where | l | = | m | . These are called equal conjugate2iameters of E . By Theorem 4(i), each ellipse inscribed in a midpoint di-agonal quadrilateral, Q , has conjugate diameters parallel to the diagonalsof Q . In particular, Theorem 4(i) applies to the unique ellipse of minimaleccentricity, E I , inscribed in Q . However, we prove(Theorem 5) a strongerresult: The equal conjugate diameters of E I are parallel to the diagonals of Q . Useful Results on Ellipses and Quadrilaterals
We now state a result, without proof, about when a quadratic equationin x and y yields an ellipse. The first condition ensures that the conic is anellipse, while the second condition ensures that the conic is nondegenerate. Lemma 1
The equation Ax + Bxy + Cy + Dx + Ey + F = 0 , with A, C > ,is the equation of an ellipse if and only if ∆ > and δ > , where ∆ = 4 AC − B and (1) δ = CD + AE − BDE − F ∆ . The following lemma allows us to express the eccentricity and center ofan ellipse as a function of the coefficients of an equation of that ellipse.
Lemma 2
Suppose that E is an ellipse with equation Ax + Bxy + Cy + Dx + Ey + F = 0 . Let a and b denote the lengths of the semi–major andsemi–minor axes, respectively, of E . Let ( x , y ) denote the center of E and let ∆ and δ be as in (1). Then b a = A + C − p ( A − C ) + B A + C + p ( A − C ) + B and (2) x = BE − CD ∆ , y = BD − AE ∆ . (3) Proof.
Let µ = 4 δ ∆ . By [7], a = 12 µ ( A + C + p ( A − C ) + B ) , (4) b = 12 µ ( A + C − p ( A − C ) + B ).3ote that µ > L Q denote the line thru the midpoints of agiven quadrilateral, Q , and we define an affine transformation, T : R → R to be the map T (ˆ x ) = A ˆ x + ˆ b , where A is an invertible 2 × Lemma 3
Let T : R → R be an affine transformation and let Q be amidpoint diagonal quadrilateral. Then Q ′ = T ( Q ) is also a midpoint diagonalquadrilateral. Lemma 4
Let T : R → R be an affine transformation and suppose that l and m are conjugate diameters of an ellipse, E . Then l ′ = T ( l ) and m ′ = T ( m ) are conjugate diameters of E ′ = T ( E ) . The following lemma shows that the scaling transformations preserve theeccentricity of ellipses, as well as the property of the equal conjugate diame-ters of an ellipse being parallel to the diagonals of Q . Lemma 5
Let T be the nonsingular affine transformation given by T ( x, y ) =( kx, ky ) , k = 0 .(i) Then E and T ( E ) have the same eccentricity for any ellipse, E .(ii) If E is an ellipse which is not a circle and if the equal conjugatediameters of E are parallel to the diagonals of Q , then the equal conjugatediameters of T ( E ) are parallel to the diagonals of T ( Q ) . Proof. (i) follows immediately and we omit the proof. To prove (ii), supposethat l and m are equal conjugate diameters of an ellipse, E , which are parallelto the diagonals of Q . By Lemma 4, l ′ = T ( l ) and m ′ = T ( m ) are conjugate4iameters of E ′ = T ( E ). Since | T ( P ) T ( P ) | = k | P P | for any two points P = ( x , y ) and P = ( x , y ), l ′ and m ′ are equal conjugate diameters of T ( E ). Since affine transformations take parallel lines to parallel lines, l ′ and m ′ are parallel to the diagonals of T ( Q ). That proves (ii).The following lemma shows when a trapezoid can be a midpointdiagonal quadrilateral. Lemma 6
Suppose that Q is a midpoint diagonal quadrilateral which is alsoa trapezoid. Then Q is a parallelogram. Proof.
