Determining center of inscribed ellipse of a pentagon

Given a convex pentagon $ABCDE$, there is a unique ellipse with center $F$ that can be inscribed in it as shown in the image below. I've written a small program to find this ellipse, and had to numerically (i.e. by iterations) solve five quadratic equations in the center $F$ coordinates and the entries of the inverse of the $Q$ matrix, such that the equation of the inscribed ellipse is

$(r - F)^T Q (r - F) = 1 $

My question is: Is it possible to determine the coordinates of the center in closed form, from the given coordinates of the vertices of the pentagon ?

Solution 1:

There is a simple geometric construction for the ellipse inscribed in a given convex pentagon $ABCDE$. One can, first of all, find tangency points $PQRST$. Draw, for instance, diagonals $AC$ and $BD$, meeting at $F$. Line $EF$ meets then side $BC$ at tangency point $P$.

This construction depends on Brianchon's theorem:

The three opposite diagonals of every hexagon circumscribing a conic are concurrent.

In fact, if $P$ is the tangency point on $BC$, we can view $ABPCDE$ as a limiting case of a hexagon circumscribed to the ellipse. Hence diagonals $AC$, $BD$ and $EP$ must intersect at the same point $F$.

Go on finding the other tangency points and remember then that the line, passing through the intersection point of two tangents to an ellipse and through the midpoint of their tangency points, also passes through the center of the ellipse. The center can thus be readily obtained, as the intersection point of $CM$ and $DN$, where $M$ is the midpoint of $PQ$ and $N$ is the midpoint of $QR$.

I implemented the above method with Mathematica, to find an explicit expression for the coordinates of the center, but the resulting expression is too large to be written here. Anyway, with the coordinates as in your figure, I get: $$ O=\left({237\over59}, {103\over59}\right). $$

Solution 2:

Disclaimer - this is not an independent answer!

I have transformed the geometric construction in Intelligenti pauca's excellent answer into algebra using homogeneous coordinates. From that, I observed the center of inellipse can be expressed as a weighted sum of vertices ($*5$) and the weights can be computed from the areas of various triangles associated with the pentagon.

In the answer below, I have thrown away the uninspiring algebra and proved the most "unexpected" relation ($*4$) by more elementary means (trigonometry).

• For any $P = (x_p,y_p), Q = (x_q,y_q)$ in the plane, let $$[PQ] \stackrel{def}{=} x_py_q - x_qy_p$$ Please note that $[PQ] = -[QP]$ and geometrically, it is twice the signed area of a triangle with vertices at $P,Q$ and origin $O$.
• For any $P, Q, R$ in the plane, let $$[PQR] \stackrel{def}{=} [PQ] + [QR] + [RQ]$$ Please note that $[PQR] = [QRP] = [RPQ]$ and geometrically, it is twice the signed area of a triangle with vertices at $P,Q,R$.

Let $A,B,C,D,E$ be the vertices of a convex pentagon (ordered counterclockwisely). Let $G$ be its center of mass and $K$ be the center of its in-ellipse.

Pick any point $P$ within the pentagon. Cut the pentagon at $P$ to form 5 triangles. We can compute $G$ as the area weighted average of the centroid of the $5$ triangles:

$$3G = \frac{\sum_{cyc} [ABP] (A+B+P)}{\sum_{cyc}[ABP]} \tag{*1}$$ RHS of $(*1)$ can be viewed as a rational function in coordinates of $P$. Since this is valid for all points within the pentagon, it is valid for all points in the plane. In particular, if we substitute $P$ by origin $O$ in $(*1)$, we obtain $$3G = \frac{\sum_{cyc}[AB](A+B)}{\sum_{cyc}[AB]}\tag{*2}$$ Substitute $P$ by $K$ in $(*1)$ and combine with $(*2)$, we can express $K$ as a weighted sum of vertices:

$$K = \frac{\sum_{cyc}[AB](A+B)}{\sum_{cyc}[AB]} - \frac{\sum_{cyc} [ABK] (A+B)}{\sum_{cyc}[ABK]}\tag{*3}$$

