When is an ellipse inscribed in a quadrilateral tangent at the midpoint of two or more sides ?

TL;DR

This paper investigates conditions under which an inscribed ellipse in a quadrilateral is tangent at midpoints of sides, revealing that such ellipses are only possible in specific quadrilaterals like trapezoids or midpoint diagonal quadrilaterals.

Contribution

It strengthens previous results by proving that non-parallelogram quadrilaterals cannot have an inscribed ellipse tangent at three midpoints, and characterizes quadrilaterals with ellipses tangent at two midpoints.

Findings

No inscribed ellipse tangent at three midpoints in non-parallelogram quadrilaterals.

Quadrilaterals with ellipses tangent at two midpoints are trapezoids or midpoint diagonal quadrilaterals.

Ellipses tangent at midpoints are highly restricted by quadrilateral geometry.

Abstract

In "Quartic Coincidences and the Singular Value Decomposition" by Clifford and Lachance, Mathematics Magazine, December, 2013, it was shown that if there is a midpoint ellipse(an ellipse inscribed in a quadrilateral, $Q$, which is tangent at the midpoints of all four sides of $Q$), then $Q$ must be a parallelogram. We strengthen this result by showing that if $Q$ is not a parallelogram, then there is no ellipse inscribed in $Q$ which is tangent at the midpoint of three sides of $Q$. Second, the only quadrilaterals which have inscribed ellipses tangent at the midpoint of even two sides of $Q$ are trapezoids or what we call a midpoint diagonal quadrilateral(the intersection point of the diagonals of $Q$ coincides with the midpoint of at least one of the diagonals of $Q$).

Equations

$$Q_{s,t}=Q(A_1,A_2,A_3,A_4),$$

$$A_1=(0,0),A_2=(0,1),A_3=(s,t),A_4=(1,0)\text{.}$$

$$G=\left\{(s,t):s,t>0,s+t>1,s\ne 1\right\}\text{.}$$

$$t^2{x}^2+(4q^2(t-1)t+2qt(s-t+2)-2st)xy+$$

$$((1-q)s+qt{)}^2{y}^2-2q{t}^{2}x-2qt((1-q)s+qt)y+q^2{t}^2=0$$

$$P_1=M{P}_1⟺\frac{qt}{(t-s)q+s}=\frac{1}{2},$$

$$P_2=M{P}_2⟺\frac{(1-q)s}{(t-1)(s+t)q+s}=\frac{1}{2}\text{and}$$

$$\frac{t(s+q(t-1))}{(t-1)(s+t)q+s}=\frac{1+t}{2}$$

$$P_3=M{P}_3⟺\frac{s+q(t-1)}{(s+t-2)q+1}=\frac{1+s}{2}\text{ and}$$

$$\frac{(1-q)t}{(s+t-2)q+1}=\frac{t}{2},$$

$$P_4=M{P}_4⟺q=\frac{1}{2}\text{.}$$

$$P_1=M{P}_1⟺\frac{q}{(1-s)q+s}=\frac{1}{2}\text{,}$$

$$P_2=M{P}_2⟺(1-q)s=\frac{s}{2}\text{,}$$

$$P_3$$

$$P_4=M{P}_4⟺q=\frac{1}{2}\text{.}$$

Full text

1 Introduction

Among all ellipses inscribed in a triangle, $T$, the midpoint, or Steiner, ellipse is interesting and well–known(see [2]). It is the unique ellipse tangent to $T$ at the midpoints of all three sides of $T$ and is also the unique ellipse of maximal area inscribed in $T$. What about ellipses inscribed in quadrilaterals, $Q$ ? Not surprisingly, perhaps, there is not always a midpoint ellipse–i.e., an ellipse inscribed in $Q$ which is tangent at the midpoints of all four sides of $Q$; In fact, in [1] it was shown that if there is a midpoint ellipse, then $Q$ must be a parallelogram. That is, if $Q$ is not a parallelogram, then there is no ellipse inscribed in $Q$ which is tangent at the midpoint of all four sides of $Q$; But can one do better than four sides of $Q$ ? In other words, if $Q$ is not a parallelogram, is there an ellipse inscribed in $Q$ which is tangent at the midpoint of three sides of $Q$ ? In Theorem 1 we prove that the answer is no. In fact, unless $Q$ is a trapezoid(a quadrilateral with at least one pair of parallel sides), or what we call a midpoint diagonal quadrilateral(see the definition below), then there is not even an ellipse inscribed in $Q$ which is tangent at the midpoint of two sides of $Q$(see Lemmas 3 and 4) .