We use proof by contradiction. So suppose that Q is a midpointdiagonal quadrilateral which is a trapezoid, but which is not a parallelogram.By affine invariance, we may assume that Q is the trapezoid with vertices(0 , , (1 , , (0 , , t ) , < t = 1. The diagonals of Q are then theopen line segments D : y = tx and D : y = 1 − x , each with 0 < x <
1. Themidpoints of the diagonals are M = (cid:18) , (cid:19) and M = (cid:18) , t (cid:19) , and thediagonals intersect at P = (cid:18)
11 + t , t t (cid:19) . Now M = P ⇐⇒ t = 1 and M = P ⇐⇒ t = 1, which contradicts the assumption that t = 1. Hence Q is not a midpoint diagonal quadrilateral. Remark 3
We use the notation Q ( A , A , A , A ) to denote the quadrilat-eral with vertices A , A , A , and A , starting with A = lower left corner andgoing clockwise. Denote the sides of Q ( A , A , A , A ) by S , S , S , and S ,going clockwise and starting with the leftmost side, S . Denote the lengthsof the sides of Q ( A , A , A , A ) by a = | A A | , b = | A A | , c = | A A | , and d = | A A | . Finally, denote the diagonals of Q ( A , A , A , A ) by D = A A and D = A A . We note here that there are two types of midpoint diagonal quadrilaterals:Type 1, where the diagonals intersect at the midpoint of D and Type 2,where the diagonals intersect at the midpoint of D . Notation:
The lines containing the diagonal line segments, D and D ,of any quadrilateral are denoted by ←→ D and ←→ D .Given a convex quadrilateral, Q = Q ( A , A , A , A ), which is not a par-allelogram, it will simplify our work below to consider quadrilaterals with aspecial set of vertices. In particular, there is an affine transformation which5ends A , A , and A to the points (0 , , (0 , , A = ( s, t ) for some s, t >
0. We thus let Q s,t denote thequadrilateral with vertices (0 , , (0 , , ( s, t ), and (1 , Q s,t is convex, s + t >
1. Also, if Q has a pair of parallel vertical sides, first rotate counter-clockwise by 90 ◦ , yielding a quadrilateral with parallel horizontal sides. Sincewe are assuming that Q is not a parallelogram, we may then also assume that Q s,t does not have parallel vertical sides and thus s = 1. Summarizing, wehave Proposition 1
Suppose that Q is a convex quadrilateral which is not a par-allelogram. Then there is an affine transformation which sends Q to thequadrilateral Q s,t = Q ( A , A , A , A ) ,A = (0 , , A = (0 , , A = ( s, t ) , A = (1 , ,with ( s, t ) ∈ G , where (5) G = { ( s, t ) : s, t > , s + t > , s = 1 } . (6)Since the midpoints of the diagonals of Q s,t are M = (cid:18) , (cid:19) and M = (cid:18) s , t (cid:19) , by Theorem 1, the center of any ellipse, E , inscribed in Q s,t mustlie on the open line segment { ( h, L Q ( h )) : h ∈ I } , where L Q ( x ) = 12 s − t + 2 x ( t − s − , (7) I = (cid:18) s , (cid:19) if s < (cid:18) , s (cid:19) if s ≥
1. .We now answer the following important question: How does one find theequation of an ellipse, E , inscribed in Q s,t and the points of tangency of E with Q s,t ? We sketch the derivation of the equation and points of tangencynow. First, since E has center ( h, L Q ( h )) , h ∈ I , one may write the equationof E in the form( x − h ) + B ( x − h )( y − L Q ( h )) + C ( y − L Q ( h )) + F = 0. (8)6hroughout the rest of the paper we denote the open unit interval by J = (0 , E is tangent to Q s,t at the points P q = ( q,
0) and P v =(0 , v ), where q, v ∈ J . Differentiating (8) with respect to x and plugging in P q and P v yields q − h = BL Q ( h )2 , (9) v − L Q ( h ) = Bh C .Plugging in P q and P v into (8) yields ( q − h ) − BL Q ( h )( q − h ) + C ( L Q ( h )) + F = 0 and h − Bh ( v − L Q ( h )) + C ( v − L Q ( h )) + F = 0. By (9), we have F = h C ( B − C ) and F = L Q ( h )4 ( B − C ). Using both expressions for F gives C = h L Q ( h ) . (10)Now by (9) again B = 2( q − h ) L Q ( h ) . (11)(9), (11), and (10) then imply that v = qL Q ( h ) h . (12)Substituting (11) and (10) into F = h C ( B − C ) yields F = q − qh . (8)then becomes( x − h ) + 2( q − h ) L Q ( h ) ( x − h )( y − L Q ( h ))+ h L Q ( h ) ( y − L Q ( h )) + q − qh = 0. (13) Remark 4
Using Lemma 1, it is not hard to show that (13) defines theequation of an ellipse for any h ∈ I . h in terms of q , which makes the final equationsimpler than expressing everything in terms of h . One way to do this is touse the following well–known theorem of Marden(see [5]). Theorem 2 ( Marden):
Let F ( z ) = t z − z + t z − z + t z − z , P k =1 t k = 1 ,and let Z and Z denote the zeros of F ( z ) . Let L , L , L be the line segmentsconnecting z & z , z & z , and z & z , respectively. If t t t > , then Z and Z are the foci of an ellipse which is tangent to L , L , and L at thepoints ζ = t z + t z t + t , ζ = t z + t z t + t , and ζ = t z + t z t + t , respectively. Applying Marden’s Theorem to the triangle ∆ A A A , where A = (cid:18) , − ts − (cid:19) ,one can show that E is tangent to Q s,t at the point (cid:18) s − h t − h + s − t , (cid:19) .Many of the details of this can be found in [2]. Hence q = s − h t − h + s − t ,which implies that h = 12 q ( t − s ) + sq ( t −
1) + 1 . (14)Substituting for h in (13) using (14) and (7), (13) becomes t x + (4 q ( t − t + 2 qt ( s − t + 2) − st ) xy + (15)((1 − q ) s + qt ) y − qt x − qt ((1 − q ) s + qt ) y + q t = 0.One point of tangency is of course given by ( q, P q and P v ). Proposition 2
Suppose that E is an ellipse inscribed in Q s,t . Then E is tangent to the four sides of Q s,t at the points q = (cid:18) , qt ( t − s ) q + s (cid:19) ∈ S , q = (cid:18) (1 − q ) s ( t − s + t ) q + s , t ( s + q ( t − t − s + t ) q + s (cid:19) ∈ S ,q = (cid:18) s + q ( t − s + t − q + 1 , (1 − q ) t ( s + t − q + 1 (cid:19) ∈ S , and q = ( q, ∈ S , q ∈ J . emark 5 It is not hard to show that each of the denominators of the q j above are non–zero. Finally we state the analogy of Proposition 2 for parallelograms. Aslightly different version was proven in [4]. We omit the details of the proof.
