Touching Circumcircles around Incentre [Self-Made]

For discussing Olympiad level Geometry Problems
User avatar
sowmitra
Posts:155
Joined:Tue Mar 20, 2012 12:55 am
Location:Mirpur, Dhaka, Bangladesh
Touching Circumcircles around Incentre [Self-Made]

Unread post by sowmitra » Sun Dec 22, 2013 6:11 pm

$I$ is the in-centre of $\triangle ABC$, and, $M,N$ the mid-points of the sides $AB,\, AC $.
$ BI \cap MN = K; \, CI \cap MN= L ; \, AL \cap BI= L_0 ; \, AK \cap CI = K_0$.
Prove that, the circumcircles of $\triangle ABL_0$ and $\triangle ACK_0$ touch each other.
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

User avatar
sowmitra
Posts:155
Joined:Tue Mar 20, 2012 12:55 am
Location:Mirpur, Dhaka, Bangladesh

Re: Touching Circumcircles around Incentre [Self-Made]

Unread post by sowmitra » Wed Feb 12, 2014 2:58 pm

Inspired from a JBMO problem.
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

Re: Touching Circumcircles around Incentre [Self-Made]

Unread post by *Mahi* » Wed Feb 12, 2014 6:38 pm

Hint:
Prove $\angle ALC = \angle AKB = 90^\circ$
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

photon
Posts:186
Joined:Sat Feb 05, 2011 3:39 pm
Location:dhaka
Contact:

Re: Touching Circumcircles around Incentre [Self-Made]

Unread post by photon » Wed Feb 12, 2014 8:08 pm

$\angle MKB=\angle KBC= \frac{1}{2}\angle B=\angle MBK $ , so $MB=MK$ . As in $\Delta AKB$ , $MA=MK=MB$ , $M$ is the center of the circumcircle of $\Delta AKB$ . M is the midpoint of AB , so it is a right-angle triangle - $\angle AKB=90^o$ . Similarly , $\angle ALC=90^o$ .
$\angle AKI+\angle ALI=180^o$ . hence quad $AKIL$ is cyclic .
Now , $\angle IAK = \angle ILK=\angle LCB = \frac{1}{2}\angle C=\angle K_0CA$ ,
$\angle IAL = \angle IKL=\angle KBC = \frac{1}{2}\angle B=\angle L_0BA$
by alternate segment theorem , $IA$ is tangent to the circumcircles of $\Delta AK_0C , \Delta AL_0B$ . As both have a common tangent to same point , they are tangent to each other .
Try not to become a man of success but rather to become a man of value.-Albert Einstein

User avatar
sowmitra
Posts:155
Joined:Tue Mar 20, 2012 12:55 am
Location:Mirpur, Dhaka, Bangladesh

Re: Touching Circumcircles around Incentre [Self-Made]

Unread post by sowmitra » Thu Feb 13, 2014 6:30 pm

Nice solution :)
It was inspired from JBMO 1997 #3.
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

Post Reply