Definition 1

A convex quadrilateral, $Q$, is called a midpoint diagonal quadrilateral(mdq) if the intersection point of the diagonals of $Q$ coincides with the midpoint of at least one of the diagonals of $Q$.

A parallelogram, p, is a special case of an mdq since the diagonals of p bisect one another. In [6] we discussed mdq’s as a generalization of parallelograms in a certain sense related to tangency chords and conjugate diameters of inscribed ellipses. We note here that there are two types of mdq’s: Type 1, where the diagonals intersect at the midpoint of $D_{2}$ and Type 2, where the diagonals intersect at the midpoint of $D_{1}$;

What about uniqueness ? If $Q$ is an mdq, then the ellipse inscribed in $Q$ which is tangent at the midpoint of two sides of $Q$ is not unique. Indeed we prove (Lemma 3) that in that case there are two such ellipses. However, if $Q$ is a trapezoid, then the ellipse inscribed in $Q$ which is tangent at the midpoint of two sides of $Q$ is unique (Lemma 4).

Is there a connection with tangency at the midpoint of sides of $Q$ and the ellipse of maximal area inscribed in $Q$ as with parallelograms ?

For trapezoids, we prove(Lemma 4) that the unique ellipse of maximal area inscribed in $Q$ is the unique ellipse tangent to $Q$ at the midpoint of two sides of $Q$. However, for mdq’s, the unique ellipse of maximal area inscribed in $Q$ need not be tangent at the midpoint of any side of $Q$.

We use the notation $Q(A_{1},A_{2},A_{3},A_{4})$ to denote the quadrilateral with vertices $A_{1},A_{2},A_{3}$, and $A_{4}$, starting with $A_{1}=$ lower left corner and going clockwise. Denote the sides of $Q(A_{1},A_{2},A_{3},A_{4})$ by $S_{1},S_{2},S_{3}$, and $S_{4}$, going clockwise and starting with the leftmost side, $S_{1}$; Given a convex quadrilateral, $Q=Q(A_{1},A_{2},A_{3},A_{4})$, which is not a parallelogram, it will simplify our work below to consider quadrilaterals with a special set of vertices. In particular, there is an affine transformation which sends $A_{1},A_{2}$, and $A_{4}$ to the points $(0,0),(0,1)$, and $(1,0)$, respectively. It then follows that $A_{3}=(s,t)$ for some $s,t>0$; Summarizing:

[TABLE]

Since $Q_{s,t}$ is convex, $s+t>1$; Also, if $Q$ has a pair of parallel vertical sides, first rotate counterclockwise by $90^{\circ}$, yielding a quadrilateral with parallel horizontal sides. Since we 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 $s\neq 1$. Note that any trapezoid which is not a parallelogram may be mapped, by an affine transformation, to the quadrilateral $Q_{s,1}$; Thus we may assume that $(s,t)\in G$, where

[TABLE]

The following result gives the points of tangency of any ellipse inscribed in $Q_{s,t}$(see [4], where some details were provided). We leave the details of a proof to the reader. Those details will also be provided in a forthcoming book [5].

Proposition 1

(i) $E_{0}$ is an ellipse inscribed in $Q_{s,t}$ if and only if the general equation of $E_{0}$ is given by

[TABLE]

for some $q\in J=(0,1)$. Furthermore, (4) provides a one–to–one correspondence between ellipses inscribed in $Q_{s,t}$ and points $q\in J$.

(ii) If $E_{0}$ is an ellipse given by (4) for some $q\in J$, then $E_{0}$ is tangent to the four sides of $Q_{s,t}$ at the points

$P_{1}=\left(0,\dfrac{qt}{(t-s)q+s}\right)\in S_{1},$**

$P_{2}=\left(\dfrac{(1-q)s^{2}}{(t-1)(s+t)q+s},\dfrac{t{\large(}s+q(t-1){\large)}}{(t-1)(s+t)q+s}\right)\in S_{2},$**

$P_{3}=\left(\dfrac{s+q(t-1)}{(s+t-2)q+1},\dfrac{(1-q)t}{(s+t-2)q+1}\right)\in S_{3}$, and $P_{4}=(q,0)\in S_{4}$.