Proposition 3
Let P be the parallelogram with vertices A = ( − l − d, − k ) , A =( − l + d, k ) , A = ( l + d, k ) , and A = ( l − d, − k ) , where l, k > , d < l . If E isan ellipse inscribed in P, then E is tangent to the four sides of P at the points q = ( − l + dv, kv ) ∈ S , q = ( − lv + d, k ) ∈ S , q = ( l − dv, − kv ) ∈ S ,and q = ( lv − d, − k ) ∈ S . Tangency Chords Parallel to the Diagonals
The following lemma gives necessary and sufficient conditions for thequadrilateral Q s,t given in (5) to be a midpoint diagonal quadrilateral. Lemma 7 (i) Q s,t is a type 1 midpoint diagonal quadrilateral if and only if s = t .(ii) Q s,t is a type 2 midpoint diagonal quadrilateral if and only if s + t = 2 . Proof.
The diagonal lines of Q s,t are ←→ D : y = ts x and ←→ D : y = 1 − x ,and they intersect at the point P = (cid:18) ss + t , ts + t (cid:19) . The midpoints of thediagonal line segments D and D are M = (cid:18) s , t (cid:19) and M = (cid:18) , (cid:19) ,respectively. Now M = P ⇐⇒ ss + t = 12 and ts + t = 12 , both of whichhold if and only if s = t . That proves (i). M = P ⇐⇒ ss + t = 12 s and ts + t = 12 t , both of which hold if and only if s + t = 2. That proves (ii).Now recall property P1 from the introduction: (P1) Each ellipse inscribedin P has tangency chords which are parallel to one of the diagonals of P.The following theorem shows that P1 completely characterizes the class ofmidpoint diagonal quadrilaterals. Theorem 3
Suppose that E is an ellipse inscribed in a convex quadrilateral Q = Q ( A , A , A , A ) . Let q j ∈ S j , j = 1 , , , denote the points of tan-gency of E with Q , and let D = A A and D = A A denote the diagonalsof Q . i) If Q is a type 1 midpoint diagonal quadrilateral, then ←→ q q and ←→ q q are parallel to D .(ii) If Q is a type 2 midpoint diagonal quadrilateral, then ←→ q q and ←→ q q are parallel to D .(iii) If Q is not a midpoint diagonal quadrilateral, then neither ←→ q q nor ←→ q q are parallel to D , and neither ←→ q q nor ←→ q q are parallel to D . Proof. Case 1: Q is not a parallelogram.Then by Proposition 1, we may assume that Q = Q s,t with diagonal lines ←→ D : y = ts x and ←→ D : y = 1 − x . Using Proposition 2, after some simplificationwe have:slope of ←→ q q = t (2 q ( t −
1) + s ) s (( t − s ) q + s ) , so that the slope of ←→ q q = ts ⇐⇒ t − q + s ( t − s ) q + s = 1 ⇐⇒ ( s + t − q = 0 ⇐⇒ s + t = 2 since q = 0 / ∈ J .slope of ←→ q q = t ( s + t − q + s , so that the slope of ←→ q q = ts ⇐⇒ s + t − q + s = 1 s ⇐⇒ ( s + t − q = 0 ⇐⇒ s + t = 2 since q = 0 / ∈ J .slope of ←→ q q = − t (2( t − q + s − t + 1)( s + t − s − t ) q − s + st + s , so that the slope of ←→ q q = − ⇐⇒ t (2( t − q + s − t + 1) = ( s + t − s − t ) q − s + st + s ⇐⇒ ( s + t −
1) ( s − t ) ( q −
1) = 0 ⇐⇒ s = t since q = 1 / ∈ J and s + t = 1.slope of ←→ q q = t ( s − t ) q − s , so that the slope of ←→ q q = − ⇐⇒ ( s − t ) q − s = − t ⇐⇒ ( q −
1) ( s − t ) = 0 ⇐⇒ s = t since q = 1 / ∈ J .Theorem 3 then follows from Lemma 7. Case 2: Q is a parallelogram.As noted in the introduction, Theorem 3 is probably known in this case,and there are undoubtedly other ways to prove it for parallelograms. UsingProposition 3, it follows easily that the slope of ←→ q q = the slope of ←→ q q = kl + d and the slope of ←→ q q = the slope of ←→ q q = kd − l . Since the diagonallines Q are ←→ D D : y = k + kl + d ( x − l − d ) and ←→ D : y = k + kd − l ( x + l − d ),and a parallelogram is a special case of a midpoint diagonal quadrilateral,that proves Theorem 3 for case 2. Note that one could map Q to the unit10quare and then use a simplified version of Proposition 3, but that does notsimplify the proof very much.Now recall that the lengths of the sides of Q ( A , A , A , A ) are de-noted by a = | A A | , b = | A A | , c = | A A | , and d = | A A | . Lemma 8
Suppose that Q = Q ( A , A , A , A ) is both a tangential and amidpoint diagonal quadrilateral. Then Q is an orthodiagonal quadrilateral. Remark 6
We actually prove more–that Q is a kite . That is, that two pairsof adjacent sides of Q are equal. Proof.
