Secondary Special Camp 2011: Geometry P 2

[Moon]

Problem 2:The incircle of triangle ABC touches the sides $BC,CA,AB$ at $A',B',C'$ respectively. Let the midpoint of the arc $AB$ of the circumcircle (not containing $C$) be $C''$, and define $A''$ and $B''$ similarly. Prove that the lines $A'A'',B'B'', C'C''$ are concurrent.

[*Mahi*]

Image:

campgeo2.png (24.34KiB)Viewed 5452 times

[*Mahi*]

$\text {Solution:}$

By alternate segment theorem ,
$\angle C'A'B'=\angle AB'C' =\angle AC'B'= 90- \frac A 2$
$\angle B'C'A'=\angle CA'B'=\angle CB'A'= 90- \frac C 2$
$\angle A'B'C'=\angle BC'A'=\angle BA'C'= 90- \frac B 2$
Now $\angle AB"C"=\angle ACC" =\angle \frac C 2$
And $\angle AC"B" =\angle ABB"=\angle \frac B 2$
So $\angle C"AB"=90 + \frac A 2$
So,$\angle C"A"B"=180-90- \frac A 2=90+ \frac A 2 =\angle C'A'B'$
Similarly we can prove that $\triangle A"B"C" \sim \triangle A'B'C' $
Now let $A"C"$ and $BA$ intersect at $X$ and $A"C"$.
In triangle $A"BX$,
$\angle XBA"=B+ \frac A 2$
$\angle BA"X= \frac C 2$
So,$\angle BXA"=90- \frac B 2$
Which means $C"A"||C'A'$
In similar way we can prove that $B"A"||B'A'$ and $B"C"||B'C'$.
So we get $\triangle A'B'C'$ homothetical to $\triangle A"B"C"$ and their center of homothety is where $A'A",B'B",C'C"$ meet ,implying $A'A",B'B",C'C"$ must meet.

[Labib]

I have a solution but I don't know how to do such a sophisticated construction in my PC....
Could anyone help, please??

[*Mahi*]

You can use geogebra ,a easy-to-use program to draw figures,to draw them.
http://www.geogebra.org/cms/en/download

[Labib]

Thanks for your help mate...

[Labib]

Here's the solution.....

2.png (257.34KiB)Viewed 5428 times

In figure $2$, $D=$circumcentre of $\triangle ABC$

If we drop a perpendicular $A''M$ from $A''$ (midpoint of arc $BC$ of the circumcircle of $\triangle ABC$) on $BC$, it would be the perpendicular bisector of $BC$.

So it will be a line through $D$.

Therefore perpendiculars dropped from $B''$ and $C''$ on $CA$ and $AB$ respectively will also be lines through $D$.

Now in order to solve this problem, we need to do some construction.

If we draw perpendiculars $A_1B'' \perp B''D $, $B_1C'' \perp C''D $, $C_1A'' \perp A''D $, They will form a triangle $\triangle A_1B_1C_1$ and all of it's similar sides will be parallel to $\triangle ABC$'s similar sides.
This triangle has the incircle $A''B''C''$.

So the triangles $\triangle ABC$ and $\triangle A_1B_1C_1$ are homothetic.

So pedal triangles inscribed in these two triagle's incircle must be homothetic.
Thus $\triangle A'B'C'$ and $\triangle A''B''C''$ are homothetic.

Therefore they have a point of homothety and so the lines $A'A''$, $B'B''$ and $C'C''$ meet at this point. [Proved]

[*Mahi*]

Please don't upload your high-res images, mobile users will be killed to see them.
When you export your image from geogebra, please select a low resolution option.

[Labib]

Sorry, I didn't know about that... :-S