# 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.

• Hmmm... I had a talk with my younger brother today, and he said that the $\mathsf{second \: problem}$ was the toughest and very few managed to solve it completely. – user9413 Jul 23 '11 at 20:06
• @chandru: Take a look at the results and see for yourself: imo-official.org/… . There are way more people who solved problem 2. – Beni Bogosel Aug 5 '11 at 11:59
• Dear Beni, I am not arguing with you. I just pointed out that, the 2nd problem was also quite tough. That's all. Lets not make a big issue of it . – user9413 Aug 5 '11 at 12:56