Since Q is tangential, there is a circle, E , inscribed in Q . Let q j ∈ S j , j = 1 , , , E with Q . Definethe triangles T = ∆ q A q and T = ∆ A A A , and define the lines L = ←→ q q and L = ←→ q q . Suppose first that Q is a type 1 midpoint diagonalquadrilateral. Then L k D by Theorem 3(i), which implies that T and T are similar triangles. Also, since E is a circle, | A q | = | A q | , which impliesthat T is isoceles. Hence T is also isoceles with b = a . In a similar fashion,one can show that c = d using the fact that L k D . Thus a + c = b + d ,which implies that Q is an orthodiagonal quadrilateral. The proof when Q isa type 2 midpoint diagonal quadrilateral is similar and we omit the details.We now prove a result somewhat similar to Lemma 8. Lemma 9
Suppose that Q = Q ( A , A , A , A ) is both a tangential and anorthodiagonal quadrilateral. Then Q is a midpoint diagonal quadrilateral. Proof.
Since Q is tangential, there is a circle, E , inscribed in Q and a + c = b + d , which implies that d = a + c − b . Since Q is orthodiagonal, a + c = b + d . Hence b + ( a + c − b ) − a − c = 0, which implies that2 ( b − c ) ( b − a ) = 0, and so a = b and/or b = c . We prove the case when a = b . Let q j ∈ S j , j = 1 , , , E with Q . Then the triangle T = ∆ q A q is isoceles since | A q | = | A q | , and thetriangle T = ∆ A A A is isoceles since a = b . Thus T and T are similartriangles, which implies that the line ←→ q q is parallel to D . By Theorem3(iii), Q is a midpoint diagonal quadrilateral. Conjugate Diameters Parallel to the Diagonals
Recall property P2 from the introduction: (P2) Each ellipse inscribed inP has a pair of conjugate diameters which are parallel to the diagonals of P.11he following theorem shows that P2 completely characterizes the class ofmidpoint diagonal quadrilaterals.
Theorem 4 (i) Suppose that Q is a midpoint diagonal quadrilateral. Then each ellipse inscribed in Q has a unique pair of conjugate diameters parallelto the diagonals of Q .(ii) Suppose that Q is not a midpoint diagonal quadrilateral. Then no ellipse inscribed in Q has conjugate diameters parallel to the diagonals of Q . Proof.
Let E be an ellipse inscribed in Q and let D and D denote thediagonals of Q . Use an affine transformation, T , to map E to a circle, E ′ ,inscribed in the tangential quadrilateral, Q ′ = T ( Q ). Let L be a diameterof E parallel to D . T maps L to a diameter, L ′ , of E ′ parallel to oneof the diagonals of Q ′ , which we call D ′ . Let D ′ be the other diagonal of Q ′ . Let L ′ be the diameter of E ′ perpendicular to L ′ , which implies that L ′ and L ′ are conjugate diameters since E ′ is a circle. By Lemma 4, T − maps L ′ to L , a diameter of E conjugate to L . To prove (i), suppose that Q is a midpoint diagonal quadrilateral. By Lemma 3, Q ′ is also a midpointdiagonal quadrilateral. By Lemma 8, Q ′ is an orthodiagonal quadrilateral,which implies that D ′ ⊥ D ′ . Since L ′ k D ′ , L ′ ⊥ L ′ , and D ′ ⊥ D ′ , L ′ must be parallel to D ′ , which implies that L is parallel to D since T − isan affine transformation. That proves (i). To prove (ii), suppose that Q is not a midpoint diagonal quadrilateral. Since Q ′ is tangential, if Q ′ were alsoan orthodiagonal quadrilateral, then by Lemma 9, Q ′ would be a midpointdiagonal quadrilateral. Hence Q ′ cannot be an orthodiagonal quadrilateral,which implies that D ′ is not perpendicular to D ′ . Now if L ′ were parallelto D ′ , then it would follow that D ′ ⊥ D ′ since L ′ k D ′ and L ′ ⊥ L ′ , acontradiction. Hence L ′ ∦ D ′ , which implies that L ∦ D since T is anaffine transformation. That proves (ii). Equal Conjugate Diameters and the Ellipse of Minimal Eccen-tricity
By Theorem 4(i), each ellipse inscribed in a midpoint diagonal quadrilat-eral, Q , has conjugate diameters parallel to the diagonals of Q . In particular,this holds for the unique ellipse of minimal eccentricity(whose existence weprove below), E I , inscribed in Q . However, in Theorem 5(ii) below, we provea stronger result for E I . Theorem 5 (i) There is a unique ellipse of minimal eccentricity, E I , in-scribed in a midpoint diagonal quadrilateral, Q . ii) Furthermore, the unique pair of conjugate diameters parallel to thediagonals of Q are equal conjugate diameters of E I . Remark 7
Suppose that Q is a type 1 midpoint diagonal quadrilateral andlet CD and CD be the equal conjugate diameters in Theorem 5(ii) parallelto the diagonals, D and D . Let ←−→ CD and ←−→ CD denote the lines containing CD and CD , respectively. Since ←−→ CD is parallel to ←→ D and ←→ D = L , ←−→ CD isparallel to L . Since L and ←−→ CD each pass through the center of E I , ←−→ CD = L .Similarly, for type 2 midpoint diagonal quadrilaterals, ←−→ CD = L . Remark 8
Theorem 5(ii) cannot hold if Q is not a midpoint diagonal quadri-lateral, since in that case no ellipse inscribed in Q has conjugate diametersparallel to the diagonals of Q by Theorem 4(ii). But Theorem 5(ii) impliesthe following weaker result: The smallest nonnegative angle between equalconjugate diameters of E I equals the smallest nonnegative angle between thediagonals of Q when Q is a midpoint diagonal quadrilateral. This was provenin [4] for parallelograms. We do not know if this property of E I can hold if Q is not a midpoint diagonal quadrilateral. Before proving Theorem 5, we need several preliminary results. We omitthe details for the proof of Theorem 5 when Q is a parallelogram. So supposethat Q is a midpoint diagonal quadrilateral which is not a parallelogram.By using an isometry of the plane, we may assume that Q has vertices(0 , , (0 , u ) , ( s, t ), and ( v, w ), where s, v, u > , t > w . To obtain this isome-try, first, if Q has a pair of parallel vertical sides, first rotate counterclockwiseby 90 ◦ , yielding a quadrilateral with parallel horizontal sides. Since we areassuming that Q is not a parallelogram, we may then also assume that Q doesnot have parallel vertical sides. One can now use a translation, if necessary,to map the lower left hand corner vertex of Q to (0 , , , (0 , u ) , ( s, t ), and ( v, w ). Note that such arotation leaves Q without parallel vertical sides. In addition, by Lemma 5with T ( x, y ) = 1 u ( x, y ), we may also assume that one of the vertices of Q is(0 , Q s,t,v,w = Q ( A , A , A , A ) , A = (0 , , (16) A = (0 , , A = ( s, t ) , A = ( v, w ),13here s, v > , t > w, s = v . (17)The sides of Q s,t,v,w , going clockwise, are given by S = (0 ,
0) (0 , u ) , S =(0 ,
1) ( s, t ) , S = ( s, t ) ( v, w ), and S = (0 ,
0) ( v, w ). By Lemma 3, Q s,t,v,w isa midpoint diagonal quadrilateral, which implies, by Lemma 6, that Q s,t,v,w is not a trapezoid since Q s,t,v,w is not a parallelogram. We find it useful todefine the following expressions, each of which depend on s, t, v , and w : f = v ( t −
1) + (1 − w ) s,f = vt − ws, (18) f = ws − v ( t − • Since Q s,t,v,w is convex, ( s, t ) must lie above ←−−−−−−→ (0 ,
1) ( v, w ) and ( v, w ) mustlie below ←−−−−−→ (0 ,
0) ( s, t ), which implies that f > f >
0. (19)Since no two sides of Q s,t,v,w are parallel, S ∦ S , which implies that f = 0. (20) M = (cid:18) v,
12 ( w + 1) (cid:19) and M = (cid:18) s, t (cid:19) are the midpoints of thediagonals of Q s,t,v,w and the equation of the line thru M and M is y = L ( x ) = t w + 1 − tv − s (cid:16) x − s (cid:17) , (21) x ∈ I = (cid:26) ( v/ , s/
2) if v < s ( s/ , v/
2) if s < v .The diagonal line segments of Q s,t,v,w are D = (0 ,
0) ( s, t ) and D = (0 ,
1) ( v, w ).Now let E be an ellipse inscribed in Q s,t,v,w and suppose that E is tan-gent to Q s,t,v,w at the points P q = (cid:16) q, wv q (cid:17) ∈ S and P r = (0 , r ) ∈ S ,0 < q < v , r ∈ J . Using these points of tangency, it is not hard to show that q = svr ( s − f ) r + f . That leads to Proposition 4 below, which gives neces-sary and sufficient conditions for the general equation of an ellipse inscribedin Q s,t,v,w . It is useful for us to emphasize the dependence of the coefficientsof the general equation on the parameter r in our notation.14 roposition 4 Suppose that (17), (19), and (20) hold.(i) Let E be an ellipse inscribed in Q s,t,v,w . Then for some r ∈ J , thegeneral equation of E is given by ψ ( x, y ) = 0 , where ψ ( x, y ) = A ( r ) x + B ( r ) xy + C ( r ) y + D ( r ) x + E ( r ) y + F ( r ) , (22) and A ( r ) = ( s + v t + w s − tvs ( w + 1) + 2 ws (2 v − s )) r +2 v (cid:0) st − ws − t v + tsw (cid:1) r + t v ,B ( r ) = − vs (2 r ( v − s ) + rs ( w + 1) + v ( t − rt − r )) , (23) C ( r ) = v s , D ( r ) = 2 srv ( − rs ( w + 1) + 2 ws + tv ( r − ,E ( r ) = − rv s , F ( r ) = r s v .(ii) Conversely, if for some r ∈ J the general equation of E is givenby ψ ( x, y ) = 0 , where (22) and (23) hold, then E is an ellipse inscribed in Q s,t,v,w . Proof.
