An Inequality for the Cevians through Circumcenter

Source

An Inequality for the Cevians through Circumcenter, source

Problem

In acute $\Delta ABC,\,$ $H\,$ is the orthocenter, $O\,$ the circumcenter; $AD,BE,CF\,$ are the altitudes, $AA_1,\,$ $BB_1,\,$ $CC_1\,$ the cevians through $O.\,$ Denote the sides of the orthic triangle $x=EF,\,$ $y=DF,\,$ $z=DE.\,$ Prove that

$\displaystyle \frac{A_1O}{OA}+\frac{B_1O}{OB}+\frac{C_1O}{OC}=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\ge\frac{3}{2}.$

Solution 1

Via van Aubel's theorem,

$\displaystyle\frac{R}{OA_1}=\frac{AO}{OA_1}=\frac{AC_1}{C_1B}+\frac{AB_1}{B_1C}.$

But $\displaystyle\frac{AC_1}{C_B}=\frac{[\Delta AOC]}{[\Delta BOC]}=\frac{\sin 2B}{\sin 2A}\,$ and, similarly, $\displaystyle\frac{AB_1}{B_1C}=\frac{\sin 2C}{\sin 2A}.\,$ It follows that

$\displaystyle\frac{OA_1}{AO}=\frac{\sin 2A}{\sin 2B+\sin 2C},\;\frac{OB_1}{BO}=\frac{\sin 2B}{\sin 2C+\sin 2A},\;\frac{OC_1}{CO}=\frac{\sin 2C}{\sin 2A+\sin 2B}.$

On the other hand, the nine-point circle is the circumcircle of the orthic triangle $DEF.\,$ Its radius is half of $R,\,$ the circumradius of $\Delta ABC\,$ and the angles at $D,E,F\,$ are $180^{\circ}-2A,\,$ $180^{\circ}-2B,\,$ and $180^{\circ}-2C,\,$ respectively. Such that, by the Law of Sines in $\Delta DEF,\,$

$\displaystyle\frac{x}{\sin (180^{\circ}-2A)}=\frac{y}{\sin (180^{\circ}-2B)}=\frac{z}{\sin (180^{\circ}-2C)}=2\frac{R}{2},$

i.e.,

$\displaystyle\frac{x}{\sin (2A)}=\frac{y}{\sin (2B)}=\frac{z}{\sin (2C)}=R.$

It follows that,

$\displaystyle\sum_{cycl}\frac{A_1O}{OA}=\sum_{cycl}\frac{\sin 2A}{\sin 2B+\sin 2C}=\sum_{cycl}\frac{x}{y+z}\ge\frac{3}{2},$

by Nesbitt's inequality.

Solution 2

$\displaystyle\begin{align}\frac{OA_1}{AA_1}&=\frac{[\Delta BOC]}{[\Delta ABC]}=\frac{\frac{1}{2}R^2\sin 2A}{\frac{1}{2}R^2(\sin 2A+\sin 2B+\sin 2C)}\\ &=\frac{\sin 2A}{\sin 2A+\sin 2B+\sin 2C}. \end{align}$

It follows that

$\displaystyle\begin{align}\frac{OA_1}{OA}&=\frac{OA_1}{AA_1-OA_1}=\frac{\sin 2A}{\sin 2A+\sin 2B+\sin 2C-\sin 2A}\\ &=\frac{\sin 2A}{\sin 2b+\sin 2C}. \end{align}$

The proof continues along the lines of Solution 1.

Acknowledgment

Leo Giugiuc has kindly posted the above problem at the CutTheKnotMath facebook page, along with a solution. The problem has been shared by Miguel Ochoa Sanchez at the Peru Geometrico facebook group. Solution 2 is by Marian Dinca.

 

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