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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.

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  • $\begingroup$ 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. $\endgroup$
    – user9413
    Commented Jul 23, 2011 at 20:06
  • $\begingroup$ @chandru: Take a look at the results and see for yourself: imo-official.org/… . There are way more people who solved problem 2. $\endgroup$ Commented Aug 5, 2011 at 11:59
  • $\begingroup$ 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 . $\endgroup$
    – user9413
    Commented Aug 5, 2011 at 12:56

2 Answers 2

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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.

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    $\begingroup$ I don't see an answer to the question there. Did you have some special answer in mind, or did you just link to a list of solutions? Also, could you explain what G8 means? To me this is the group of eight :) $\endgroup$
    – t.b.
    Commented Jul 31, 2011 at 8:04
  • $\begingroup$ If you look at post number 18 in the link, you'll find a related idea, but there is no proof to it. $\endgroup$ Commented Aug 5, 2011 at 9:08
  • $\begingroup$ I really apologise for not being clear. IMO problems are shortlisted as Algebra(A),Geometry(G),Combinatorics(C) and Number Theory(NT). There are 6-8 problems per topic.So, a Geometry problem may be designated as G1,G2..,G8(depending on how many problems are shortlisted).In this case ,problem 6 was designated as G8(Geometry 8) meaning a super hard problem(As the number increases, so does the level of difficulty) $\endgroup$
    – Eisen
    Commented Aug 26, 2011 at 15:22
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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

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