Using Lemma 1, it is not hard to show that ψ ( x, y ) = 0 defines theequation of an ellipse for any r ∈ J . Using standard calculus techniques, it isalso not difficult to show that the ellipse defined by ψ ( x, y ) = 0 is inscribedin Q s,t,v,w for any r ∈ J . The converse result, that any ellipse inscribed in Q s,t,v,w has equation given by ψ ( x, y ) = 0 for some r ∈ J can be proven ina similar fashion to the proof of Proposition 2. We leave the details to thereader.The following lemma gives necessary and sufficient conditions for Q s,t,v,w to be a midpoint diagonal quadrilateral. Lemma 10
Suppose that (17), (19), and (20) hold.(i) Q s,t,v,w is a type 1 midpoint diagonal quadrilateral if and only if vt = ( w + 1) s . (24) (ii) Q s,t,v,w is a type 2 midpoint diagonal quadrilateral if and only if ( t − v = ( w − s . (25)15 roof. ←→ D has equation y = ts x . Using (21), ←→ D = L ⇐⇒ w + 1 − tv − s = ts and (26) t − s w + 1 − tv − s = 0.It follows easily that (26) holds if and only if (24) holds, which proves (i).The proof of (ii) follows in a similar fashion. Proof of Theorem 5Proof.
We assume first that Q is a tangential quadrilateral. Then Q isan orthodiagonal quadrilateral by Lemma 8, and so the diagonals of Q areperpendicular. Also, there is a unique circle, Φ, inscribed in Q , which impliesthat Φ is the unique ellipse of minimal eccentricity inscribed in Q since Φhas eccentricity 0. Since any pair of perpendicular diameters of a circle areequal conjugate diameters, in particular the unique pair which are parallelto the diagonals of Q are equal conjugate diameters of Φ, and Theorem 5holds. So assume now that Q is not a tangential quadrilateral. It sufficesto assume that Q = Q s,t,v,w and that (17), (19), and (20) hold. Let E be an ellipse inscribed in Q s,t,v,w . By Proposition 1, the general equationof E is given by ψ ( x, y ) = 0, where ψ is given by (22) and (23) for some r ∈ J . Let a and b denote the lengths of the semi–major and semi–minoraxes, respectively, of E . Now b a is really a function of r ∈ J if we allow E to vary over all ellipses inscribed in Q s,t,v,w . By (2) in Lemma 2, b a = G ( r ),where G ( r ) = A ( r ) + C ( r ) − q ( A ( r ) − C ( r )) + ( B ( r )) A ( r ) + C ( r ) + q ( A ( r ) − C ( r )) + ( B ( r )) . Since the squareof the eccentricity of E equals 1 − b a , it suffices to maximize b a . Letting O ( r ) = A ( r ) + C ( r ) , (27) M ( r ) = ( A ( r ) − C ( r )) + ( B ( r )) ,we have G ( r ) = O ( r ) − p M ( r ) O ( r ) + p M ( r ) . Since Q is not a tangential quadrilateral, itfollows easily that Q s,t,v,w is also not a tangential quadrilateral. If M ( r ) = 016or some r ∈ J , then A ( r ) − C ( r ) = 0 and B ( r ) = 0, which impliesthat the ellipse inscribed in Q s,t,v,w corresponding to r is a circle. But thatcontradicts the assumption that Q s,t,v,w is not a tangential quadrilateral.Thus M ( r ) = 0 for all r ∈ J . Since M is non–negative, it follows that M ( r ) > , r ∈ J . (28)Define the quartic polynomial N ( r ) = O ( r ) − M ( r ).After some simplification, N factors as N ( r ) = 16 s v r (1 − r ) (( s − v ) r + v )(( s − v ) r + f ), (29)and N has roots r = 0 , r = 1 , r = f v − s , r = vv − s . (30)Note that r = r ⇐⇒ f = 0, which cannot hold by (20), and r = 0 = r since v = 0 = f by (17) and (19). r = 1 ⇐⇒ f = 0, which cannothold by (19), and r = 1 ⇐⇒ s = 0, which cannot hold by (17). Thus allroots listed in (30) are distinct . A simple computation yields G ′ ( r ) = p ( r )( O ( r ) + p M ( r )) p M ( r ) , (31)where the quartic polynomial p is given by p ( r ) = 2 M ( r ) O ′ ( r ) − O ( r ) M ′ ( r ). (32)To finish the proof of Theorem 5, the following lemmas will be usedto show that p has a unique root in J . Lemma 11 N ( r ) > on J . Proof.
