An Inequality on Circumscribed Quadrilateral


An Inequality on Circumscribed Quadrilateral, problem

Solution 1

Assume the incircle is defined by $x^2+y^2=1.\,$ Consider the tangency points of $AB,\,$ $BC,\,$ $CD,\,$ $DA\,$ with the circle as $M(1,0),\,$ $N(\cos 2t,\sin 2t),\,$ $S(\cos 2u, \sin 2u),\,$ and $Q(\cos 2v,\sin 2v),\,$ respectively, with $0\lt t\lt u\lt v\lt\pi\,$ and $t,u-t,v-u\lt\displaystyle\frac{\pi}{2},\,$ $v\lt\displaystyle\frac{\pi}{2}.\,$ It has been shown many times before that $\displaystyle A\left(1,\frac{\sin t}{\cos t}\right),\,$ $\displaystyle B\left(\frac{\cos (u+t)}{\cos (u-t)},\frac{\sin (u+t)}{\cos (u-t)}\right),\,$ $\displaystyle C\left(\frac{\cos (v+u)}{\cos (v-u)},\frac{\sin (v+u)}{\cos (v-u)}\right),\,$ $D\displaystyle\left(1,\frac{\sin v}{\cos v}\right).\,$ From here we deduce that

$\begin{align} s &= \tan t+\tan (u-t)+\tan (v-u)-\tan v\\ &= \tan t+\tan (u-t)+\tan (v-u)+\tan (\pi-v). \end{align}$

Since $y=\tan x\,$ is strictly convex on $\left(0,\displaystyle\frac{\pi}{2}\right),\,$ Jensen's inequality yields

$\displaystyle \begin{align} &\tan t+\tan (u-t)+\tan (v-u)+\tan (\pi-v)\\ &\qquad\qquad\qquad\ge 4\tan\frac{t+(u-t)+(v-u)+(\pi -v)}{4}\\ &\qquad\qquad\qquad=4\tan\frac{\pi}{4}=4, \end{align}$

with equality iff $t=u-t=v-u=\pi -v=\frac{\pi}{4},\,$ i.e. when $ABCD\,$ is a square.

Solution 2

We rely on the diagram below to introduce and explain the notations.

An Inequality on Circumscribed Quadrilateral, proof

From the diagram, $\alpha+\beta+\gamma+\delta=\pi\,$ and

$s=R(\tan\alpha + \tan\beta +\tan\gamma +\tan\delta).$

As in Solution 1, we use Jensen's inequality:

$\displaystyle s\ge 4R\tan\frac{\alpha+\beta+\gamma+\delta}{4}=4R\tan\frac{\pi}{4}=4R,$

with equality when $\alpha=\beta=\gamma=\delta,\,$ i.e., when $ABCD\,$ is a square.


Let $A_1,A_2,\ldots,A_n\,$ be a tangential polygon with inradius $R\,$ and semiperimeter $s.\,$ Show that

$s\ge Rn\cot\displaystyle\frac{(n-2)\pi}{2n}.$

When does the equality hold?

Let $M_k\,$ be the projections of the incenter $I,\,$ onto $A_kA_{k+1},\,$ $k=1,2,\ldots,n,\,$ $A_{k+1}\equiv A_1.\,$ Clearly, $A_kM_k=R\cot\displaystyle\frac{A_k}{2}\,$ so that $s=R\displaystyle\sum_{k=1}^n\cot\frac{A_k}{2}.$

But the cotangent function is strictly convex on $\displaystyle\left(0,\frac{\pi}{2}\right),\,$ implying, by Jensen's inequality,

$\displaystyle\begin{align} R\sum_{k=1}^{n}\cot\frac{A_k}{2}&\ge Rn\cot\left(\frac{1}{n}\cdot\sum_{k=1}^{n}\frac{A_k}{2}\right)\\ &=Rn\cot\frac{(n-2)\pi}{2n}, \end{align}$

with equality only if $A_k=\displaystyle\frac{(n-2)\pi}{n},\,$ for all $k,\,$ and, from here, iff $A_kM_k=R\displaystyle\cot\frac{(n-2)\pi}{2n},\,$ for all $k=1,2,\ldots,n,\,$ i.e., when the polygon $A_1,A_2,\ldots,A_n\,$ is regular.

We mention in passing that the function $\displaystyle y=x\cot\frac{(x-2)\pi}{2x}\,$ is strictly decreasing for $x\ge 4,\,$ while $y(4)=4.$


In his proof of a curious result by Igor Sharygin, Leo Giugiuc made use of a theorem that in multiple place on the web is simply referred to as Ivanova (1976), without actual reference to the source. As a response to my inquiry, Leo had simply sent me a proof (Solution 1). Solution 2 is along the same lines but more of a geometric than a trigonometric nature. The source reference is still missing. Meanwhile, Leo came up with the above generalization.


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