Remark 1

Using Proposition 1, it is easy to show that one can always find an ellipse inscribed in a quadrilateral, $Q$, which is tangent to $Q$ at the midpoint of at least one side of $Q$, and this can be done for any given side of $Q$.

For the rest of the paper we work with the quadrilateral $Q_{s,t}$ defined above. The following lemma gives necessary and sufficient conditions for $Q_{s,t}$ to be an mdq.

Lemma 1

(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 diagonals of $Q_{s,t}$ are $D_{1}$: $y=\dfrac{t}{s}x$ and $D_{2}$: $y=1-x$, and they intersect at the point $P=\left(\dfrac{s}{s+t},\dfrac{t}{s+t}\right)$; The midpoints of $D_{1}$ and $D_{2}$ are $M_{1}=\left(\dfrac{s}{2},\dfrac{t}{2}\right)$ and $M_{2}=\left(\dfrac{1}{2},\dfrac{1}{2}\right)$, respectively. Now $M_{2}=P\iff\dfrac{s}{s+t}=\dfrac{1}{2}$ and $\dfrac{t}{s+t}=\dfrac{1}{2}$, both of which hold if and only if $s=t$; That proves (i); $M_{1}=P\iff\dfrac{s}{s+t}=\dfrac{1}{2}s$ and $\dfrac{t}{s+t}=\dfrac{1}{2}t$, both of which hold if and only if $s+t=2$. That proves (ii).

The following lemma shows that affine transformations preserve the class of mdq’s. We leave the details of the proof to the reader.

Lemma 2

Let $T:R^{2}\rightarrow R^{2}$ be an affine transformation and let $Q$ be a midpoint diagonal quadrilateral. Then $Q^{\prime}=T(Q)$ is also a midpoint diagonal quadrilateral.

The following result shows that among non–parallelograms, the only quadrilaterals which have inscribed ellipses tangent at the midpoint of two sides are the mdq’s.

Lemma 3

*Let $Q$ be a convex quadrilateral in the $xy$ plane which is **not *a trapezoid.

(i) There is an ellipse inscribed in $Q$ which is tangent at the midpoint of two sides of $Q$ if and only if $Q$ is a midpoint diagonal quadrilateral, in which case there are two such ellipses.

(ii) There is no ellipse inscribed in $Q$ which is tangent at the midpoint of three sides of $Q$.

Proof. By Lemma 2 and standard properties of affine transformations, we may assume that $Q=Q_{s,t}$, the quadrilateral given in (1) with $(s,t)\in G$; The midpoints of the sides of $Q_{s,t}$ are given by $MP_{1}=\left(0,\dfrac{1}{2}\right)\in S_{1}$, $MP_{2}=\left(\dfrac{s}{2},\dfrac{1+t}{2}\right)\in S_{2}$, $MP_{3}=\left(\dfrac{1+s}{2},\dfrac{t}{2}\right)\in S_{3}$, and $MP_{4}=\left(\dfrac{1}{2},0\right)\in S_{4}$; Now let $E_{0}$ denote an ellipse inscribed in $Q_{s,t}$, and let $P_{j}\in S_{j},j=1,2,3,4$ denote the points of tangency of $E_{0}$ with the sides of $Q_{s,t}$; By Proposition 1(ii),

[TABLE]

Equations (5) and (10) each have the unique solutions $q_{1}=\dfrac{s}{s+t}$ and $q_{4}=\dfrac{1}{2}$, respectively. The system of equations in (6) and (7) has unique solution $q_{2}=\dfrac{s}{t^{2}+st+s-t}$, and the system of equations in (8) and (9) has unique solution $q_{3}=\dfrac{1}{s+t}$; It is trivial that $q_{1},q_{3},q_{4}\in J=(0,1)$; Since $(s,t)\in G$, $t\left(s+t-1\right)>0$, which implies that $q_{2}\in J$. We now check which pairs of midpoints of sides of $Q_{s,t}$ can be points of tangency of $E_{0}$; Note that different values of $q$ yield distinct inscribed ellipses by the one–to–one correspondence between ellipses inscribed in $Q_{s,t}$ and points $q\in J$.

(a) $S_{1}$ and $S_{2}$: $q_{1}=q_{2}\iff\dfrac{s}{\allowbreak t\left(s+t-1\right)+s}-\dfrac{s}{s+t}=$

$-\allowbreak\dfrac{st(s+t-2)}{\left(ts+t^{2}+s-t\right)\left(s+t\right)}=0\iff s+t=2$.

