# A concave quadrilateral cannot be circumscribed by an ellipse nor by a parabola.

It's intuitive that a concave quadrilateral cannot be circumscribed by an ellipse nor by a parabola.

I can prove it by analytic geometry:

Without loss of generality, let $$L_1\equiv m_1x -y +r_1=0$$, $$L_2\equiv m_2x -y +r_2=0$$, $$L_3\equiv m_3x -y +r_3=0$$, $$L_4\equiv m_4x -y +r_4=0$$ be the equations of lines $$AB$$, $$BC$$, $$CD$$, $$DA$$ of a concave quadrilateral ABCD.

Since ABCD is a concave quadrilateral, by virtue of the theorem stated and proved in The concave quadrilateral and the slopes of its sides, we have

$$(m_1-m_2)(m_2-m_3)(m_3-m_4)(m_4-m_1)<0$$

Besides that, all the conics which circumscribe the concave quadrilateral ABCD can be given by the equation $$kL_1L_3+L_2L_4=0$$ (except the degenerate conic consisting of the pair of concurrent lines $$AB$$ and $$CD$$). This degenerate conic is a degenerate hyperbole because a concave quadrilateral cannot have parallel opposite sides.

Therefore all the conics circumscribing the quadrilateral ABCD (except the mentioned degenerate hyperbole) are given by the equation $$k(m_1x -y +r_1)(m_3x -y +r_3)+(m_2x -y +r_2)(m_4x -y +r_4)=0,$$ $$(m_1m_3k +m_2m_4)x^2-((m_1+m_3)k+(m_2+m_4))xy+(k+1)y^2+...=0$$

The type of this circumscribing conic is given by

$$\delta=(km_1m_3+m_2m_4)(k+1)-\frac 14(k(m_1+m_3)+(m_2+m_4))^2,$$

This circumscribing conic is an ellipse if $$\delta>0$$, an hyperbole if $$\delta<0$$, a parabola if $$\delta=0$$.

Developing $$\delta$$ we get

$$4\delta= 4m_1m_3k^2+4m_2m_4k+4m_1m_3k+4m_2m_4-((m_1+m_3)^2k^2+2(m_1+m_3)(m_2+m_4)k+(m_2+m_4)^2),$$ $$4\delta=-(m_1-m_3)^2k^2+2[2m_2m_4+2m_1m_3-(m_1+m_3)(m_2+m_4)]k-(m_2-m_4)^2,$$

$$4\delta=-(m_1-m_3)^2k^2+2[(m_1-m_3)(m_2-m_4)+2(m_1-m_4)(m_3-m_2)]k-(m_2-m_4)^2$$

Now let $$\psi$$ be the discriminant of this second degree equation in k: $$\psi=4[(m_1-m_3)(m_2-m_4)+2(m_1-m_4)(m_3-m_2)]^2-4(m_1-m_3)^2(m_2-m_4)^2,$$ $$\frac 14 \psi=[(m_1-m_3)(m_2-m_4)+2(m_1-m_4)(m_3-m_2)]^2-(m_1-m_3)^2(m_2-m_4)^2,$$ $$\frac 14 \psi=4(m_1-m_3)(m_2-m_4)(m_1-m_4)(m_3-m_2)+4(m_1-m_4)^2(m_3-m_2)^2,$$ $$\frac 1{16} \psi=(m_1-m_3)(m_2-m_4)(m_1-m_4)(m_3-m_2)+(m_1-m_4)^2(m_3-m_2)^2,$$ $$\frac 1{16} \psi=(m_1-m_4)(m_3-m_2)[(m_1-m_3)(m_2-m_4)+(m_1-m_4)(m_3-m_2)],$$ $$\frac 1{16} \psi=(m_1-m_4)(m_3-m_2)(m_1-m_2)(m_3-m_4),$$

$$\frac 1{16} \psi=(m_1-m_2)(m_2-m_3)(m_3-m_4)(m_4-m_1)$$

Hence this discriminant $$\psi$$ is negative and since $$-(m_1-m_3)^2<0$$, $$\delta$$ is always negative. Consequently a concave quadrilateral can only be circumscribed by hyperboles, therefore it cannot be circumscribed by an ellipse nor by a parabola,

QED.

Is there another formal proof of it?

I have no idea about the parabola, I'll only cover the case of ellipse.

Since "concaveness" is preserved under linear transform, we can use a linear transform to map the ellipse to the unit circle centered at origin. The question becomes "can we inscribe a concave quadrilateral into that unit circle?".

Let $$ABCD$$ be a concave simple quadrilateral. Since it is concave, one of its vertices, say $$D$$, lies in the interior of the triangle $$ABC$$. This means we can find $$\alpha, \beta, \gamma > 0$$ such that $$D = \alpha A + \beta B + \gamma C\quad\text{ and }\quad\alpha+\beta+\gamma + 1$$

Let $$\lambda = \alpha+\beta$$ and $$\mu = \frac{\alpha}{\lambda}$$, we have $$\lambda,\mu \in (0,1)$$ and $$(\alpha, \beta, \gamma) = (\lambda\mu,\lambda(1-\mu),1-\lambda)$$

Let $$\varphi(P)$$ be the squared distance of point $$P$$ from origin. When $$A,B,C$$ lies on the unit circle, we have $$\varphi(A) = \varphi(B) = \varphi(C) = 1$$. In order to inscribe $$ABCD$$ into the unit circle, we also need $$\varphi(D) = 1$$.

As a function of $$P$$, $$\varphi(P)$$ is strictly convex over the Euclidean plane. This implies

\begin{align} \varphi(D) &= \varphi(\alpha A + \beta B + \gamma C) = \varphi(\lambda(\mu A + (1-\mu)B) + (1-\lambda) C)\\ &< \lambda \varphi(\mu A + (1-\mu)B) + (1-\lambda) \varphi(C)\\ &< \lambda( \mu \varphi(A) + (1-\mu)\varphi(B)) + (1-\lambda) \varphi(C)\\ &= \lambda( \mu + (1-\mu) ) + 1-\lambda\\ &= 1 \end{align} As a result, $$D$$ falls inside the interior of unit disk and cannot lie on the unit circle.

Update

I find an argument that works for both parabola and ellipse. The key is the concept extreme point of a convex set. There are several easy to verify facts.

1. If $$D$$ lies in the interior of triangle $$ABC$$, then $$D$$ is not an extreme point for the triangle (which is a convex set).
2. If $$P \in X \subset Y$$ for convex sets $$X$$ and $$Y$$, then $$P$$ is an extreme point for $$Y$$ implies $$P$$ is an extreme point for $$X$$.

3. The region bounded by an ellipse is convex and the ellipse is the set of extreme points for this region.

Let's say we have an ellipse $$\mathcal{E}$$ circumscribe a concave quadrilateral $$ABCD$$. Since $$ABCD$$ is concave, one of its vertices, say $$D$$ lies in the interior of triangle $$ABC$$.

• By $$(1)$$, $$D$$ is not an extreme point for triangle $$ABC$$.
• By $$(2)$$, $$D$$ is not an extreme point for the region bounded by $$\mathcal{E}$$ (since this region contains triangle $$ABC$$).
• By $$(3)$$, $$D$$ doesn't lie on $$\mathcal{E}$$.

This contradict with the assumption that $$\mathcal{E}$$ circumscribe $$ABCD$$.

For parabola, the argument is similar. A parabola divides the plane into two components. One component of it is convex while the other concave. It is easy to see if a parabola $$\mathcal{P}$$ circumscribe an quadrilateral $$ABCD$$, then the interior of the edges $$AB$$, $$BC$$, $$CD$$, $$DA$$ and body of quadrilateral lies inside the convex component.

We can replace $$(3)$$ by an alternate fact.

1. The parabola is the set of extreme points for the corresponding convex component.

Essentially same argument like before tell us if $$ABCD$$ is concave, then the point $$D$$ (the one lies inside the triangle formed by other points) cannot lies on $$\mathcal{P}$$. A contradiction again!

• Jensen's inequality...interesting. – MrDudulex Oct 21 '18 at 22:09