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.

  • $\begingroup$ 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 Jul 23 '11 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. $\endgroup$ – Beni Bogosel Aug 5 '11 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 Aug 5 '11 at 12:56

This link may be useful.There are quite a few solutions there.


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.

  • 1
    $\begingroup$ 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. Jul 31 '11 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. $\endgroup$ – Beni Bogosel Aug 5 '11 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 Aug 26 '11 at 15:22

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.