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India TST:Day2 Problem 1

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#1 h May 16, 2009, 9:04 AM • 2 Y

Y by Adventure10, Mango247

Let $\gamma$ be circumcircle of $\triangle ABC$ .Let $R_a$ be radius of circle touching $AB,AC$ & $\gamma$ internally.Define $R_b,R_c$ similarly.
Prove That $\frac {1}{aR_a} + \frac {1}{bR_b} + \frac {1}{cR_c} = \frac {s^2}{rabc}$ .

This post has been edited 1 time. Last edited by djmathman, Jan 13, 2020, 8:23 PM

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Ahiles

#2 h May 16, 2009, 12:31 PM • 2 Y

Y by Adventure10, Mango247

Let $B'$ and $C'$ b tangency points of $\gamma$ with $AB$ and $AC$ . It' s wellknown that $I$ is midpoint of $B'C'$ . Let $O_A$ be the center of $\gamma$ . Then

$IB' = AI \tan \dfrac{A}{2} = \dfrac{AI \sin \dfrac{A}{2} }{\cos \dfrac{A}{2}} = \dfrac{r }{\cos \dfrac{A}{2}}$

$R_a = \dfrac{IB'}{\cos \dfrac{A}{2}} = \dfrac{r }{\cos^2 \dfrac{A}{2}} = \dfrac{r}{\frac {p(p - a)}{bc}} = \dfrac{r \cdot bc}{p(p - a)}$

Then

$\sum \dfrac{1}{aR_a} = \sum\dfrac{p(p - a)}{r abc} = \dfrac{p}{rabc}\sum (p - a) = \dfrac{p^2}{rabc}$

Hm.... Am I right?

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#3 h Aug 30, 2010, 12:52 PM • 2 Y

Y by Adventure10, Mango247

Ahiles wrote:

It' s wellknown that $I$ is midpoint of $B'C'$

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=31739&start=0

The radius of the $A-$ mixtilinear incircle $R_a=r\sec^2\frac{A}{2}=\frac{r}{\frac{1+\cos A}{2}}=\frac{2r}{1+\cos A}$

So, $\frac{1}{aR_a}=\frac{1+\cos A}{2ra}$

and hence

${\sum\frac{1}{aR_a}=\frac{1}{2r}\sum\frac{1+\cos A}{a}=\frac{1}{2r}\sum\frac{1+\frac{b^2+c^2-a^2}{2bc}}{a}}$

$\Longrightarrow \sum\frac{1}{aR_a}=\frac{1}{4r}\sum\frac{(b+c)^2-a^2}{abc}=\frac{1}{4r}\sum\frac{4s(s-a)}{abc}$

$\Longrightarrow \sum\frac{1}{aR_a}=\frac{s}{rabc}\sum(s-a)=\frac{s^2}{rabc}$ as required.

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sayantanchakraborty

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sayantanchakraborty

#4 h Mar 10, 2014, 8:00 PM • 2 Y

Y by Adventure10 and 1 other user

Using the Casey theorem for A,B,C and circle internally tangent to arc BC(not containing A) we get
$AX=\frac{2bc}{a+b+c}$ and analogous expressions for others.
So $LHS=\frac{a+b+c}{2abc}(\sum\frac{1}{tan\frac{A}{2}})=\frac{a+b+c}{2abc}\frac{1}{tan\frac{A}{2} tan \frac{B}{2} tan \frac{C}{2}}=\frac{r^2}{sabc}$ .

Please point out if I am wrong anywhere....

Bye..

Sayantan....

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aops29

#5 h Jan 13, 2020, 8:06 PM • 4 Y

Y by Pluto1708, djmathman, Adventure10, Mango247

The right hand side of the equation should be
$\frac{s^2}{rabc}$ Mods, please fix ~~I couldn't solve this because I was trying to prove the other way around...~~

Fixed, thanks! ~dj

This post has been edited 1 time. Last edited by djmathman, Jan 13, 2020, 8:23 PM

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