EGMO 2025 ??% Speedrun

(It's Glen again.)

Last week, I wrote about my experience test-solving EGMO 2025/3. As it turned out, I had some free time at EGMO (the night before Day 1, i.e. after we made the mark scheme for Problem 3, and the morning of Day 1 itself) to try the other problems, so I figured I'd write about them as well. I will try to be a little more brief than last week, so the post doesn't turn into a saga.

Problem 1

(EGMO 2025/1) For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$ for all $1 \leqslant i \leqslant m-1$

Disclaimer: this was a pretty embarrassing case of wrong reasoning leading to the right answer. Try to spot where I went wrong! (Hint: there are many such places.)

• Let's try small cases: $N=3,4$ work. $N=5$ fails. $N=6$ works.
• If $N$ is odd, then the sequence starts with $N=1,2,\ldots$, so $3|N$.
• $N=3^k$ works: any two consecutive $c_i$ are $1\pmod3$ and $2\pmod3$ in some order.
• If $N$ works, then any power of $N$ works: the only extra case to check is $N-1,N+1$, whose sum is clearly not coprime with $N$.
• Actually, by the same reasoning, if $N$ works and $p|N$, then $pN$ works.
• So, any $2^a 3^b$ works. Since $10$ works, then any $2^a 5^b$ where $a>0$ works.
• Suppose $N$ even, and $3,5$ don't divide $N$. Then the sequence starts with $1,3,5,7$ or $9$. In the former case, $5+7$ is coprime with $N$, which is bad.
• So $7|N$. Try $N=14$: this works. In general, $2p$ works, since in the middle case we have $p-2,p+2$ which add up to $2p$ which is a multiple of $p$.
• Here's a guess: any power of $2$ or $3$, and any $2^a p^b$, $p\ge 3$ prime, $a\ge 1$.
• Let's suppose $3,5|N$, then the sequence starts with $1,2,4,7,\ldots$, so $11|N$.
• Try $N=165$: we have $1,2,4,7,13,17,19,23,29,\ldots$ this fails because $23+29=52$ is coprime to $165$.
• Oh, actually that was $N=330$ because I forgot all the even numbers. Remove the $2$.
• But here's an idea: note that $23,29$ skips over $25,27$, which have the primes $3,5$ between them. Maybe we could generalise this?
• If $N$ even and has more than one odd prime factor, we want to pick two consecutive odds $n,n+2$ which, between them, contain all the prime factors of $N$. Then, $n-2,n+4$ should work. Surely we can just apply the Chinese Remainder Theorem.
• If $N$ odd, then we can just do the same thing, but with two consecutive numbers $n,n+1$.
• Let's try to make this rigorous: partition the odd prime factors of $N$ into two nonempty sets, and let the products of these sets be $P,Q$. By Chinese Remainder Theorem, there's some $1\le n\le N$ such that $n \equiv 0 \pmod{P}$ and $n \equiv -1 \pmod{Q}$. Clearly $n$ can't be $1$ (or we'd fail the first condition) or $n-2$ or $n-1$ (or we'd fail the second condition). So $n-1,n+2$ are both coprime to $N$ (being $\le 2$ away from a multiple of $p$ for all primes $p|N$), and their sum is $2n+1$ which is $1$ away from $2n$ and $2(n+1)$, and so coprime to $N$.
• In the even case, the same thing probably works.
• Wait, no it doesn't. I forgot that odd plus odd is always even, so all even numbers work. Oops.
• Final answer: all even numbers $>2$ and all powers of $3$.

Really, this is a lesson in not trying to go too fast. I like this problem: it has a cool solution set, and isn't too trivial (I certainly found it harder than, say, IMO 2023/1). Approximate time taken: 15-20 minutes.

Problem 2

(EGMO 2025/2) An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$. 

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots, $ there are infinitely many positive integers $n$ with $a_n = b_n$.

• Rewriting the condition: $\frac{a_1+\cdots + a_{a_n}}{a_n} = a_n$.
• Put $n=1$: $a_1 \le \frac{a_1+\cdots + a_{a_1}}{a_1} = a_1$, with equality iff $a_1=1$. That's nice.
• Put $n=2$. If $a_2=2$, then the LHS is $<a_2$.
• If $a_2=3$, then $a_3=5$. The next two terms are $7,9$ or $6,10$.
• Can $a_2=4$? I guess we can make $a_3=5,a_4=6$? It feels like there should be some bound here.
• Using the strictly increasing property, $a_2 = \frac{a_1+\cdots+a_{a_2}}{a_2} \ge \frac{a_1 + a_2 + (a_2+1) + \cdots + (a_2+a_2-2)}{a_2} = \frac{1 + (a_2-1)\left(\frac{3a_2-2}{2}\right)}{a_2}$ which rearranges to $a_2^2 - 5a_2 + 4 \le 0$, so we indeed have $a_2=3$ or $4$.
• In the $a_2=4$ case, we must start with $1,4,5,6,9,11$ and then the next three add to $45$ and then the next two add to $40$.
• This looks like it really wants to be the odd integers.
• Actually, if we had started from $1,3,5,7,9,11$ instead, the possibilities from here on out would be the same.
• Guess: $(b_n)$ is the sequence $1,3,5,7,\ldots$.

I am stuck for a while here, so I try Problem 4 for a bit before coming back.

• Let's rearrange the condition: $a_1 + \cdots + a_{a_n} = a_n^2$. I'm not sure why I didn't do that sooner; that looks a lot better.
• Replace $n$ with $n+1$, and subtract the previous equation to get $a_{a_n+1} + \cdots + a_{a_{n+1}} = a_{n+1}^2 - a_n^2$. Not sure how that might be useful.
• Ok, regardless, there are infinitely many $k$ such that $a_1+\cdots+a_k = k^2$, so resembling the odd numbers is reasonable.
• Aha, if $a_{n+1} = a_n+1$, then the difference of square thing above gives $a_{a_n+1} = 2a_n+1$, which is what we wanted.
• So we're left with the case where that holds for finitely many $n$, i.e. it's not true for sufficiently large $n$.
• This probably fails some sort of bound? Consider the last time this happens. Wait, we can't necessarily guarantee it happens at least once, but I guess we can fix it by setting $a_0=0$ and saying that it happens at $n=0$. Then, our sequence is $a_1,\ldots,a_{a_n}$, which add to $a_n^2$. This is followed by $a_{a_n+1} = 2a_n+1$ and this is $<a_{a_{n+1}}$ by our assumption.
• Oh, but now our sequence goes up by at least two each time, so it's $a_1,\ldots,a_{a_n},2a_n+1,\ge 2a_n+3,\ge 2a_n+5,\ldots$ so for each $N>a_n+1$, we have $a_1+\cdots+a_N \ge N^2$, with equality iff all the $\ge$ above are actually $=$ up to the $N^\text{th}$ term. Since this holds for $N$ of arbitrary size, then we must have equality at all points, so this sequence eventually agrees with $(b_n)$.

That was a cool dichotomy. You either have some nice condition which implies that $(a_n)$ agrees with $(b_n)$ at infinitely many points, or failing that, $(a_n)$ is eventually the same as $(b_n)$. I like that. It's a pretty nice sequence problem, actually. Approximate time taken: 30-45 minutes.

Problem 3

Refer to last week's post. I'm claiming that my original bash solution plus the initial synthetic observations took the first 20-30 minutes, so the overall time taken for Day 1 is 1-1.5 hours.

That's gone better than my previous speedruns, thanks to the bashable geometry.

Problem 4

(EGMO 2025/4) Let $ABC$ be an acute triangle with incentre $I$ and $AB \neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \neq B$ and $Q \neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \parallel RB, AB \parallel QR, AC \parallel SP$, and $AP \parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R, S, T$, and $I$ are concyclic.

(That random extra point was constructed for stupid reasons that will become clear later.)

• Oh no, I probably have to angle-chase. From the parallel lines, $\angle STR = \angle PAQ = \angle A + \frac{180^\circ - \angle A}2 = 90^\circ + \frac{\angle A}2$.
• That's a nice angle. I think it's equal to the angle at $I$? $\angle CIB  = 180^\circ - \frac{\angle B}2 - \frac{\angle C}2 = 90^\circ + \frac{\angle A}2$. Cool.
• So we want $\angle BIR = \angle CIS$. It would be nice if $\triangle BIR \sim \triangle CIS$.
• But it looks like the external angles at $\angle R$ and $\angle S$ are different? One is $\frac{\angle B}2$ and the other is $\frac{\angle C}2$. Maybe we could try to make up the difference? (At this point I draw that angle labelled $\frac{\beta-\gamma}2$, which isn't even in the right place to make up this difference.)
• Whatever. I'm just going to bash this - it's a standard configuration for complex numbers, except I have forgotten how to do the setup. I think $A,B,C$ should be $-u^2,-v^2,-w^2$ and the midpoint arcs should be $uv,vw,uw$, but I'm not sure if I got the signs right. Sanity check: if this is the case, then the incentre is $uv+vw+uw$, and the difference between this and the midpoint of arc $BC$ not containing $A$ is $uv+uw = u(v+w)$. On the other hand, the difference between that and $A$ is $uv+vw+uw+u^2 = (u+v)(u+w)$. In either case, dividing by the conjugate gives $u^2vw$. Okay, this is probably correct.
• So we have $a=-u^2, b=-v^2, c=-w^2, p = uw, q = uv$, and so $s=uw-w^2+u^2$ and $r$ is similar.
• $IS$ is then $vw+uv+w^2-u^2 = (w+u)(v+w-u)$. When dividing by the corresponding expression for $IR$, the $v+w-u$ terms cancel and we have $\frac{w+u}{v+u}$.
• On the other hand, $\frac{AP}{AQ}$ is given by $\frac{u^2+uw}{u^2+uv} = \frac{u+w}{u+v}$, which is equal to the expression above. We wanted to show both of these had the same argument, but actually they're just equal, so we're done.

So I made a very silly mistake: I got my lines mixed up, because they were too parallel. This led me to think that $ABR$ and $BCS$ are straight lines, which should obviously not be the case. My two hypothesised similar triangles were in fact similar, and it was in fact not hard to show this. Moral of the story: draw a good diagram! By the way, note that the complex bash in fact shows that $\triangle RIS \sim \triangle BAP$, this is cool, I guess. Approximate time taken: 10-15 minutes.

Problem 5

(EGMO 2025/5) Let $n > 1$ be an integer. In a configuration of an $n \times n$ board, each of the $n^2$ cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate $90^{\circ}$ counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of $n$, the maximum number of good cells over all possible starting configurations.

I have just learnt from AoPS that the person who proposed this problem was Melek Güngör, who happened to also coordinate this problem at EGMO. That must have been an interesting experience.

• Why has Turbo reappeared.
• Can we even draw a loop? Well, if $n$ is odd then the answer is no, by the checkerboard colouring: Turbo must alternate between black and white squares but there are more of one colour than the other.
• We're left with the case where $n$ is even.
• If $n=2$, there's only one possible loop. Suppose the top left square is good. If we travelled anticlockwise, then all arrows point down. If we travelled clockwise, then the top left and bottom right squares must have right arrows and the other two have left arrows. By inspection, in either case there's only one good square.
• Is the answer just $\frac{n^2}4$? I guess this is always possible: pick our favourite cycle through all the squares (e.g. travel through the first row, then zigzag backwards through all the columns), and put arrows so that they're pointing in correct direction when we land on them. Then, anything that's $0 \pmod4$ from the original good square will also be a good square.
• Why are we not done? This seems pretty forced. I guess the only possibility that remains is that we can start on some other random square and trace out a different path. Is that possible?
• I try to draw some examples for $n=4$, but that doesn't really help.
• Wait, corners are somewhat forced: we can only exit them in two directions. This means that we can only land on a corner in half the possible turns (i.e. two possible remainders mod $4$). Each of these determines at most one cycle (possibly shifted by a multiple of $4$), so there are at most two possible valid cycles. I guess this gives us an upper bound of $\frac{n^2}2$.
• Suppose both of these cycles exist. Then, at each corner, one cycle goes anticlockwise and one cycle goes clockwise. Since cycles can't self-intersect, probably one cycle has to be clockwise at all corners and the other has to be anticlockwise at all corners.
• So I guess both cycles have opposite turning number? Maybe the arrows denote different framings of the unknot in a plane, so maybe some sort of knot invariant? Okay never mind I'm overthinking this.
• Let's look at the $n=2$ case again: anticlockwise has all arrows in the same direction while clockwise has arrows in two different directions. Maybe this generalises?
• Consider the anticlockwise case. Then, the first two squares on the first row must have arrows in the same direction. (And likewise for the other corners.)
• On the other hand, for the clockwise case, the first two squares on the first column must have arrows in the opposite direction from each other (and likewise for the other corners).
• So in the clockwise cycle, we enter the top left corner, go right, and then leave the board(!) which is bad, contradiction.

There turns out to be a subtlety in this question that most coordinators (myself included) missed: one needs to mention that the cycle has length divisible by $4$, so after completing the cycle, the arrows point in the same direction as the beginning. Why is this needed, you might ask. (And indeed, I did ask this when I was translating the Chinese scripts and the coordinators for this problem asked me whether this point was written down.) This is in fact needed for both the lower bound and the upper bound: for the lower bound, we need it so that when we start, say, $4$ spaces after the original good cell, we know that the loop won't randomly diverge from the original loop when we get back to the original square. On the other hand, for the upper bound, we need it in the step where we say that a cell and a direction determines at most one loop. As it turned out, (if I'm understanding the marking scheme correctly,) everyone who omitted this was capped at 5/7, which explains the score distribution for this problem (image modified from the EGMO website):

As it stands, my bullet-point solution is probably worth like 4/7 or 5/7, depending on whether my brief explanations are accepted. But then again, my solution in a contest would probably be better written. Approximate time taken: 15-20 minutes.

Problem 6

(EGMO 2025/6) In each cell of a $2025 \times 2025$ board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to $1$, and the sum of the numbers in each column is equal to $1$. Define $r_i$ to be the largest value in row $i$, and let $R = r_1 + r_2 + ... + r_{2025}$. Similarly, define $c_i$ to be the largest value in column $i$, and let $C = c_1 + c_2 + ... + c_{2025}$ What is the largest possible value of $\frac{R}C$?

• Another grid?!
• Small cases: $n=1$ and $n=2$ both have $\frac{R}C = 1$.
• For $n=3$ we can do something like $$\begin{matrix}\frac13 & \frac23 & 0 \\ \frac13 & \frac16 & \frac12 \\ \frac13 & \frac16 & \frac12 \end{matrix}$$ for $\frac{R}C= \frac{\frac53}{\frac32} = \frac{10}9$, ew. Matrices are hard to type, so the rest will be in words. Sorry.
• We could also put a $1$ in the top-left corner, but then we just get $\frac{R}C=1$.  
• For $n=4$, we could have two columns of $\frac14$s, then a $\frac12,\frac16,\frac16,\frac16$ and finally a $0,\frac13,\frac13,\frac13$ for $\frac{R}C = \frac98$.
• Alternatively, we could have a column of $\frac14$s, a $\frac34,\frac1{12},\frac1{12},\frac1{12}$ and two of $0,\frac13,\frac13,\frac13$ for $\frac{21}{20}$, which is worse.
• We could also put a $1$ in the top-left corner and reuse the $n=3$ construction to get $\frac{16}{15}$, which is also worse.
• Maybe, in general, we should have $n-2$ columns of $\frac1n$s, then a column with $\frac2n$ and stuff, and finally a column with $0$ followed by $\frac1{n-1}$s? That works out to $\frac{\frac2n + 1}{\frac2n + \frac1{n-1} + \frac{n-2}n} = 1 + \frac{n-2}{n^2}$, which doesn't look great.
• Oh, we could also distribute the last two columns more evenly: in the $n=4$ case, we can have two columns of $\frac14$s, one that's $\frac12,\frac12,0,0$ and one that's $0,0,\frac12,\frac12$. This gives $\frac43$, which beats everything above.
• More generally, we could have $n-2$ columns of $\frac1n$s, a column of $\frac2n$s and $0$s, and a column of $0$s and $\frac2n$s. This gives $\frac{2n}{n+2}$, which is at least increasing in $n$.
• But this only works for even $n$ and $2025$ is odd. We could instead have a column with $\frac{n-1}2$ $\frac2n$s, $\frac{n-1}2$ $0$s and one $\frac1n$ and in the next column, swap the $0$s and the $\frac2n$s. This gives $\frac{2n-1}{n+2}$.
• We could also just put a $1$ in the bottom right and fill the rest of the square as in the $n-1$ case. That gives $\frac{3n}{2n+2}$, which is significantly worse.
• We could also have one column with $\frac{n-1}2$ $\frac2n$s and the other $\frac{n+1}2$ squares equal; in the last column we have $\frac{n-1}2$ $0$s and $\frac{n+1}2$ $\frac2{n+1}$s. This evaluates to $\frac{(2n-1)(n+1)}{(n+3)n}$, which turns out to be slightly better than the other construction.
• This is way too gross to be the actual answer. We can probably do better.
• We're getting values that are asymptotically $2$, because our "big" cells are approximately twice the size of our "regular" cells. 
• How big can the "big" cells be? Well, they can't be too big because then you can't fit very many of them in one column, which is bad. There's probably a nice balance in the middle by taking the square root or something...also $2025$ is a square number, so that's a little suspicious.
• Let's try $n-\sqrt{n}$ columns of $\frac1n$, and in each of the remaining columns, we have $\sqrt{n}$ $\frac1{\sqrt{n}}$s and the rest $0$.

• This gives $\frac{\frac{n}{\sqrt{n}}}{\frac{n}{\sqrt{n}}+\frac{n-\sqrt{n}}{\sqrt{n}}} = \frac{\sqrt{n}}{2-\frac1{\sqrt{n}}}$, which looks a lot better than what we had previously (it goes to $\infty$ now). When $n=2025$, this is $\frac{45}{2-\frac1{45}} = \frac{2025}{89}$.
• This is way too nice to not be the right construction.
• Let's try to bound this. The construction should tell us what stuff should look like. WLOG reshuffle the columns so that all the columns which contain a cell which is the largest of its row are on the right. Then, reshuffle the rows so that the largest cells in each row are arranged in blocks like in the construction. Let $a_i$ be the largest element in each of these blocks, and $b_i$ the size of the block.

• Some obvious (in)equalities:
• $R \le a_1b_1 + \cdots + a_kb_k \le k$.
• $C \ge \frac{n-k}n + a_1 + \cdots + a_k$.
• $a_ib_i \le 1$ for all $i$.
• $b_1+\cdots + b_k = n$.
• Can we deduce the bound from these? I don't really see it.
• In the intuitive case that $a_ib_i=1$ for all $i$, Cauchy-Schwarz tells us that $(b_1+\cdots + b_k)(a_1+\cdots+a_k) \ge k^2$, so $\frac{R}C \le \frac{k}{\frac{n-k}n + k^2} = \frac{n}{\frac{n}k - 1 + nk}$. The denominator of that is $\ge 2\sqrt{n}-1$ by AM-GM, so that gives us the bound we want.
• So it looks like we are somewhat morally correct, but we need to somehow get to $a_ib_i=1$ for all $i$ or something similar.
• I personally believe that all inequalities can be done with smoothing (though possibly with some amount of pain if the intended solution is some magical nonsense), so let's try that. We want to make our configuration as regular as possible.
• Within each row, we may WLOG make all the first $n-k$ values equal. This doesn't change $R$, and it sets $C$ to the sum of the $a_i$ plus $1$ minus the smallest possible sum of the rightmost $k$ cells across all rows, which is that smallest $C$ can be if we fix the last $k$ columns.
• Within each column, we may WLOG average the first $b_1$ cells, and then the next $b_2$ cells, etc. Then, the maxima within each row stay in the same place. $R$ is unchanged under this, while $C$ decreases. We're now left with a bunch of blocks of equal cells:

• We want the shaded rectangles to each sum to $1$. Is there some way to force this? This is the same thing as saying that everything on the right side minus the shaded rectangles is $0$. If there are two of them which aren't in the same row and column, we can decrease them while increasing the corresponding diagonal entries - surely that should increase $\frac{R}C$?
• Well, effectively this increases $R$ by some weighted average of $b_i+b_j$ times that it increases $C$, which is...not great since we don't know how big the $b_i$ are. In fact if the $b_i$ are too small it'd be better to do the opposite and decrease the diagonal entries, which is weird.
• We could instead try to use the stuff on the left. Suppose the big rectangles have values $c_1,\ldots,c_k$ where WLOG $c_1 \ge c_2 \ge \cdots \ge c_k$ by rearranging rows/columns. Then if the bottom block has nonzero stuff on the right, we can decrease that along with one of the $c_{<k}$, and correspondingly increase $c_k$ and the relevant box on the right. This should keep $R$ and $C$ constant but "moves the $c_i$ closer to each other", so I guess we need to say something like we want to pick a configuration that maximises $\frac{R}C$ which also minimises the total moment of inertia of all the $c_i$ or something. 
• A maximum value can be achieved by some sort of compactness argument. I was going to link a blog post about this, but I realised that this hasn't been written about? The legal values of the grid form a closed and bounded (so compact) subset of $\mathbb{R}^{n^2}$ and $\frac{R}C$ is continuous so a maximum must be achieved. The set of the points achieving the maximum value is a closed subset, so compact, so we can also minimise the moment of inertia or any reasonable continuous thing.
• There is also the slight problem of ties, but I think this might be resolved if we also insist that $k$ has to be the smallest possible? This is getting really messy.
• I think this smoothing approach should let us make all the off-diagonal blocks $0$ in each of the rows $j$ where $c_j < c_1$. Except maybe the top-most of these. Argh.
• Even then, I'm not sure what happens in the top however many rows where $c_j = c_1$.
• Let's go back and try to bound directly. The averaging stuff means that we instead have $C \ge c_1 + a_1 + \cdots + a_k$, which is nicer?
• As I noted while constructing the (hypothesised) equality case, $\frac{R}C$ is big when all the $\frac{a_i}{c_i}$ are big, so maybe it makes sense to compare them.
• We can loosen the bound on $C$ to $$C \ge \sum (b_i-1)c_i + a_i,$$ so now $$\frac{R}C = \frac{\sum (b_i-1)a_i + a_i}{\sum (b_i-1)c_i + a_i}.$$
• Maybe we can use the fact that if $p_1,\ldots,p_k,q_1,\ldots,q_k > 0$ then $\frac{p_1+\cdots+p_k}{q_1+\cdots+q_k}$ is between $\min\left\{\frac{p_i}{q_i}\right\}$ and $\max\left\{\frac{p_i}{q_i}\right\}$. So we want to upper-bound $\frac{a_ib_i}{a_i + (b_i-1)c_i}$.
• For fixed $a_i,b_i$, this is maximised when $c_i=0$, so it's $\le b_i$, which is...not great since we don't know how big the $b_i$ are. Can $c_i$ actually be $0$? Intuitively no but we haven't done anything to rule it out.
• Oh, but we loosened the equality just now. We just need to upper-bound $\frac{a_ib_i}{a_i + (b_i-1)c_1}$, and $c_1$ can't be $0$; it's at least $\frac1n$. So this is $\le \frac{a_ib_i}{a_i + \frac{b_i-1}n}$.
• When is this maximised? If we flip the fraction, we want to minimise $\frac1{b_i} + \frac{b_i-1}{na_ib_i}$, so we want $a_i$ to be as large as possible, but $a_i \le \frac1{b_i}$. So this is $\le \frac1{\frac1{b_i} + \frac{b_i-1}n}$.
• So we have $$\frac{a_ib_i}{a_i + (b_i-1)c_1} \le \frac1{\frac1{b_i} + \frac{b_i-1}n} \le \frac1{\frac2{\sqrt{n}} - \frac1n}$$ where the last step is by AM-GM...hold on, isn't that exactly the bound we wanted?
• Sanity check: in the equality case, all the $\frac{a_ib_i}{a_i + (b_i-1)c_1}$ are equal, and $c_1=\frac1n$, and $a_ib_i=1$, so it should actually check out.
• Great, so we're done since $$\frac{R}C \le \frac{\sum a_ib_i}{\sum a_i + (b_i-1)c_1} \le \max\left\{\frac{a_ib_i}{a_i + (b_i-1)c_1}\right\} \le \frac1{\frac2{\sqrt{n}} - \frac1n}.$$ At last!

This took me way too long. I spent too much time staring at the picture trying to make various blocks go up and down, but that turned out to be too hard to control. I'd be curious if that could be done. It turned out that the key idea was to return to the insight that motivated the construction in the first place. Also, I didn't really realise it, but all the fancy stuff about averaging isn't necessary: the bounds I used all come from the four (in)equalities that I listed at the start of trying to prove the bound. That's a really cool question: I love the construction, and the method of bounding is actually quite nice. Approximate time taken: 3-3.5 hours.

I think that puts me within the time limit for Day 2, but not by much. This certainly went worse than Day 1.

General reflections

On the problems

I was pleasantly surprised by the high quality of the problems here - I thought they were all pretty nice, especially Problems 2 and 6. Also, they weren't too easy, except maybe Problem 4 which dies pretty quickly to either bash or synthetic (so as long as you're decent in one of the two paradigms, you're good to go). The solution to Problem 5 was somewhat disappointing to me because you just have to look at the corners - it would be cool if there were some sort of global argument, but even so, I think it was cute. I liked that all of the solutions were reasonably motivatable and that none of them were too bashy (even the geometry bashes were short), which made for a generally enjoyable experience.

If EGMO was not on your radar for practice problems, it should be, especially if you want to train on medium-difficulty (IMO P2/P5) problems. The difficulty levels for this year's EGMO (to me, at least), seem to be about ~IMO P1 for EGMO P1/P4, easy-ish medium (where medium = IMO P2/P5) for EGMO P2/P5, medium for EGMO P3, and hard-ish medium for EGMO P6. So that's four good problems at the IMO medium level.

On EGMO participation

I wonder why Singapore doesn't send a team to EGMO. Yes, I know we participate in CGMO, but I do think that EGMO has much better problems (in terms of quality and difficulty), and would be much better from the social aspect. I suppose it's because of financial reasons, though a quick scroll through this year's list of countries would show some countries whose Olympiad programmes have presumably fewer resources than us. Based on some countries' team shirts, it seems that girls' math olympiad teams are a thing that trading companies like to sponsor, so that could be an option as well. Also, Singapore has been sending a team to the informatics counterpart, so surely it's possible.

I guess it could also be a manpower thing, but who wouldn't want a trip to Europe? If it comes to it, I'm sure there are many x-men who are based on Europe *coughs and gestures vaguely at self* who would be happy to be the Leader or something.