IMO 2026 Day 1 Livesolve

 

Glen here.

Unfortunately due to other commitments, I was unable to be at IMO this year, but I did have time today to attempt the problems from Day 1. I'll be writing about my thought process in solving these problems, but I did two of these (the non-geometries) in my head so certain ideas might have been forgotten. Also, I might be a little terse because I want to go to sleep.

Problem 1

(IMO 2026/1) There are $2026$ integers greater than $1$ written on a blackboard, not necessarily different. In a move, Confucius chooses two integers $m>1$ and $n>1$ from different places on the blackboard and replaces these two integers with

$$\gcd(m,n) \quad \text{ and } \quad  \frac{\mathrm{lcm}(m,n)}{\gcd(m,n)}.$$

He continues to make moves while it is possible to do so.

(a) Prove that, regardless of the choices of Confucius, after finitely many moves, exactly one integer $M$ on the blackboard is greater than $1$.

(b) Prove that the value of $M$ does not depend on the choices of Confucius.

• Ok, so monovariant. The product of those two things is the lcm, which is $\le mn$ with equality iff the gcd is $1$.

• If the gcd is $1$, we have created a $1$. Yay.

• So the monovariant we want is just product of everything minus number of $1$s.

• On to (b). You probably get the lcm of all the numbers or something? Does the lcm change? Oh yeah it probably does if the things aren't coprime.

• Let's try $4,8$. We get $2,4$...oh this is just the Euclidean algorithm on the exponents.

• Ok, so we should just be getting the product of $p^{\mathrm{gcd}(v_p(n_i))}$ over all primes. Just use the usual Euclidean algorithm proof that $\mathrm{gcd}(a,b) = \mathrm{gcd}(a,b-a)$ and that $\mathrm{gcd}(a_1,\ldots,a_n) = \mathrm{gcd}(\mathrm{gcd}(a_1,a_2),a_3,\ldots,a_n)$.

I think solving this took me less time than reading the question. There's been a trend of putting really easy problems as P1 in the IMO to make getting HMs easier so as to encourage students. That's a sentiment that I would agree with if it didn't effectively make the IMO a 5-question paper for students in medal range.

Problem 2

(IMO 2026/2) Let $ABC$ be a triangle and let points $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Let points $K$ and $L$ be chosen inside triangles $BMC$ and $BNC$, respectively, such that $K$ lies inside the angle $LBA$, $L$ lies inside the angle $ACK$, and 

$$\angle KBA= \angle ACL, \angle LBK= \angle LNC,  \angle LCK= \angle BMK$$ Let $O$ be the circumcentre of triangle $AKL$. Prove that $OM = ON$.

•  What is this undrawable diagram. Why are there so many clauses to avoid configuration issues.

• After some effort I finally come up with a diagram. (I'll probably scan this in when I have time.)

• The trick is that you can draw ray $BK$ and $CL$ first such that the first angle condition is satisfied, then choose $K$ and $L$ on each ray to make the other two angle conditions work.

• Surely the first thing to do is to simplify those angle conditions.

• Extend $BK$ to meet $AC$ at $X$ and $CL$ to meet $AB$ at $Y$. Then the first angle condition tells us that $B,C,X,Y$ concyclic.

• Aha, then midpoint theorem plus Reim tell us that $M,N,X,Y$ concyclic. Thank you Ker Yang.

• What about the other two conditions? How unique are $K,L$? How could we even construct them?

• Ok, $\angle MKX = \angle KCA$. This is some alternate segment theorem thing? Something like $MK$ being tangent to $(XKC)$. That doesn't really help us draw $K$...

• As we move along ray $BX$, $\angle MKX$ increases while $\angle KCA$ decreases, so it does seem like we have a unique $K$. How can we classify this?

• I see some Stewart's theorem attempt on my rough work. It looks wrong. I think I was trying to solve for the length of $BK$ and got a quadratic.

• Could we try inversion? Maybe inverflect to swap $B,N$ and $C,M$? Or invert to swap $B,Y$ and $C,X$? No, all the angle conditions become even more cursed.

• I stare at the diagram uselessly for a while.

• Wait, $\angle CNL = \angle XBL$ gives $B,L,N,X$ concyclic, and similarly $C,K,M,Y$ concyclic. Ok, this is a much nicer characterisation of $K,L$. We can just define them as $(CMY)\cap BX$ and $(BNX) \cap CY$. Well, not quite - we have to choose the intersection that lies inside the triangle...ew.

