An Inequality with Circumradii And Distances to the Vertices

Problem

An Inequality with Circumradii And Distances to the Vertices - problem

Solution 1

Denote $MA=x,\,$ $MB=y,\,$ $MC=z;\,$ $\angle BMC=2\alpha,\,$ $\angle AMC=2\beta,\,$ $\angle AMB=2\gamma.\,$ Clearly, $0\lt\alpha,\beta,\gamma\lt\displaystyle \frac{\pi}{2}\,$ and $\alpha +\beta +\gamma=\pi.$

From the Law of Cosines in $\Delta MBC,\,$ $BC=\sqrt{y^2-2yz\cos 2\alpha+z^2}.\,$ By the Law of Sines in $\Delta MBC,\,$ $\displaystyle R_a=\frac{\sqrt{y^2-2yz\cos 2\alpha+z^2}}{2\sin 2\alpha}.\,$ As $y^2+z^2\ge 2yz,\,$

$\displaystyle \begin{align} R_a &=\frac{\sqrt{y^2-2yz\cos 2\alpha+z^2}}{2\sin 2\alpha}\\ &\ge\frac{\sqrt{2yz-2yz\cos 2\alpha}}{2\sin 2\alpha}\\ &=\frac{2\sqrt{yz}\sin\alpha}{2\sin 2\alpha}=\frac{\sqrt{yz}}{2\cos\alpha}, \end{align}$

such that $\displaystyle \frac{1}{R_a}\le\frac{2\cos\alpha}{\sqrt{yz}}.\,$ We have

$\displaystyle \frac{MB\cdot MC}{R_a}=\frac{yz}{R_a}\le\frac{2yz\cos\alpha}{\sqrt{yz}}=2\sqrt{yz}\cos\alpha.$

Similarly,

$\displaystyle \frac{MC\cdot MA}{R_b}\le 2\sqrt{zx}\cos\beta\\ \displaystyle \frac{MA\cdot MB}{R_c}\le 2\sqrt{xy}\cos\gamma.$

Thus,

$\displaystyle \frac{MB\cdot MC}{R_a}+\frac{MC\cdot MA}{R_b}+\frac{MA\cdot MB}{R_c}\\ \qquad\qquad\qquad\qquad\le 2\sqrt{yz}\cos\alpha+2\sqrt{zx}\cos\beta+2\sqrt{xy}\cos\gamma.$

Suffice it to show that

$2\sqrt{yz}\cos\alpha+2\sqrt{zx}\cos\beta+2\sqrt{xy}\cos\gamma\le x+y+z.$

But, according to the famous Wolselholme's inequality,

For real $x,y,z\,$ and $\alpha+\beta+\gamma=\pi,$

$yz\cos\alpha +zx\cos\beta +xy\cos\gamma\le x^2+y^2+z^2.$

Replacing here $x,y,z\,$ with $\sqrt{x},\sqrt{y},\sqrt{z}\,$ completes the proof.

Solution 2

Based on the following configuration and the formula $abc=4RS,$

An Inequality with Circumradii And Distances to the Vertices - Dan Sitaru's proof

$\displaystyle\begin{align} \frac{MB\cdot MC}{R_a}&=\frac{MB\cdot MC}{\displaystyle MB\cdot MC\cdot BC\cdot\frac{1}{4[\Delta MBC]}}\\ &=4\frac{[\Delta MBC]}{BC}\\ &=2 MD, \end{align}$

such that, by similarity,

$\displaystyle\begin{align} &\frac{MB\cdot MC}{R_a}+\frac{MC\cdot MA}{R_b}+\frac{MA\cdot MB}{R_c}\\ &\qquad\qquad\qquad\qquad =2(MD+ME+MF)\le MA+MB+MC, \end{align}$

by the Erdös-Mordell inequality.

Acknowledgment

The problem due to Leo Giugiuc and Kadir Altintas has been posted by Leo Giugiuc at the CutTheKnotMath facebook page, with their solution (Solution 1) communicated privately. Solution 2 is by Dan Sitaru.

 

|Contact| |Up| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62378432

Search by google: