a product of distances inequality of circumscribed polygon

Question

I learned the following theorem from Roger Johnson's book Advanced Euclidean Geometry §101g:

Theorem: If a polygon is inscribed in a circle, and A is a point of the circle, the product of the distances from A to the sides of the polygon, equals the product of the distances from A to the tangents of the circle at the vertices of the polygon.

Is the following true:
If the point A is inside the circle, the first product is always less than the second product, regardless of the convexity of the polygon.
The inequality also holds for concave polygons (Coloring is produced by Geometer's Sketchpad: colour the point $A$ with purple if the first product is less than the second product)

Answer 1

Your inequality is true.

Let $\mathcal{C}$ be any circle of radius $r$. For any $P, Q \in \mathcal{C}$, let

• $\ell_P$ and $\ell_Q$ be the tangent lines of $\mathcal{C}$ at $P$ and $Q$,
• $\ell_{PQ}$ be the line passing through $P$ and $Q$.

Choose a Cartesian coordinate system for the Euclidean plane so that $\mathcal{C}$ is centered at origin with $P, Q$ position symmetrically respect to the $x$-axis. i.e. for some suitably chosen $\theta \in (0,\pi)$, $P, Q$ are positioned at $(r\cos\theta, r\sin\theta)$ and $(r\cos\theta,-r\sin\theta)$.

In this coordinate system, the equations for the lines are

$$\begin{array}{rc} \ell_P :& \cos\theta x + \sin\theta y - r = 0\\ \ell_Q :& \cos\theta x - \sin\theta y - r = 0\\ \ell_{PQ} :& x - r\cos\theta = 0 \end{array}$$

Let $\mathcal{D}$ be the open disk bounded by $\mathcal{C}$. If $A = (u,v) \in \mathcal{D}$, we will have $r^2 > u^2 + v^2$. The distances of $A$ to the lines will be given by the formulae:

$$\begin{align} d(A,\ell_P) &= r - ( \cos\theta u + \sin\theta v )\\ d(A,\ell_Q) &= r - ( \cos\theta u - \sin\theta v )\\ d(A,\ell_{PQ}) &= | u - r\cos\theta | \end{align}$$ With a little bit of algebra, we find $$\begin{align} d(A,\ell_P)d(A,\ell_Q) - d(A,\ell_{PQ})^2 &= (r - \cos\theta u)^2 - v^2\sin^2\theta - (u- r\cos\theta)^2\\ &= \sin^2\theta(r^2 - u^2-v^2)\\ &> 0\end{align}$$ As a result,

$$\sqrt{d(A,\ell_P)d(A,\ell_Q)} > d(A,\ell_{PQ})$$

Let $P_1,P_2,\ldots,P_n$ be any $n$ points on $\mathcal{C}$ and use $P_0$ as an alias of $P_n$. Substitute $(P,Q)$ by following $n$ pairs of points $(P_n,P_1), (P_1,P_2),\ldots,(P_{n-1},P_n)$, your inequality follows:

$$\prod_{k=1}^n d(A,\ell_{P_k}) = \prod_{k=1}^n \sqrt{d(A,\ell_{P_{k-1}})d(A,\ell_{P_k})} > \prod_{k=1}^n d(A,\ell_{P_{k-1}P_k})$$

If you replace $\mathcal{D}$ by the closed disk $\bar{\mathcal{D}}$, the $>$ in above inequality becomes $\ge$.