• The definitions of $K,L$ look pretty symmetric. Can we do some inverflection to swap them? I try some possibilities but they don't quite work out.

• Let's go back to the choosing an intersection thing. That's a little weird. If I had to bash this, I'd set $B,C,X,Y$ on a circle, solve for $A$ by intersecting lines and thus $M,N$, and then get $K,L$ by intersecting circles with lines. But the definitions of $K,L$ don't depend on each other, and so the roots should be able to be chosen independently. This means that changing either $K$ or $L$ to the other intersection wouldn't change the question...does this mean that $A,K,L$ and the other two intersections $K',L'$ are all on the same circle?

• I draw a couple more diagrams and that seems plausible.

• How do I prove this? No idea.

• Let's try something else. Maybe midpoints means harmonics? I don't quite see where I can project stuff.

• I stare at the diagram some more.

• Oh, I can't believe I missed this. Let $Z=BX \cap CY$. Then $ZK\times ZK' = ZY\times ZC = ZX \times ZB = ZL \times ZL'$, so $K,L,K',L'$ concyclic. Ok, now I have faith that my guess was correct.

• How do I show that $A$ is on the circle as well??

• Let me list down the circles that I have so far: $(BCXY),(MNXY),(BLNX),(CKMY)$. $A$ is the radical centre of the last three. $Z$ is the radical centre of the first and the last two. Not sure how this helps. I guess $AZ$ is the radical axis of the last two.

• Ok, it seems like we know a lot about powers of points. Maybe we can try to get more stuff from those. Isn't the power of a point like linear or something? So maybe if we can get the power of the point of $Z,B,C$ with respect to $(KLK'L')$ then we can show that the power of the point $A$ is $0$.

• Well, these are $ZB\times ZX$, $BM\times BY$, $CN\times CX$ respectively. So now the power of the point $A$ is some linear combination of those? How do I express $A$ as a linear combination of $Z,B,C$?

• Wait, hold on. The power of the point isn't linear. What's linear is the difference of the powers of points with respect to two different circles.

• Ok, there's an obvious choice for the second circle; take $(BCXY)$ so that the $Z$ term cancels out. Let $F(P)$ be the power of the point $P$ to $(KLK'L')$ minus the power to $(BCXY)$. So $F(Z) = 0$, $F(B) = BM \times BY$, $F(C) = CN\times CX$. 

• Now how do we find $A$ in terms of $Z,B,C$? Ok, it might be easier to write $Z$ in terms of $A,B,C$. Aha, $Z$ is just $\left(1: \frac{AY}{YB} : \frac{AX}{XC}\right)$ in barycentric coordinates, so if we want $F(A) = -AY\times AB$ (minus the power of the point to $(BCXY)$) then we need to show that $$-AY\times AB + \frac{AY}{YB}(BM\times BY) + \frac{AX}{XC}(CN\times CX) = 0$$

• Indeed, this follows from rewriting $AY\times AB$ as $AY\times AM + AX\times AN$. (I got stuck a little here because I initially wrote $AY\times AB$ as $AY\times YB$. Oops.)

• Great, so $A$ actually is on the circle. 

• Ok, in theory this is now bashable: we can intersect the perpendicular bisectors of $BX$ and $XN$ to get the circumcentre of $(BXN)$, then drop a perpendicular to get the perpendicular bisector of $KK'$. Intersect this with the perpendicular bisector of $MN$ and check that it's symmetric or something.

• But surely I could just finish this synthetically.

• How do I show $OM = ON$? Ok, that's the same thing as $M,N$ having the same power with respect to the circle. Omg, we can use the same trick again.

• The power of $A$ with respect to $(AKLK'L')$ is $0$. The power of $B$ is $BK\times BK' = BM\times BY$. The power of $C$ is $CN\times CX$, similarly.

• Ah, so if we compare this to the powers of $A,B,C$ with respect to $(MNXY)$, we get exactly the result. Amazing.

I actually love this question. There's barely any angle chase; everything happens with lengths. You actually get mileage out of algebraic intuition. And the first half acts as a tutorial to the second half. Excellent.

Problem 3

(IMO 2026/3) Let $n$ be a positive integer. Liu Bang and Xiang Yu have a stick of length $1$ and want to divide it between themselves. Liu Bang marks at most $n$ points on the stick, and then Xiang Yu marks at most $n$ points on the stick. The marked points are distinct. Then, the stick is cut at all marked points, creating a number of pieces. Afterwards, they take turns claiming any unclaimed piece of the stick, with Liu Bang going first. Each player's goal is to maximise the total length of their own pieces. For each $n$, determine the largest value $c$ such that Liu Bang may guarantee a total length of at least $c$, regardless of Xiang Yu's play.

