IMO 2011 problem 6 Geometry The is year's IMO problem 6 was a geometry problem that only 6 participants managed to solve completely. The problem is formulated like this:

Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $\ell$ be a tangent line to $\Gamma$, and let $\ell_a$, $\ell_b$ and $\ell_c$ be the lines obtained by reflecting $\ell$ in the lines $BC$, $CA$ and $AB$, respectively
Show that the circumcircle of the triangle determined by the lines $\ell_a$, $\ell_b$ and $\ell_c$ is tangent to the circle $\Gamma$.

A few solutions were found to this problem: using inversions, complex numbers, angle chasing, etc. My question is if we can reduce the problem to a simpler one in the following way:

Can we construct a triangle $\Delta$ for which $\Gamma$ is the incircle and $\Gamma_1$ is the 9 point circle? Of course, the answer should be yes if the circles are tangent and the radius of $\Gamma_1$ is greater than the radius of $\Gamma$. In this way we just apply a well known theorem of Feuerbach which says that the incircle and 9 point circle are tangent. How could we construct the triangle $\Delta$, starting from $ABC$?

This was my first idea when I saw the problem but didn't manage to finalize it.
 A: This link may be useful.There are quite a  few solutions there.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2365045&sid=0cdd97cc9547c2079a4ba23c56ba8f74#p2365045
In fact,this was the toughest problem at the IMO 2011.It was G8 on the Shortlist,meaning a hard problem. The IMO committee actually ended up misjudging the difficulty of the problems,as evident from the way they were numbered on the Shortlist.
A: Solution of problem 6 IMO 2011:
I use the method of analytic geometry.
Starting with the unit circle and 3 arbitrary points A,B C on its circumference, I found after laborious computations the equation of the second circumscribed circle.
It is possible to construct the equation which is the tangent-condition of the 2 circles.
Substitution completes the proof.
In the case of a isosceles triangle the centre M of the second circle describes a limacon of Pascal, which degenerates in a circle in the case of a equilateral triangle
Then that circle coincides the original unit circle