First define the linear function of r, L ( r ) = ( s − v ) r + f . L (0) = f > L (1) = f > L > J . Similarly, ( s − v ) r + v > J . By (29), N > J since r (1 − r ) > J .17 emma 12 O ( r ) > on J . Proof. If O ( r ) = 0 for some r ∈ J , then N ( r ) = − M ( r ) ≤ M ( r ) ≥ J . That contradicts Lemma 11. Hence O ( r ) is nonzero on J .Since O (0) = ( s + t ) v >
0, that proves Lemma 12.Now assume that Q s,t,v,w is a type 1 midpoint diagonal quadrilateral.In [2] we proved that there is a unique ellipse of minimal eccentricity in-scribed in any convex quadrilateral, Q . The uniqueness for midpoint diag-onal quadrilaterals would then follow from that result. However, the proofhere, specialized for midpoint diagonal quadrilaterals, is self–contained, usesdifferent methods, and does not require the result from [2]. Use (24) tosubstitute s ( w + 1) v for t in Proposition 4. Some simplification then yields A ( r ) = s (4 w ( v − s ) r − vwr + s ( w + 1) ), B ( r ) = 2 sv (2 ( s − v ) r +2 vr − s (1 + w ) ), C ( r ) = s v , D ( r ) = 2 rs v ( w − E ( r ) = − rs v , and F ( r ) = r s v . Using (27) and (32), it then follows that p ( r ) = − v s (2 ( s − v ) r + v ) α ( r ), where α ( r ) = 2 ( s − v ) (cid:0) v + w + 1 (cid:1) r +2 v (cid:0) v + w + 1 (cid:1) r − s ( v +( w +1) ). (33)(17), (19), and (20) now become s, v > s − v > α (0) = − s ( v + ( w + 1) ) < α (1) = s ( v + ( w − ) >
0, whichimplies that α has precisely one root in J since α is a quadratic. By (31), (28),and Lemma 12, G is differentiable on J . The linear function 2 ( s − v ) r + v is nonzero at r = 0 since v > r = 1 since 2 s − v >
0. Thus theother factors of p are nonzero and hence p has precisely one root, r ∈ J ,which is also the unique root of G ′ ( r ) in J by (31). Since G ( r ) > J , G (0) = G (1) = 0, and G is positive in the interior of I and vanishes at theendpoints of I , G ( r ) must yield the global maximum of G on J . That provesTheorem 5(i). Note that the equation of E I is obtained by letting r = r inProposition 4. To prove Theorem 5(ii), by Theorem 4(i), E I has conjugatediameters, CD and CD , parallel to the diagonals, D and D , of Q s,t,v,w . ←→ D has equation y = ts x = w + 1 v x and ←→ D has equation y = 1 + w − v x .Since Q s,t,v,w is type 1, ←→ D = L , the line thru the midpoints of D and D ,and thus L has equation y = w + 1 v x . It then follows that ←−→ CD = L (seeRemark 7) and hence ←−→ CD has equation y = w + 1 v x . For convenience, we18et β = ( s − v ) r + v , ζ = ( s − v ) r + s .By Proposition 4, B ( r ) E ( r ) − C ( r ) D ( r )4 A ( r ) C ( r ) − B ( r ) = 12 sv ( s − v ) r + v . Thus by (3)of Lemma 2, E I has center ( x , y ), where x = B ( r ) E ( r ) − C ( r ) D ( r )4 A ( r ) C ( r ) − B ( r ) =12 svβ . Since E I has center ( x , L ( x )) by Newton’s Theorem and L (cid:18) svβ (cid:19) =12 w + 1 v svβ = 12 s ( w + 1) β , E I has center (cid:18) svβ , s ( w + 1) β (cid:19) . Since ←−→ CD passes through the center of E I and has the same slope as ←→ D , which is w − v , ←−→ CD has equation y − s ( w + 1) β = w − v (cid:18) x − svβ (cid:19) , which simplifies to y = w − v x + sβ . Proof.
Now suppose that CD intersects E I at the two distinct points P = ( x , y ) = (cid:18) x , w + 1 v x (cid:19) and P = ( x , y ) = (cid:18) x , w + 1 v x (cid:19) . Since P and P lie on E I , by Proposition 4, A ( r ) x j + B ( r ) x j y j + C ( r ) y j + D ( r ) x j + E ( r ) y j + I ( r ) = 0 , j = 1 ,
2. Substituting y j = w + 1 v x j yields A ( r ) x j + B ( r ) x j (cid:18) w + 1 v (cid:19) + C ( r ) x j (cid:18) w + 1 v (cid:19) + D ( r ) x j + E ( r ) (cid:18) w + 1 v (cid:19) x j + I ( r ) =0 , j = 1 ,
2, which simplifies to( A ( r ) + (cid:18) w + 1 v (cid:19) B ( r ) + (cid:18) w + 1 v (cid:19) C ( r )) x j + (34)( D ( r ) + (cid:18) w + 1 v (cid:19) E ( r )) x j + I ( r ) = 0 , j = 1 , x and x are two distinct real roots of any quadratic, then x − x = √ τa ⇒ ( x − x ) = τa , τ = discriminant and a = leading coefficient. For the specific quadraticin the variable x j given in (34) we have τ = ( D ( r ) + (cid:18) w + 1 v (cid:19) E ( r )) − F ( r )( A ( r ) + (cid:18) w + 1 v (cid:19) B ( r ) + (cid:18) w + 1 v (cid:19) C ( r ))and a = A ( r ) + (cid:18) w + 1 v (cid:19) B ( r ) + (cid:18) w + 1 v (cid:19) C ( r ).Now applying Proposition 4 and simplfying yields τ = 16 r s v (1 − r ) ζ,a = 4 r sβ .Hence ( x − x ) = τa = v s (1 − r ) ζβ . Now y − y = w + 1 v ( x − x ) ⇒ ( x − x ) + ( y − y ) = ( x − x ) (cid:18) w + 1 v (cid:19) ! , which implies that( x − x ) + ( y − y ) = (35) (cid:18) w + 1 v (cid:19) ! v s (1 − r ) ζβ .Similarly, suppose that CD intersects E I at the two distinct points P =( x , y ) = (cid:18) x , w − v x + sβ (cid:19) and P = ( x , y ) = (cid:18) x , w − v x + sβ (cid:19) .Since P and P lie on E I , we have A ( r ) x j + B ( r ) x j y j + C ( r ) y j + D ( r ) x j + E ( r ) y j + I ( r ) = 0 , j = 3 ,