• Well, LB can always get at least $\frac12$. That's probably not the answer.

• Optimal play means they will just greedily take the biggest piece, so once all the cuts are made everything else is forced.

• Let's try $n=1$. Maybe something like $1:2$? I think that guarantees LB $\frac23$ regardless of what XY does. If LB cuts nearer to the middle, then XY can just not cut and get $>\frac13$. If he cuts further away from the middle, then XY can cut the bigger piece in half and again get $>\frac13$. Cool.

• Now, $n=2$. Something like $1:2:2$? It seems like LB can get $\frac35$ that way? If XY cuts both the bigger pieces then LB ends up with exactly $\frac35$. Cutting one of the big ones twice seems even worse. Conversely XY always gets at least $\frac25$ no matter what LB does. I can't remember what reason I had for this at the time but it was probably wrong.

• So in general we just do $1:2:\cdots:2$? That can't be right. What went wrong?

• Oh hold on, XY can just halve the smallest piece, which guarantees him $\frac12$, which is optimal for him. 

• Ok, so in general, if there are two pieces of the same size, then XY can halve everything else, which guarantees him $\frac12$. LB wants to cut everything into different sizes.

• Oh also, if LB doesn't use all his cuts, then LB can just halve everything and guarantee $\frac12$. So LB has to create $n+1$ differently-sized things.

• What other restrictions are there?

• Let's try $n=2$ again. Maybe LB can cut it in a $1:2:3$ ratio. That's three different lengths. Oh wait, but now XY can just cut the $3$ into a $1$ and a $2$ and then he guarantees $\frac12$.

• In general, if LB cuts the stick in the middle, then XY can just mirror the cuts and guarantee $\frac12$.

• But also, nothing is stopping us from rearranging the pieces. So if we can rearrange the pieces such that a cut is in the middle, then that's bad.

• More generally, if we can rearrange the pieces such that a cut is close to the middle, then XY can guarantee losing by at most that small amount.

• Can we try to interpolate this with the other thing about having distinct lengths? If we can find two collections of pieces whose sums of lengths are the same, then we can match up the cuts on both sides, and cut everything else into half, guaranteeing $\frac12$.

• More generally, if two collections of pieces have similar total lengths, say differing by $d$, then by putting these two side by side and mirroring cuts and halving everything else, XY guarantees losing by at most $d$.

• There's some technicality along the lines of what if one cut is beyond the length of the shorter thing. But then we could just remove the final piece of the longer thing and get a lower $d$.

• So now we want all possible sums of lengths to be as far away from each other as possible. Probably something something binary.

• Let's try $1:2:4$ for $n=2$. After some case whacking I am convinced that LB guarantees a $\frac47$.

• I think this method gives an upper bound? There are $2^{n+1}$ possible subsets, so XY guarantees losing by at most $\frac1{2^{n+1}-1}$ by the argument above. So the winning margin goes to $0$ as $n$ gets large. Interesting.

• Ok, I am convinced that this is the correct bound. The $1:2:4$ case felt super tight; there were so many different ways for XY to end up with $\frac37$. 

• So now we probably want to show that if LB does $1:2:\cdots:2^n$ then he can win by at least $\frac1{2^{n+1}-1}$. Ok, let's scale everything up so the lengths are actually $1,2,\ldots,2^n$.

• I feel like I should actually try to prove the $1:2:4$ case properly. LB needs to get at least $4$. If XY doesn't cut the $4$ then he wins. If he cuts the $4$ into $2,2$ then LB ends up with $4$. Actually, even if he cuts the $4$ into like $2.5,1.5$ then LB still ends up with $4$. And if he additionally cuts the $2$ the result is still the same. Why are there so many equality cases...

• Maybe we can try to perturb everything to a "nice" equality case?

• There's probably some continuity + compactness thing that we can do to say that an equality case exists. (I think about it further and actually verify that, but I'm not going to type it out because it doesn't end up being useful.)

• Look at all the lengths of the pieces in the equality case. We probably want to perturb the cuts. This can always be done unless one of the lengths involved is tied with another length or something like that. So now we have a bunch of ties...now what?

• If XY cuts the biggest piece exactly once, then we're done by induction.

• But what happens if he cuts it a lot of times?

