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Once upon a time on the IMO


About the author: Tigran Sloyan is the founder of MathFights. He graduated from MIT with degrees in Mathematics and Computer Science and later worked at Oracle and Google.

In this episode of A Problem and I, I would like to walk you through my thought process and my emotions while solving the IMO 2006 problem number 3. Why this problem in particular? Because it is supposedly quite difficult, it is solvable without any advanced olympiad math knowledge, and because it had a big impact on my professional success.

The Problem

Determine the least real number M such that the inequality

\mid ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2) \mid \leq M(a^2+b^2+c^2)^2

holds for all real numbers a,b and c .

Before we talk about the solution let me give you some context.

The Context

As most of you probably know, IMO is conducted over two days and participants get 3 questions over 4 hours and 30 minutes each day, where \#1 is generally the easiest one and \#3 is generally the hardest one.

I first saw this question during the contest itself on July 12, 2006 and was somewhat happy and sad at the same time. I was happy because I was generally pretty good at solving inequality type questions. However, I was unhappy that it was the \#3 which meant that it was most likely very difficulty. In any case, after quickly solving question \#1 , I skipped \#2 (it was combinatorics and I never shined at solving those) and started with this one.

The First Step

In general it's very common in inequality problems to have the equality case be at a=b=c , however it was obvious that that wasn't going to be the case in this problem. So my next thought was to set a=b and see what happens.

After you set a=b and substitute b with a , the left hand side becomes

\mid ac(a^2-c^2)+ca(c^2-a^2) \mid = 0

This meant that the right hand side is also 0 when a=c or b=c . Which led me to believe that the left hand side was divisible by (a-b)(b-c)(c-a) . And after opening the parentheses and some identity juggling I realized the following

\mid (a-b)(b-c)(c-a)(a+b+c) \mid = \mid ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2) \mid

Thus I had reduced the original inequality to the following one with some simple modifications (I later found out that for this step the contestants were given 1 point out of 7):

\mid (a-b)(b-c)(c-a)(a+b+c) \mid \leq M(a^2+b^2+c^2)^2

The Substitution

Since most of the standard inequalities that I knew and was thinking of using were only true for non-negative real numbers, I decided to use the symmetry of the inequality and as they say "without loss of generality" assume that a\geq b \geq c and x=a-b, y=b-c and z=a+b+c , where x,y,z are non-negative real numbers (note that z could potentially be negative but since changing a,b,c to -a,-b,-c doesn't change our inequality, we could assume that z is non-negative). Then solving for a,b and c I found that

a=\frac{2x+y+z}{3}, b= \frac{y+z-x}{3}, c =\frac{z-x-2y}{3}

And plugging these values in I ended up with the following.

 xy(x+y)z \leq M(\frac{(2x+y+z)^2+(y+z-x)^2+(z-x-2y)^2}{9})^2

Yeah, I know, my first thought was -- "this is worse than what we have started with....maybe doing the substitution was a bad idea...", but I still had hope that maybe opening the parenthesis on the right hand side would clear things up a little bit instead of making it worse. So I rolled up my sleeves and started opening the right hand side's parenthesis, and fortunately for me, it got much simpler thereafter. Here is what the above inequality reduces to after some modifications:

 xy(x+y)z \leq \frac{M}{9}(2x^2+2y^2+2yx+z^2)^2

The Breakthrough

This looked quite simple to me and I was sure that I could figure out a way to find M . But I was stuck at this point for a few moments considering which path to take next. I was seriously considering introducing differential over z here and looking for maxima and minima, but that was a very ugly path and I felt that there should be a simpler one.

The breakthrough came when I realized that since xy\leq \frac{(x+y)^2}{4} then xy(x+y)\leq \frac{(x+y)^3}{4} and 2x^2+2y^2+2xy = 2(x+y)^2-2xy \geq \frac{3}{2}(x+y)^2 . This means exactly that we can increase the left hand side of our inequality and decrease the right hand side by setting x=y=\frac{x+y}{2}=\frac{z_{0}}{2} . Thus finding M reduces to the question of finding M for which the following is true for any non-negative real z and z_0 .

z_{0}^3z \leq \frac{M}{9}(3z_{0}^2+2z^2)^2

The Last Push

At this point I again started to think about using calculus to finish this off, but I was sure that there must be a better way, and then it hit me.

The most well known inequality for any olympiad student is that the geometric mean of any non-negative real numbers is less than or equal to the arithmetic mean of the same numbers, i.e.


and for n=4 this would be


and plugging in x_1=x_2=x_3=z_0^2 and x_4=2z^2 we get that


which is equivalent to




I thought that this was a very strange answer for an IMO question because I was expecting something "nicer". So I had to make sure there was an equality case.

In the Arithmetic-Mean - Geometric-Mean (AM-GM) inequality, the equality case holds when x_1=x_2=x_3=x_4 , i.e. 2z^2=z_0^2=(2x)^2=(2y)^2 , or z^2 = 2x^2 = 2y^2 . Plugging these in into the original equations I found that one of the equality cases was at

 (a,b,c) = (2,2+3\sqrt{2},2-3\sqrt{2})

And after a few more checks of the steps I took, I was confident that this was correct.

The Aftermath

The best part of this story however, started after the 4.5 hours were up and I could go check my "weird" answer with somebody else. I knew the guy sitting two tables behind me -- he got a perfect score on IMO 2005 -- so once the time was up I walked straight up to him and asked for the answer to \#3 and his reply was - "I didn't solve it".

My breath stopped for a moment and I thought to myself -- "I definitely did something wrong...". And then I kept asking around, mostly targeting gold medalist geniuses from different countries that I knew from previous years, and their answer was always the same. I was almost sure that I must have done something wrong when I run into a guy from the Iranian team who told me that he got \frac{9\sqrt{2}}{32} for M . I was so depressed by then that it took me about 2 minutes to realize that that's the same answer :). And when I did, I started jumping around being absolutely sure that my "weird" answer was the correct one.

I later found out that only about 20 students out of the 600 participants solved this question(no wonder I couldn't find anyone to check my answer with). And solving this question helped me get my first ever silver medal on the IMO.

  • Daniel Liu

    Wow, that problem is really cool! I like the equality case!

    Great job on being one of only 20 in the world who solved the problem.

    • tigransloyan

      thanks! I was lucky though ;)

      • Negato

        No, that was skills! ;)

  • Jeff Chen

    Wow...One thing that I noticed is that you had a lot of gut feelings. You felt that there were simpler steps, that there was a good solution, etc....If we have those gut feelings, should we trust them or should we ignore them? This question highly relates to the CD round in MCs.

    • tigransloyan

      Thanks for the comment Jeff . In general it's a good idea to trust your gut, at least to give it a shot(but don't over do it -- especially if you are sitting in a competition, you don't want to waste your time chasing a gut feeling). But you also have to make sure that your gut feelings are more or less educated guesses. For example, the reason I felt that there must be a non-calculus solution to this is that 99% of IMO problems don't involve calculus in the solution. Or the reason that I decided to try AM-GM on this was that from seeing and solving a lot of AM-GM solvable inequalities, I could tell that this was kind of in the same ball park, etc. etc. What do you think? Have you had similar situations where you just knew it had to work out that way?

  • Daniel Hwang

    Tiks2pro. Awesome!