(b) $S_{1}$ and $S_{3}$: $q_{1}=q_{3}\iff\dfrac{1}{s+t}-\dfrac{s}{s+t}=\allowbreak-\dfrac{s-1}{s+t}=0$, which has no solution since $s\neq 1$.

(c) $S_{2}$ and $S_{3}$: $q_{2}=q_{3}\iff\dfrac{1}{s+t}-\dfrac{s}{\allowbreak t\left(s+t-1\right)+s}=\allowbreak\dfrac{(t-s)\left(s+t-1\right)}{\left(ts+t^{2}+s-t\right)\left(s+t\right)}\allowbreak=0\iff s=t$.

(d) $S_{1}$ and $S_{4}$: $q_{1}=q_{4}\iff\dfrac{1}{2}-\dfrac{s}{s+t}=\allowbreak-\dfrac{1}{2}\dfrac{s-t}{s+t}=0\iff s=t$.

(e) $S_{2}$ and $S_{4}$: $q_{2}=q_{4}\iff\dfrac{1}{2}-\dfrac{s}{\allowbreak t\left(s+t-1\right)+s}=\allowbreak\dfrac{1}{2}\dfrac{(s+t)\left(t-1\right)}{ts+t^{2}+s-t}=0$, which has no solution since $s+t\neq 0$ and $t\neq 1$.

(f) $S_{3}$ and $S_{4}$: $q_{3}=q_{4}\iff\dfrac{1}{2}-\dfrac{1}{s+t}=\allowbreak\dfrac{1}{2}\dfrac{s+t-2}{s+t}=0\iff s+t=2$.

That proves that there is an ellipse inscribed in $Q_{s,t}$ which is tangent at the midpoints of $S_{1}$ and $S_{2}$ or at the midpoints of $S_{3}$ and $S_{4}$ if and only if $s+t=2$, and there is an ellipse inscribed in $Q_{s,t}$ which is tangent at the midpoints of $S_{2}$ and $S_{3}$ or at the midpoints of $S_{1}$ and $S_{4}$ if and only if $s=t$; Furthermore, if $s\neq t$ or if $s+t\neq 2$, then there is no ellipse inscribed in $Q_{s,t}$ which is tangent at the midpoint of two sides of $Q_{s,t}$. That proves (i) by Lemma 1. To prove (ii), to have an ellipse inscribed in $Q_{s,t}$ which is tangent at the midpoint of three sides of $Q_{s,t}$, those three sides are either $S_{1}$, $S_{2}$, and $S_{3}$; $S_{1}$, $S_{2}$, and $S_{4}$; $S_{1}$, $S_{3}$, and $S_{4}$; or $S_{2}$, $S_{3}$, and $S_{4}$; By (a)–(f) above, that is not possible.

For trapezoids inscribed in $Q$ we have the following result.

Lemma 4

Assume that $Q$ is a trapezoid which is not a parallelogram. Then

(i) There is a unique ellipse inscribed in $Q$ which is tangent at the midpoint of two sides of $Q$, and that ellipse is the unique ellipse of maximal area inscribed in $Q$.

(ii) There is no ellipse inscribed in $Q$ which is tangent at the midpoint of three sides of $Q$.

Proof. Again, by affine invariance, we may assume that $Q=Q_{s,1}$, the quadrilateral given in (1) with $t=1$; Note that $0<s\neq 1$; Now let $E_{0}$ denote an ellipse inscribed in $Q_{s,1}$. Letting $MP_{j}\in S_{j},j=1,2,3,4$ denote the corresponding midpoints of the sides and using Proposition 1(ii) again, with $t=1$, we have

[TABLE]

[TABLE]

[TABLE]

[TABLE]

The unique solution of the equations in (12) and in (15) is $q=\dfrac{1}{2}\in J$; The unique solution of the equation in (11) is $q=\dfrac{s}{1+s}\in J$, and the unique solution of the system of equations in (14) and (14) is $q=\dfrac{1}{1+s}\in J$; We now check which pairs of midpoints of sides of $Q_{s,1}$ can be points of tangency of $E_{0}$:

(a) $q=\dfrac{1}{2}$ gives tangency at the midpoints of $S_{2}$ and $S_{4}$.

(b) $S_{1}$ and $S_{2}$ or $S_{1}$ and $S_{4}$: $\dfrac{s}{1+s}=\dfrac{1}{2}\iff s=1$.

(c) $S_{3}$ and $S_{2}$ or $S_{3}$ and $S_{4}$: $\dfrac{1}{1+s}=\dfrac{1}{2}\iff s=1$.

(d) $S_{1}$ and $S_{3}$: $\dfrac{s}{1+s}=\dfrac{1}{1+s}\iff s=1$.

Since we have assumed that $s\neq 1$, the only way to have an ellipse inscribed in $Q_{s,1}$ which is tangent at the midpoint of two sides of $Q_{s,1}$ is if those sides are $S_{2}$ and $S_{4}$ and $q=\dfrac{1}{2}$. That proves that there is a unique ellipse inscribed in $Q_{s,1}$ which is tangent at the midpoint of two sides of $Q_{s,1}$. Now suppose that $E_{0}$ is any ellipse with equation $Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0$, and let $a$ and $b$ denote the lengths of the semi–major and semi–minor axes, respectively, of $E_{0}$. Using the results in [8], it can be shown that $a^{2}b^{2}=\dfrac{4\delta^{2}}{\Delta^{3}}$, where $\Delta=4AC-B^{2}$ and $\delta=CD^{2}+AE^{2}-BDE-F\Delta$; By Proposition 1(i), then, with $t=1$ and after some simplification, we have $a^{2}b^{2}=f(q)=\dfrac{s}{4}q\left(1-q\right)$; $q=\dfrac{1}{2}$ clearly maximizes $f(q)$ and thus gives the ellipse of maximal area inscribed in $Q_{s,1}$. That proves the rest of (i). (ii) now follows easily and we omit the details.

Remark 2

It can be shown (see [6]) that if $Q$ is a trapezoid which is not a parallelogram, then $Q$ cannot be an mdq. Thus the only quadrilaterals Lemmas 3 and 4 have in common are parallelograms.

Since a convex quadrilateral which is not a parallelogram either has no two sides which are parallel, or is a trapezoid, the following theorem follows immediately from Lemma 3(ii) and Lemma 4(ii).

Theorem 1

Suppose that $Q$ is a convex quadrilateral which is not a parallelogram. Then there is no ellipse inscribed in $Q$ which is tangent at the midpoint of three sides of $Q$.

2 Examples** **

(1) Let $Q$ be the quadrilateral with vertices, $(0,0),(0,1),(2,4)$, and $(1,1)$; It follows easily that $Q$ is a type 1 midpoint diagonal quadrilateral. The ellipse with equation $10\left(x-\dfrac{2}{3}\right)^{2}-10\left(x-\dfrac{2}{3}\right)\left(y-\dfrac{4}{3}\right)+4\left(y-\dfrac{4}{3}\right)^{2}=\dfrac{5}{3}$ is tangent to $Q$ at $\left(0,\dfrac{1}{2}\right)$ and at $\left(\dfrac{1}{2},\dfrac{1}{2}\right)$, the midpoints of $S_{1}$ and $S_{4}$, respectively. The ellipse with equation $54\left(x-\dfrac{4}{5}\right)^{2}-54\left(x-\dfrac{4}{5}\right)\left(y-\dfrac{8}{5}\right)+16\left(y-\dfrac{8}{5}\right)^{2}=\dfrac{27}{5}$ is tangent to $Q$ at$\allowbreak\left(1,\dfrac{5}{2}\right)$ and at $\allowbreak\left(\dfrac{3}{2},\dfrac{5}{2}\right)$, the midpoints of $S_{2}$ and $S_{3}$, respectively. One can show that neither of these ellipses is the ellipse of maximal area inscribed in $Q$.

(2) Let $Q$ be the trapezoid with vertices $(0,0),(0,1),(2,1)$, and $(1,1)$; The ellipse with equation

$\left(x-\dfrac{5}{4}\right)^{2}-3\left(x-\dfrac{5}{4}\right)\left(y-\dfrac{1}{2}\right)+\dfrac{25}{4}\left(y-\dfrac{1}{2}\right)^{2}=1$ is tangent to $Q$ at $\left(2,1\right)$ and at $\left(\dfrac{1}{2},0\right)$, the midpoints of $S_{2}$ and $S_{4}$, respectively.