IMO 2025 Livesolve (Day 1)

(Drew here.) As a retired contestant, I decided it would be fun to attempt the IMO 2025 paper and see how well I would do on it! Let's get started.

Problem 1

(IMO 2025/1) A line in the plane is called sunny if it is not parallel to any of the $x$–axis, the $y$–axis, or the line $x+y=0$. 

Let $n\ge3$ be a given integer. Determine all nonnegative integers $k$ such that there exist $n$ distinct lines in the plane satisfying both of the following:

for all positive integers $a$ and $b$ with $a+b\le n+1$, the point $(a,b)$ lies on at least one of the lines; and

exactly $k$ of the $n$ lines are sunny.

To begin, I decided to try the $n=3$ case, as that's the smallest one. The points form a right triangle with $3$ points on each edge, and a sunny line is a line not parallel to any of the edges of the main triangle.

We can shift the points a bit to instead form an equilateral triangle, and a sunny line is any line not parallel to any of the sides of the equilateral triangle.

Importantly, the side of the equilateral triangle has $n$ points: what happens if we have a line through these points?

The line through these points is not sunny, so the remaining $n-1$ points are still covered by $k$ sunny lines. This means if $(k,n)$ is a valid solution for $n\geq 4$, then $(k,n-1)$ is a valid solution, too.

The other case is that none of the $n$ lines is the longest side of the equilateral triangle. In this case, each of the $n$ points on the baseline must have exactly one line passing through it.

For this to hold for all $3$ sides seems difficult: a configuration of lines would want to be along the sides. But it's possible for $n=3$ by drawing the medians. Intuitively this makes sense, because the vertices of the equilateral triangle can "double up" and act as the point of intersection for both its edges.

But if $n\geq4$, there are too many other points that we need to pass through.

There are $2n-3$ other points on the perimeter, and if $n\geq4$, the inequality $2n-3>n$ holds. This means that, by pigeonhole, one of the $n$ lines passing through the baseline (yellow) must also pass through two of the remaining perimeter points.

But any line through two of these perimeter points would either pass through the vertex opposite the baseline (which means it's the longest side of the equilateral triangle), or pass through two points from opposite sides, in which case it intersects the yellow line outside the triangle (e.g. by Menelaus).

It remains to resolve the case $n=3$.

$k=0,1,3$ work by the constructions in the image, and $k=2$ fails by brute forcing all the possibilities.

Time taken: 30 minutes

Post-solve thoughts

Using the ideas I have here, there is a cleaner way to do the problem. In case 2, there are $3n-3$ points on the perimeter, and each line that's not the longest edge can only pass through $2$ points. So $3n-3\leq 2n$, which only holds for $n=3$, and for $n=3$ equality must hold.

Through a vertex of the triangle, the only line that passes through exactly two points is the median, which shows this construction is the only one in this case.

And this also aids in the brute force, since the second case doesn't admit a solution with $k=2$. And if we're in the first case, we can solve the problem for $n=2$ instead (even if the problem explicitly states $n\geq 3$!), which is trivial.

In fact, the solution presented here characterises all possible configurations. For the $n=3$ case, either we have $2$ parallel non-sunny lines passing through $3$ and $2$ points with an arbitrary line through the last point, or the $3$ medians. Then, add non-sunny lines to the outside to create a triangle that's $1$ length larger, all the way until $n$.

Overall, I enjoyed this problem; even if I have overcomplicated some of the details, the core of the problem is still obvious: $n$ collinear points must either have a line right through or $n$ different lines passing through, then induct down to a smaller value of $n$.

Problem 2

(IMO 2025/2) Let $\Omega$ and $\Gamma$ be circles with centres $M$ and $N$, respectively, such that the radius of $\Omega$ is less than the radius of $\Gamma$. Suppose $\Omega$ and $\Gamma$ intersect at two distinct points $A$ and $B$. Line $MN$ intersects $\Omega$ at $C$ and $\Gamma$ at $D$, so that $C, M, N, D$ lie on $MN$ in that order. Let $P$ be the circumcentre of triangle $ACD$. Line $AP$ meets $\Omega$ again at $E\neq A$ and meets $\Gamma$ again at $F\neq A$. Let $H$ be the orthocentre of triangle $PMN$.

Prove that the line through $H$ parallel to $AP$ is tangent to the circumcircle of triangle $BEF$.

AB isn't even perpendicular to CD...

Here's my original diagram for the problem. I don't really like the current formulation, it seems more natural to phrase in terms of triangle $ACD$, since $P$ is the circumcenter and $B$ is the reflection of $A$ over $CD$.

