Reflection on the IMO 2021

I talk about out my experience during the competition and my thoughts while solving the problems. This was the 62nd IMO held online from Saint Petersburg.

Olympiad mathematics - IMO

Reflection on the IMO

November 12, 2021

This year's IMO was weird.. Sadly it was online due to Covid.

I learned one lesson through it. To solve most* IMO P1/P4 questions I should ask myself two questions: What is enough to finish? What would be helpful if it were true?

I will be discussing only P1,P4 and P5 since those are the ones I tried the most on the competition.

Some statistics

Problem No. Mean Points Number of full solves (7/7)
Problem 1 4.394 286
Problem 4 3.817 309
Problem 5 2.152 175

By the above data, it seems that this year's P1, P4, P5 weren't as easy as say 2019:

Problem No. Mean Points Number of full solves (7/7)
Problem 1 5.179 382
Problem 4 3.736 257
Problem 5 3.567 250
Even more striking are the mean and full solves on P1 and P4 in 2014 for example:
Problem No. Mean Points Number of full solves (7/7)
Problem 1 5.348 370
Problem 4 5.189 378

Problem 1

IMO 2021 P1

Let $n \geqslant 100$ be an integer. Ivan writes the numbers $n, n+1, \ldots, 2 n$ each on different cards. He then shuffles these $n+1$ cards, and divides them into two piles. Prove that at least one of the piles contains two cards such that the sum of their numbers is a perfect square.

Small discussion

My train of thought on the IMO was: induction? = quick no, since you can't guarantee anything will hold after moving from $n$ to $n+1$. The next thing I did was phrase it as a game:

Rephrased problem

Boris and Ivan play a game. Boris chooses a positive integer $n$. Ivan writes the numbers $n, n+1, \ldots, 2 n$ each on different cards. Ivan wins if he can split these $n+1$ cards into two piles such that in each pile there aren't two cards that sum to a perfect square. For which $n$ does Boris have a winning strategy?

Phrasing combinatorial questions as games is very useful for me! After I phrased it like this, it immediately came to me: When couldn't Ivan achieve his goal? The most logical way of trying to achieve his goal is to start greedy ordering the cards in each pile in such a way that he doesn't end up with two cards summing up to a perfect square.

When couldn't he do this? If for example he had to place a card with value $a$ in one of the two piles but there already were two cards with value $b$ in the first pile and $c$ in the second pile such that $a+b$ and $a+c$ are both perfect squares.

What if Ivan couldn't fix this situation? A fix would be to place $c$ with $b$ into the first pile, but this wouldn't work if $b+c$ is also a perfect square. If Ivan had this situation in his $n+1$ numbers, he couldn't win.

ok now what?

Well I need to see how often such a triple occurs. I attempted to solve the system $a+b=x^2$, $b+c=y^2$, $c+a=z^2$. What I mean by "solved" is: $b=x^2-a$ and $c=z^2-a$ so $b+c=(x^2-a)+(z^2-a)=y^2$ so $a=\frac{z^2+x^2-y^2}{2}$ and similarily: $(a,b,c)=(\frac{z^2+x^2-y^2}{2},\frac{x^2+y^2-z^2}{2},\frac{y^2+z^2-x^2}{2})$. Now I realised that if I pick $(x,y,z)$ I would get a corresponding triple of $(a,b,c)$ so I picked consecutive $x,y,z$ since those would lead to the mutually "closest" $a,b,c$.

Now I got $(a,b,c)=(\frac{x^2-4x}{2},\frac{x^2+2}{2},\frac{x^2+4x}{2})$ which means $x=2k$ (even) and then I got the triple $(a,b,c)=(2k^2-4k,2k^2+1,2k^2+4k)$. So if for some $k$ all three of these numbers belong in my $n+1$ numbers I'm done. Proving such numbers exist for $n\geqslant 100$ is straightforward. I struggled getting the bound as tight as the problem desired. I got $n \geq 106$ in the first 1.5h of the competition but couldn't manage to figure out if my calculations were faulty or I was missing something. "Surely", I thought, "this is the way to solve the problem so I must have missed something while solving the system of equations..". Turns out I could have just found examples for $n=100,\ldots,106$ which would have been much easier than trying to fix the quadratics. Lost one point because of this. Luckily, it wasn't a deciding point for the bronze medal.

Problem 4

IMO 2021 P4

Let $\Gamma$ be a circle with centre $I$, and $A B C D$ a convex quadrilateral such that each of the segments $A B, B C, C D$ and $D A$ is tangent to $\Gamma$. Let $\Omega$ be the circumcircle of the triangle $A I C$. The extension of $B A$ beyond $A$ meets $\Omega$ at $X$, and the extension of $B C$ beyond $C$ meets $\Omega$ at $Z$. The extensions of $A D$ and $C D$ beyond $D$ meet $\Omega$ at $Y$ and $T$, respectively. Prove that

\[A D+D T+T X+X A=C D+D Y+Y Z+Z C.\]

For some reason, this took me over 3 hours to solve. I think I just got scared of the weird problem statement. It also didn't help that my diagram led me to believe it was harder than it actually was (I trusted my diagram too much and convinced myself that $XT \neq YZ$ since it was off by 7 milimeters in my diagram. Never draw diagrams with pens, always use pencils/graphite with narrow tip).

First I reduced it to a sum of three lines by tangent segments: $AD=AS+SD=AP+DR$ similarily $CD=DS+CQ$ thus it was enough to prove $PX+XT+TR = QZ + ZY+ YS$.

At first I didn't think this was really useful. After playing with angles for a long time I realised $\angle TIX = \angle YIZ$ which actually does imply $XT = YZ$ and I knew I was done once I got this.

Something trivial I didn't notice at first was that since $AI$ is the exterior angle bisector of $\angle XAY$ we must have $I$ is the midarc of $XY$ with $A$ hence $XI=YI$ and similarily $TI=ZI$ so we must have $TZ||XY$ and hence $TX=YZ$ (by symmetry). This was much faster than the angle chase I figured out.

Now we're almost done since $\triangle YSI \cong \triangle XPI$ and so $YS=XP$ similarily $ZQ=TR$ and hence the conclusion!

Problem 5

IMO 2021 P5

Two squirrels, Bushy and Jumpy, have collected 2021 walnuts for the winter. Jumpy numbers the walnuts from 1 through 2021, and digs 2021 little holes in a circular pattern in the ground around their favourite tree. The next morning Jumpy notices that Bushy had placed one walnut into each hole, but had paid no attention to the numbering. Unhappy, Jumpy decides to reorder the walnuts by performing a sequence of 2021 moves. In the $k$-th move, Jumpy swaps the positions of the two walnuts adjacent to walnut $k$.

Prove that there exists a value of $k$ such that, on the $k$-th move, Jumpy swaps some walnuts $a$ and $b$ such that $a < k < b$.

So since P4 was not that easy for me, did not have enough time to fiddle enough with this problem which is sad. Now, nearly 6 months later, I sat down and solved it. It took me about 2.5 hours, which is not usually the case for me with P2/P5 problems on the IMO.