# a product of distances inequality of circumscribed polygon

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)

• What is your exact question? Commented Jul 13, 2020 at 16:18

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 $$\ge$$.

• You can't base a proof on arbitrary coordinates. Commented Jul 22, 2020 at 12:31
• Euclidian geometry is affine geometry. Euclidian space is an affine space. Your proof could be correct, but it's not right. And this is not so clear because you translated it into a non affine space with some additional structure, . In euclidian geometry for example, there is not preffered origin. Commented Jul 23, 2020 at 8:33
• did you delete your comment? Commented Jul 24, 2020 at 7:50