• Here's an annoying equality case in the $1:2:4:8$ case: XY can just cut the $8$ into $2,2,4$. But also this is the same as halving the $8$ and the $4$, so maybe we can somehow smooth the case where the biggest piece is cut into many pieces? Something like: perturb so that one piece is half the size, then swap everything else with whatever happens to the second-largest piece. But what if the second-largest piece is also cut into a bunch of pieces? Oh no...

• Maybe we can add XY's cuts one by one and see how the result changes? Ok, I think everything changes super wildly, so that's not very productive.

• Maybe in the case where XY uses all $n$ cuts we can show that LB's last free piece has length at least $1$? I guess if XY were only allowed to cut things into integer lengths that would be true. But we can't assume that; recall the $1:2:2$ case.

• Maybe we can try to smooth the size of the smallest piece? Like increase it to $1$ or decrease it to $0$? There are a bunch of cases...and I think we get stuck if e.g. the two smallest pieces are length $\frac12$ and both came from the $1$.

• There's also this annoyance about whether XY uses up all his cuts. Like somehow using $n$ and $n-1$ cuts feel a little different because in the former LB gets an extra free piece.

• Maybe we can consider them as the same thing by just adding extra pieces of length $0$. I guess this is the same as allowing cuts to coincide (which is what I did for the compactness argument earlier to work).

• Can we assume that XY made $n$ or $n-1$ cuts? If he used $n-2$ then I think he could have WLOG cut the smallest piece into thirds and gotten something better. Not sure if this helps.

• Maybe we can try to loosen LB's strategy in the final stage and find some way to pair stuff such that he always gets $\ge 2^n$, even if he's not choosing optimally. Don't see how to do this.

• Insert many failed attempts at smoothing and induction here.

• Let's rephrase this: we have some $a_1\ge a_2 \ge \ldots \ge a_{2n+2} =  0$ such that the $a_i$ can be grouped into sets with sums $1,2,\ldots,2^n$. We want to show that $a_1-a_2+a_3-a_4+\cdots \ge 1$.

• Random observation: this regrouping condition reminds me of SMO Open 2016/5.

• Now I'm picturing pairs of points on the number line which are all close together: we want $a_1,a_2$ to be close together, and then $a_3,a_4$, etc.

• Intuitively, it seems that $a_1,a_2$ should be roughly $2^{n-1}$, $a_3,a_4$ should be roughly half that, etc. But there's also some leeway here - see the many equality cases in the $1:2:4$ case.

• Maybe we can break the number line into intervals and count how many $a_i$ are in each interval? Then something something parity? But then things from neighbouring intervals could still be close to each other if they are at the boundary of the interval.

• Maybe there's just some direct inequality thing. I don't know.

• Let's look at $a_1$. It has to be at most $2^{n-1}+1$. (And equality is possible: e.g. in the $1:2:4$ case we could cut the $4$ into $1,3$.) If it's around that size then we must have $a_2$ coming from one of the largest two pieces. This then forces $a_3$ to be on the order of $2^{n-2}$, and maybe this sort of looks like we can do some sort of global bound?

• I want to say that $a_2 \le 2^{n-1}$, $a_4 \le 2^{n-2}$, etc. Then adding everything together, we get the sum of the $a_{2k}$ is at most $1+2+4+\ldots+2^{n-1}$, which is exactly what we want.

• But this breaks in the case where we cut the biggest pieces into a bunch of tiny little pieces, even if that looks extremely suboptimal.

• But if we have done that then $a_2,a_4$ and the like are even further from the equality case, which seems bad.

• Maybe something like $a_2+a_4+\cdots+a_{2k}$?

• This feels a bit weird because if we set $k=n+1$ then we just get the inequality we're trying to prove.

• Oh, we can do some sort of extremal thing. So pick the smallest $k$ for which $a_2+a_4+\cdots+a_{2k} > 2^{n-1} + 2^{n-2} + \cdots + 2^{n-k}$. Then we must have $a_{2k} > 2^{n-k}$. Now all the $a_1,\ldots,a_{2k}$ can only have come from pieces of size $2^{n-k+1}$ or larger, which have total size $2^n + \cdots + 2^{n-k+1} < 2(a_2+\cdots + a_{2k}) \le a_1 + \cdots + a_{2k}$. (Doing this in my head took up all my RAM.) Contradiction. Yay!

• So now setting $k=n+1$, we're done. Phew.

This is more of an inequality question than a combinatorics problem. The inequality part of this took me far longer than any of the combinatorial ideas. Blech. I am slightly disappointed.

Okay, I'm sleepy. Sorry if this got incoherent. Not sure when I'll be free to attempt Day 2 but hopefully that will be soon.