I don't have a compass/ruler with me so I just used the gridlines to help a little with the accuracy of the diagram. (I haven't done geo on paper in a long time, I don't even know where my compass and ruler are.)

5/2 * 5/3 is a calculation for the orthocenter.

But for the purposes of this blogpost, I'll work with a computer generated diagram instead.

Sketches plug again.

Evidently we have to handle $\triangle BEF$ and its circumcircle, so let's start with that. Since $ACEB$ is cyclic,  we can angle chase $$\angle BEF=\angle BEA=\angle BCA=2\angle DCA.$$ 

(I'm using directed angles here. $\angle XYZ$ is the counterclockwise rotation from line $XY$ to line $YZ$, taken mod $180^\circ$. To angle chase with this, I'm using the facts $\angle XYZ=\angle WYZ$ if $W,X,Y$ are collinear in no particular order, and $\angle XYZ=\angle XWZ$ if $W,X,Y,Z$ are collinear in no particular order. The last important fact is $\angle XYZ+\angle YZX+\angle ZXY=0$. I find that this makes angle chasing much less messy.)

So we know the angles of $\triangle BEF$ in terms of the angles of $\triangle ACD$, which is always welcome. Using this I want to create a useful similarity, so I artificially double the angles $\angle BCD$ and $\angle BDC$. If $J$ is the point such that $B$ is the incenter of $\triangle JCD$, then $\triangle JCD\sim\triangle BEF$. Similarly, if $I$ is the point such that $A$ is the incenter of $\triangle ICD$, then $\triangle ICD$ is also similar to those other two triangles.

($I$ is labelled in my handdrawn diagram, but I wrote down $J$ because $\triangle JCD$ is directly similar to $\triangle BEF$. This means that they have the same orientation on the plane, i.e. there is a spiral similarity / a composition of a rotation and a dilation that brings $\triangle JCD$ to $\triangle BEF$.

"Uh", I wrote on the paper. Finding the center of the spiral similarity didn't seem easy. But after staring at the diagram a while longer, I figured it would be nice if I had a similar triangle including $B$ instead of $J$. And indeed, $\triangle BMN$ is also similar to $\triangle BEF$ via one extra line of angle chasing.

I don't enjoy angle chasing.

There are three mysteries in the problem statement: what is $\triangle BEF$ (which we have just answered), what is the tangency point, and how is $H$ involved. Let's tackle the second mystery with the help of the spiral similarity.

Let $Q$ be the desired tangency point, assuming the problem statement is true. (This is not well defined at the moment, but we are just trying to find facts about $Q$ that we will prove later.) Since the tangent to $(BEF)$ at $Q$ is parallel to $EF$, $Q$ must be one of the arc midpoints of $EF$. The diagram suggests that it's the midpoint of the arc opposite $B$.

Since $Q$ lies on $(BEF)$, we can extend the similarity earlier: let $R$ be the point such that $BEFQ\sim BMNR$. This yields a triple of similar triangles: $\triangle BEM\sim\triangle BFN\sim\triangle BQR$. And since $BM=ME$, we should have $BR=RQ$.

I'm pretty close to identifying a better definition for $Q$ now. $\angle BQR=\angle BEM=\angle MBE$, which is the angle of rotation for the spiral similarity. Since the similarity brings $\triangle BMN$ to $\triangle BEF$, the angle of rotation is $\angle MBE$, but it's also the angle from $MN$ to $EF$.

Now it would be really nice if $QR$ was tangent to $(BEF)$, as that would make $QR$ parallel to $EF$. And since $\angle BQR$ is the angle of rotation from $BQ$ to $QR$, and is equal to the angle of rotation from $MN$ to $EF$, this would also mean that $BQ$ is parallel to $MN$.

At this point a lot of equivalent statements went through my head. Notably, $Q$ would have to be the intersection of $(BMN)$ and $(BEF)$ and $ME$ and $NF$. (I didn't notice any of these from my diagram, so it was quite a facepalm moment when I realised Q was the intersection after all.)

Some statements which should be true. I just need to find the order in which to prove them now.

Finally we can proceed. Let $Q=ME\cap NF$, which lies on $(BEF)$ and $(BMN)$ by the spiral similarity config (or just angle chase).

Now we prove $Q$ is the midpoint of arc $EF$, by angle chasing to show $\angle EFQ=\angle QEF$.

$$\begin{align*}\angle QEF&=\angle MEA\\&=\angle EAM\\&=\angle PAM\\&= \angle MCP\\&=\angle DCP\\&=\angle PDC\\&=\angle PDN\\&=\angle NAP\\&=\angle NAF\\&=\angle AFN\\&=\angle EFQ.\end{align*}$$

Phew, that was long. But we're almost done understanding $Q$. Bring back $R$, still defined in the same way; we get $R$ is the midpoint of arc $MN$, so $R$ lies ok the perpendicular bisector of $MN$. And $RB=RQ$ as well! So $BQ$ is another chord whose perpendicular bisector passes through $R$, proving $MN\parallel BQ$.

And by the angle chasing we had above, since $\angle BQR$ is the angle of rotation from $MN$ to $EF$, we have $QR\parallel EF$, and $QR$ is tangent to $(BEF)$ because of the spiral (or because $Q$ is the arc midpoint).

The third mystery is all that remains: why is $HQ$ parallel to $EF$? We've somehow not touched $H$ in this entire process. This is surprisingly not hard! $H$ is the incenter of $\triangle MNQ$—I wildly conjectured this even before getting the characterisation of $Q$, so I might have gotten lucky.

But it was motivated: $\triangle MNH\sim\triangle DCA$ by simple angle chasing. Since we made $A$ the incenter of $\triangle ICD$ earlier, it's reasonable to do it to $H$ too. But now we have enough information to show $H$ is the incenter, since we only need to double angles $\angle EMN$ and $\angle FNM$. I'll leave this to the reader as it is quite repetitive.

Finally, $QHMN\sim IADC$ by construction, so $QH\parallel IA$, and $P$ being the circumcenter of $ACD$ lies on $IA$. And we're finally done.

The final diagram.

Time taken: 70 minutes

Post-solve thoughts

I'm surprised I came up with this solution and this fast, which is pure angle chasing and similarities, given these are my weaknesses.

The solving process was draw tangency point → conjecture stuff about it → find the right order to prove the conjectures → finish, which is very common for tangency problems where the point is unknown.

Drawing the tangency point is pretty easy with an accurate diagram. If it doesn't visually satisfy the things that should be true, maybe draw it better, which I needed to.

To conjecture stuff about the point, an accurate diagram definitely helps. Without a ruler and compass I had to resort to the gridlines for an accurate position of $B$ and $P$. It's important to work backwards here, since you don't need to prove it yet. But if you want to claim something that's not proven, make sure it's consistent with everything else known first.

Finding the right order is usually the step that trips me up, because I run into cyclic arguments. This time, however, everything worked out conveniently, because I chose a good definition for $Q$. (Choosing $Q$ to be the midpoint of arc doesn't work even though that's what I wanted to start with, because there are two indistinguishable arc midpoints!)

Then in the finish, I admit I got lucky on knowing $MNQ$ has incenter $H$, but it's not hard to angle chase even without guessing.

My advice for this is to know which angles can be chased using the angles in the reference triangle, and calculate them when needed. I don't like it when my diagram is full of angles.

AoPS has enlightened me on some stuff. Inverting at $A$ helps a bit, $CEB$ and $DFB$ become collinear, $B$ and $P$ swap roles, etc. Which makes me even more surprised that I didn't do that (I didn't know what to do with $H$). And also there are shorter solutions that what I did, with much less angle chasing. Coord bashing works as well, which I contemplated a couple minutes in, but rejected because a synthetic solution would have more content.

$P$ is in fact the $A$-excenter of $\triangle AMN$, using this information we can delete $C$ and $D$ and take $\triangle AMN$ as the reference triangle instead.

Question for reader: What if we did the same with $P'$ being the incenter of $\triangle AMN$? What happens to $H'$, the orthocenter of $\triangle P'MN$?

Problem 3

(IMO 2025/3) Let $\mathbb{N}$ denote the set of positive integers. A function $f\colon\mathbb{N}\to\mathbb{N}$ is said to be bonza if $$f(a)~~\text{divides}~~b^a-f(b)^{f(a)}$$ for all positive integers $a$ and $b$.
Determine the smallest real constant $c$ such that $f(n)\leqslant cn$ for all bonza functions $f$ and all positive integers $n$.

After solving the first two problems in 100 minutes, I was ready for the last problem. I've done well in the recent divisibility FEs that I've tried, so let's see how this one goes.

Let $P(a,b)$ denote the assertion that $f(a)\mid b^a-f(b)^{f(a)}$. $P(a,a)$ quickly resolves to $f(a)\mid a^a$. This gives $f(1)=1$ (which doesn't help when substituted anywhere else) and $f(2)\in\{1,2,4\}$. I checked the largest value of $f(2)$ as shown in the following image:

$f(a)\mid a^a$ is rather restrictive when $a$ is a prime power. $f(p)\mid p^p$ so $f(p)=p^k$ for some $0\leq k\leq p$.

$P(p^k,b)$ gives a divisibility for $p^k$. Unfortunately flt does not give such a strong statement about $p^k$ dividing stuff, so the best I could do is $p\mid b-f(b)$ for all $b$ as long as $k>0$. But that's still huge information especially if many primes combine.

One more thing to note is that if the problem statement is true ($c$ exists), then every prime $p>c$ satisfies $f(p)\in\{1,p\}$.

 Needing to rewrite flt is a sign that my NT has rusted.

$p\mid b-f(b)$ if $f(p)\neq1$ makes $f(q)=1$ impossible if $p>q$ (both primes). So once $f(q)=1$ for a prime $q$, the same holds for all larger primes.

At this point I experimented with other uses of $f(p)=1$. I soon came to the realisation that if I can pin down $f(3)=1$, then $f(\text{any odd prime})=1$.

Indeed, this works out easily as in the image: if $f(3)\neq1$ then we get a contradiction by choosing a large enough $2$ mod $3$ prime. So $f(p)=1$ for all odd primes $p$.

Something even stronger can be shown. If an odd prime $p$ divides $f(n)$, then $P(n,p)$ shows $p\mid f(n)\mid p^n-1^{f(n)}$, which is a contradiction. So $f(n)$ is a power of $2$ that divides $n^n$, in particular it equals $1$ for $n$ odd.

After this observation, if I sub $P(a,b)$ where $a$ is odd, I don't get any information. If I do so with $b$ odd and $a$ even, I get $f(a)\mid b^a-1$ for all odd $b$. This is related to the Carmichael function, but I don't remember that anymore. Instead for a couple values of $a$, I set $b=3$ and got a bound on $f(a)$.

Similar analysis shows that $f(2^n)\mid 2^{n+2}$ for $n\geq2$ because $\nu_2(3^{2^n}-1)=n+2$ by factoring (or LTE, but I forgot about even that). And the exact same logic applies to $f(n)$ if $n$ is even, showing that $f(n)\leq 4\times 2^{\nu_2(n)}\leq 4n$, so $c=4$ works in this case.

Finally I decided to conjecture that the following function $f$ works:

And it can be checked to work by verifying these cases:

This forces $c\geq 4$, and therefore resolves the case where there exists a prime $p$ such that $f(p)=1$.

In the other case, $p\mid f(p)$ for all primes $p$, and so...

This case is nonexistent?!

Yeah $f(n)=n$ is the only solution in the other case. Wow. It's done.

Time taken: 60 minutes

Post-solve thoughts

That was a very underwhelming problem because of how I finished it. I probably should have done that case first...

The difficulty in the problem lies in... where? I think it's in me being rusty at NT.

The first part about limiting the values of $f(a)$ then choosing $a$ to be prime is very standard; $f(p)$ finding in NT FEs is common. "NT is about primes", I quote.

Then the choice to focus on $f(3)=1$ was deliberate, because it's both easier to resolve and not any weaker than the fact that $f(p)=1$ for all odd primes $p$ (which I conjectured to be true).

Finally, the construction of $f$ evaded me for a while because I did not remember LTE, but after that calculation was done, I chose $f$ to be the largest possible value for each input, and it just worked!

(Additional note: $f(n)=2^{\nu_2(n)+2}$ for even $n$ is NOT a solution because $f(2)=8$ does not divide $2^2$. Verifying FE solutions is not something that should be done solely in fear of losing marks, but also because of possible mishaps like this.)

Apparently the function with $f(4)=16$ and the rest of the function equal to $1$ if odd and $2$ on every other even works, and ignores most of the LTE stuff I did.

So I believe this problem could be made harder by adding "sufficiently large" in multiple places. With the same setup but "for every $a,b>20^{25}$" and "$f(n)\leqslant cn$ for all sufficiently large $n$" being used in the definition of a bonza function, what's the new minimum value of $c$? I think $c=4$ is still the best, and proving it requires a little more care or insight to handle.

(P.S. after writing this, I saw that Evan Chen has classified all bonza functions for the original problem!)

Post Paper Thoughts

Well, that was easier than expected for an IMO paper. I felt like I could have gotten lucky at times, but given the short total time, I figured it would not be too hard for people to solve all three. And I'm confirmed by the statistics - over a hundred people scored a 7 on problem 3, and over a sixth of the contestants scored at least 6, an absurd number for what's supposed to be the hardest problem of the day.

This livesolve is supposed to be a reflection of my thoughts while doing the problem and ideally should showcase some dead ends, but somehow I never encountered any large difficulty where I had to backtrack, so it might seem just like a solution walkthrough...

Day 2 coming soon™️ with harder problems :)