4. Substituting y j = w − v x j + sβ and simplifyingyields ( A ( r ) + w − v B ( r ) + (cid:18) w − v (cid:19) C ( r )) x j + (36) (cid:18) sβ B ( r ) + 2 sβ w − v C ( r ) + D ( r ) + w − v E ( r ) (cid:19) x j + (cid:18) sβ (cid:19) C ( r ) + (cid:18) sβ (cid:19) E ( r ) + I ( r ) = 0 , j = 3 , x j given in (36) we have τ = (cid:18) sβ B ( r ) + 2 sβ w − v C ( r ) + D ( r ) + w − v E ( r ) (cid:19) − A ( r ) + w − v B ( r ) + (cid:18) w − v (cid:19) C ( r )) (cid:18) sβ (cid:19) C ( r ) + (cid:18) sβ (cid:19) E ( r ) + I ( r ) ! .The leading coefficient is a = A ( r ) + w − v B ( r ) + (cid:18) w − v (cid:19) C ( r ).Applying Proposition 4 again and simplfying yields τ = 16 r s v ( r − ζ β ,a = 4 s (1 − r ) ζ .Thus ( x − x ) = τa = 16 r s v ( r − ζ β (4 s (1 − r ) ζ ) = r sv β . Now ( y − y ) = (cid:18) w − v (cid:19) ( x − x ) ⇒ ( x − x ) +( y − y ) = (cid:18) w − v (cid:19) ! ( x − x ) ,which implies that ( x − x ) + ( y − y ) = (37) (cid:18) w − v (cid:19) ! r sv β . L and L are equal conjugate diameters if and only if | P P | = | P P | ⇐⇒ ( x − x ) + ( y − y ) = ( x − x ) + ( y − y ) . Using (35) and (37), | P P | = | P P | ⇐⇒ (cid:18) w + 1 v (cid:19) ! v s (1 − r ) ζβ = (cid:18) w − v (cid:19) ! r sv β ⇐⇒ (cid:18) w + 1 v (cid:19) ! ( v s (1 − r ) ζ ) β − (cid:18) w − v (cid:19) ! r sv β = 0 ⇐⇒ (cid:18) w + 1 v (cid:19) ! ((1 − r ) ζ ) − (cid:18) w − v (cid:19) ! r β = 0 ⇐⇒
21 ( s − v ) ( v + w + 1) r + 2 v ( v + w + 1) r − s ( v + ( w + 1) ) v = 0 ⇐⇒ α ( r ) = 0, where α is given by (33). Since r is a root of α by definition,that completes the proof of Theorem 5(ii). Example
Consider the quadrilateral, Q , with vertices (0 , , (0 , , (8 , , Q = Q s,t,v,w with s = 8, t = 4, v = 6, and w = 2, which satisfy(17), (19), and (20). Q is a type 1 midpoint diagonal quadrilateral since vt = ( w + 1) s . The diagonal lines of Q are ←→ D : y = 12 x , which is also theequation of L , and ←→ D : y = 1 + 16 x .First, let E be the ellipse inscribed in Q corresponding to r = 37 . ByProposition 4(i) the equation of E is 33 x − xy +196 y +28 x − y = − E has center (cid:18) , (cid:19) and the points of tangency of E with Q aregiven by q = (cid:18) , (cid:19) , q = (cid:18) , (cid:19) , q = (cid:18) , (cid:19) , and q = (cid:18) , (cid:19) . • As guaranteed by Theorem 3, the slope of ←→ q q = slope of ←→ q q = 16 =slope of D . • Suppose that DI and DI are diameters of E which are parallel tothe diagonals D and D , respectively, and suppose that DI intersects E at the two distinct points P and P , while DI intersects E at the twodistinct points P and P . The equations of ←−→ P P and ←−→ P P are thus y −
74 = 12 (cid:18) x − (cid:19) and y −
74 = 16 (cid:18) x − (cid:19) , respectively. To determine thecoordinates of P and P , substitute the equation of DI into the equation of E . That yields 33 x − x (cid:18)
74 + 12 (cid:18) x − (cid:19)(cid:19) + 196 (cid:18)
74 + 12 (cid:18) x − (cid:19)(cid:19) +28 x − (cid:18)
74 + 12 (cid:18) x − (cid:19)(cid:19) = −
36, which has solutions x = 72 ± √ P = 14 (7 − √ ,
1) and P = 14 (7 + √ , P =14 (2 , (cid:0) − √ , − √ (cid:1) and P = 14 (2 , (cid:0) √ , √ (cid:1) . The familyof lines which are parallel to ←−→ P P have equation y = 12 x + b . Substituting22nto the equation of E yields 33 x − x (cid:18) x + b (cid:19) +196 (cid:18) x + b (cid:19) +28 x − (cid:18) x + b (cid:19) = −
36. Solving for x gives x = 72 − b ± √ − b , and thusthe chords of E which are parallel to ←−→ P P have midpoints (cid:18) − b, − b (cid:19) ,which each lie on ←−→ P P . Hence DI and DI are the pair of conjugatediameters of E which are parallel to the diagonals of Q , as guaranteed byTheorem 4.Second, let E I be the unique ellipse of minimal eccentricity inscribed in Q . By (33), α ( r ) = 164 r + 492 r −
360 and the unique root of α in J is r = −
32 + 2782 √
41. By Proposition 4(i), after some simplification, the equa-tion of E I is 3(427 − √ x + 8( −