# Points on Incircle: Another Look

### Problem

In usual notations in $\Delta ABC,$ for $P\;$ on the incircle

$\displaystyle 5r\le\frac{PA^2}{h_a}+\frac{PB^2}{h_b}+\frac{PC^2}{h_c}\le\frac{5}{2}R.$

### Hint

In usual notations in $\Delta ABC,$ for $P\;$ on the incircle

$\displaystyle\frac{PA^2}{bc}+\frac{PB^2}{ca}+\frac{PC^2}{ab}=\frac{2R+r}{2R}.$

Demonstration of the Hint

\displaystyle\begin{align} \frac{PA^2}{bc}+\frac{PB^2}{ca}+\frac{PC^2}{ab}&=\frac{1}{abc}(aPA^2+bPB^2+cPC^2)\\ &=\frac{1}{4SR}(2pr^2+4SR)\\ &=\frac{1}{4SR}(2Sr+4SR)\\ &=\frac{r+2R}{2R}. \end{align}

### From the Hint to the Problem

The identity in the Hint is equivalent to

$\displaystyle\frac{1}{2SR}(aPA^2+bPB^2+cPC^2)=\frac{r+2R}{2R}$

so that $\displaystyle\frac{1}{2S}(aPA^2+bPB^2+cPC^2)=\frac{r+2R}{2}.$

It follows that

\displaystyle\begin{align}\frac{PA^2}{h_a}+\frac{PB^2}{h_b}+\frac{PC^2}{h_c}&=\frac{1}{2S}\left(aPA^2+bPB^2+cPC^2\right)\\ &=\frac{2R+r}{2}. \end{align}

Due to Euler's formula, $R\ge 2r,\;$ $\displaystyle 5r\le\frac{r+2R}{2}\le\frac{5}{4}R.$

### Acknowledgment

The problem, due to Nguen Viet Hung (Vietnam), has been posted to the CutTheKnotMath facebook page by Dan Sitaru (Romania), with a graphical hint by Kevin Soto Palacios (Peru).

This is problem JP.018 (Junior problem #18) from the Romanian Mathematical Magazine.