It turns out the ratios among $[ABK], [BCK], \ldots$ can be computed from the areas of various triangles associated with the pentagon. More precisely, let

$$ \lambda_A = [ABE][ACD],\;\; \lambda_B = [BCA][BDE], \ldots $$ (definition of $\lambda_C,\lambda_D,\lambda_E$ can be derived from that of $\lambda_A$ by cyclic permuting the vertices). We have

$$\frac{[ABK]}{\lambda_A+\lambda_B} = \frac{[BCK]}{\lambda_B+\lambda_C} = \frac{[CDK]}{\lambda_C+\lambda_D} = \frac{[DEK]}{\lambda_D+\lambda_E} = \frac{[EAK]}{\lambda_E+\lambda_A}\tag{*4}$$

Combine these with $(*3)$, we obtain a formula we seek.

$$ \bbox[border: 1px solid blue;padding: 1em;]{ K = \frac{\sum_{cyc}[AB](A+B)}{\sum_{cyc}[AB]} - \frac{\sum_{cyc} (\lambda_A + \lambda_B) (A+B)}{2\sum_{cyc}\lambda_A}}\tag{*5}$$

What's remain is to justify $(*4)$. Since under affine transform, the ratio of areas remains invariant. It suffices to verify $(*4)$ when the inellipse is the unit circle centered at origin and $A$ lies on the +ve $x$-axis.

Let $A_1,B_1,C_1,D_1,E_1$ be points on the inellipse touching the pentagon edges $CD$, $DE$, $EA$, $AB$, $BC$ respectively. Let $$2\alpha = \angle C_1OD_1, 2\beta = \angle D_1OE_1, 2\gamma = \angle E_1OA_1, 2\delta = \angle A_1OB_1, 2\epsilon = \angle B_1OC_1$$

It is not hard to see the vertices are located at $$ \begin{align} A &= \frac{1}{\cos\alpha}(1, 0)\\ B &= \frac{1}{\cos\beta}(\cos(\alpha+\beta),\sin(\alpha+\beta))\\ C &= \frac{1}{\cos\gamma}(\cos(\alpha+2\beta+\gamma),\sin(\alpha+2\beta+\gamma))\\ D &= \frac{1}{\cos\delta}(\cos(\alpha+2\epsilon+\delta),-\sin(\alpha+2\epsilon+\delta))\\ E &= \frac{1}{\cos\epsilon}(\cos(\alpha+\epsilon),-\sin(\alpha+\epsilon)) \end{align}$$ With this, it is not hard to verify

$$[AB] = \tan\alpha + \tan\beta, [BC] = \tan\beta + \tan\gamma,\ldots$$

Furthermore, $$ \begin{align}[ABE] &= (\tan\alpha + \tan\beta)(\tan\alpha + \tan\epsilon)\sin(2\alpha)\\ &= 2\frac{\sin(\alpha+\beta)\sin(\alpha+\epsilon)}{\cos\alpha\cos\beta\cos\epsilon}\sin\alpha\\ [ACD] &= (\tan\gamma + \tan\delta)\left(1 - \frac{\cos(\alpha+2\beta+2\gamma)}{\cos\alpha}\right)\\ &= 2\frac{\sin(\gamma+\delta)}{\cos\alpha\cos\gamma\cos\delta} (\sin(\alpha+\beta+\gamma)\sin(\beta+\gamma)\\ &= 2\frac{\sin(\gamma+\delta)\sin(\delta+\epsilon)\sin(\beta+\gamma)}{\cos\alpha\cos\gamma\cos\delta} \end{align}$$ This leads to $$\lambda_A = [ABE][ACD] = 4\tan\alpha \prod_{cyc}\frac{\sin(\alpha+\beta)}{\cos\alpha}$$ There are similar formulas for $\lambda_B,\lambda_C$ and at the end, we get

$$\frac{\lambda_A + \lambda_B}{[AB]} = \frac{\lambda_A + \lambda_B}{\tan\alpha + \tan\beta} = 4\prod_{cyc}\frac{\sin(\alpha+\beta)}{\cos\alpha}$$ Notice RHS is invariant under cyclic permutation of vertices. This justifies $(*4)$ in this special case and hence in general.