problem 6 of Indian Mathematical Olympiad 1989

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problem 6 of Indian Mathematical Olympiad 1989

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perpendicular bisector

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Source: Plane Geometry (Incircle of a triangle)

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1581 posts

makar

#1 h Sep 1, 2009, 3:32 PM • 2 Y

Y by Adventure10, Mango247

Triangle $ABC$ has incentre $I$ and the incircle touches $BC, CA$ at $D, E$ respectively. Let $BI$ meet $DE$ at $G$ . Show that $AG$ is perpendicular to $BG$ .

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843 posts

shoki

#2 h Sep 1, 2009, 6:40 PM • 1 Y

Y by Adventure10

see http://www.mathlinks.ro/viewtopic.php?t=298621

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599 posts

#3 h Sep 1, 2009, 7:04 PM • 1 Y

Y by Adventure10

I was also pretty sure that the innitial question was wrong....
Alternative.
Let $ED$ intersect $AB$ at $K$ .
The $(A,L,B,K)$ (note that $AD,CL,BE$ are concurent at the georgonne point) is a harmonic 4-tumple.( $L$ is the intersection of the incircle with the $AB$ ).
By well known lemma in order to prove that $AG$ is perpendicular to $BG$ it suffices to prove that $GB$ is bisector of $\angle LGK$ ,which can be proved by an easy angle chasing

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39 posts

#4 h Sep 2, 2009, 3:21 AM • 2 Y

Y by Adventure10, Mango247

$AG$ perpendicular to $BG$ .(not $BC$ )

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saif

#5 h Sep 2, 2009, 3:24 PM • 2 Y

Y by Adventure10, Mango247

we need to prove that $IGEA$ is cyclic
it's easy because
$\angle EGI=\angle B/2 + \angle EDB=\angle B/2 +\angle B/2 +\angle A/2+ \angle C=180^{o}-\angle A/2$
so it's done

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1581 posts

makar

#6 h Sep 2, 2009, 4:29 PM • 2 Y

Y by Adventure10, Mango247

It is $AG$ perpendicular to $BG$
sorry for the error

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1581 posts

makar

#7 h Sep 11, 2009, 12:50 PM • 2 Y

Y by Adventure10, Mango247

is there any alternate solution possible?

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1896 posts

#8 h Sep 12, 2009, 6:48 PM • 2 Y

Y by Adventure10, Mango247

Let AG intersect BC at H.
Then $\frac {AB}{BH} = \frac {AG}{GH}$ by the angle bisector theorem.
By Menelaus, $\frac {DC}{DH} \cdot \frac {HG}{GA} \cdot \frac {AE}{EC} = 1$ . But $DC = EC$ , so $\frac {AG}{GH} = \frac {AE}{HD}$ .
So $\frac {AB}{BH} = \frac {AE}{HD} = \frac {AB - AE}{BH - HD} = \frac {AB - AE}{BD}=1$ , since $AB=AF+FB=AE+BD \implies AB-AE=BD$ . So $AB = BH$ , so $BG \perp AH$ .

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madhusudan kale

23 posts

madhusudan kale

#9 h Jan 2, 2011, 3:12 PM • 1 Y

Y by Adventure10

saif wrote:

we need to prove that $IGEA$ is cyclic
it's easy because
$\angle EGI=\angle B/2 + \angle EDB=\angle B/2 +\angle B/2 +\angle A/2+ \angle C=180^{o}-\angle A/2$
so it's done

also,
$\angle EGI=\angle B/2 + \angle EDB=\angle IDB+\angle IDE$
but $\angle IDB=90^{o}$ and $\angle IDE=\angle C/2$
thus $\angle EGI=\angle B/2 +90^{o}+\angle C/2$
which is equal to $180^{o}-\angle A/2.$
Done.

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9767 posts

jayme

#10 h Jan 2, 2011, 3:18 PM • 1 Y

Y by Adventure10

Dear Mathlinkers,
old problem but always fruitfull...
See,
http://perso.orange.fr/jl.ayme An unlikely concurrence, revisited and generalized vol. 4 p. 5
Sincerely
Jean-Louis

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607 posts

#11 h Nov 29, 2013, 10:03 AM • 1 Y

Y by Adventure10

Let incircle touch $AB$ at $F$ .just prove $FIGE$ is cyclic. $BG$ is perpendicular bisector of $FD$ ,so $FG=GD$ , $FI=ID$ and $IG$ is common.so $\angle IFG=\angle IDG$ ,also $\angle IDG=\angle IEG$ ,hence $FIGE$ is cyclic.

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geniusramanujan

26 posts

geniusramanujan

#12 h Jan 27, 2015, 10:17 AM • 1 Y

Y by Adventure10

we need to prove AGEI IS cyclic
IAB+ABI=AIG
A/2+B/2=(180-C)/2=DEC=AEG
THEREFORE AIG=AEG
AGEI IS CYCLIC
SO WE ARE DONE

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lifeismathematics

1188 posts

lifeismathematics

#13 h Nov 6, 2022, 10:47 AM • 2 Y

Y by Mango247, Mango247

goal is to prove AIGE cyclic

This post has been edited 1 time. Last edited by lifeismathematics, Nov 6, 2022, 10:47 AM

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