Oh actually it is already available.

somebodyyouusedtoknow wrote: Where did you get this problem? The official site. https://www.apmo-official.org/static/problems/apmo2023_prb.pdf

i have read this statement five times and i still have no clue what it means. reportedly some members of the usa tst group last year felt the same way edit: after a bit of discussion, I think the question is asking us to arrange the $n$ squares so that each touches two others, but only at a vertex (so you basically form a union of cycles)

I believe they should have defined touch more precisely... From the way I understand it now I think touching is only at the vertices but then its trivial from greatest to least

"Suppose that no two squares touch, except possibly at their vertices. Show that it is possible to arrange these squares in a way such that every square touches exactly two other squares." Is this what the problem is trying to say? "Show that it is possible to arrange these squares such that: a) no two squares touch, except possibly at their vertices, and b) exactly two vertices of each square are shared with a vertex of another square."

I believe two people on the USA IMO team had issues with the wording. One of them even sent a protest.

Wording issues are awful, but at least we know now that these IMO team members were not trying to throw the problem We can find a construction for $n=6$ which actually generalizes readily when you increase the length of each side by the same constant—the squares are 1,4,5,2,3,6 in that order. Therefore we may induct from $n$ to $n+6$ by adding this construction for $n+1,\ldots,n+6$ independent of the construction for $n$. As such, we just have to construct for $n=5,7,8,9,10$ ($n=6$ is done already), which is not that bad if you essentially try to place the squares of side length $n$ and $n-1$ opposite from each other, and then divide up the rest of the squares into two groups with the same side length sum and as close a number of squares as possible. This won't work for $n=7,8$, since the rest of the squares have odd side length sum, so you take $n$ and $n-2$ instead. (This approach doesn't generalize very well, because it relies on the numbers being small to avoid squares colliding in the middle of the configuration, but for $n \leq 10$ it's fine) My constructions were (for $n=5,7,8,9,10$ in order): 12534, 1472653, 13874652, 125694378, 1894527(10)63. The interpretation of these orderings of squares is left as an exercise.

This is the diagram for $n = 5,6,7,8,9,10,11$ We can use induction as @above has shown

similar to official solution at here is apmo generally easier than say, like rmm or other contests because i've seen that trend in the past few years

S.Das93 wrote: similar to official solution at here is apmo generally easier than say, like rmm or other contests because i've seen that trend in the past few years Basically, APMO isn't a hard contest. For sure RMM is much much harder than this one! But note that APMO has a $5$ problems with just $4$ hours. However, RMM has $6$ problems in two days with totally $9$ hours. So, when you consider all these facts both have a same difficulty. While, if you just compare the problems individually, RMM is harder than APMO.

M11100111001Y1R wrote: S.Das93 wrote: similar to official solution at here is apmo generally easier than say, like rmm or other contests because i've seen that trend in the past few years Basically, APMO isn't a hard contest. For sure RMM is much much harder than this one! But note that APMO has $5$ problems with just $4$ hours. However, RMM has $6$ problems in two days with totally $9$ hours. So, when you consider all these facts both have the same difficulty. While, if you just compare the problems individually, RMM is harder than APMO.

carefully wrote: Let $n \geq 5$ be an integer. Consider $n$ squares with side lengths $1, 2, \dots , n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices. Show that it is possible to arrange these squares in a way such that every square touches exactly two other squares. IDK if I understand this problem rightly. First let's divide the cases into when $n\equiv2, 0(mod4)$ and when it is a odd number.. When $n\equiv2(mod4)$, then we could make a $2*{n/2}$ diagonal rectangle and make the 4 vertices as $1, n, 2, n-1$. Because we want every square to meet 2 other squares, we can consider a shape like a rectangle. So, we have to arrange the left $n-4$ numbers that are $3, 4, ..., n-3, n-2$. We would like to divide these numbers into 2 equal size groups so that the sum of the two groups has a difference of 1. This is easy if we divide it by $(3, n-2, 5, n-6...... ), (4, n-3, 6, n-7, ......)$ then we can put it in and we are done with this case. When $n\equiv0(mod4)$, make the $4$ vertices as $n, n-1, n-2, n-3$ and divide the rest $n-4$ numbers into the frontest and last(like a rainbow) and make as many pairs at $5~n-4$ and pair the last 4 numbers as $(1, 3), (2, 4)$ then we will get 2 pairs that have a difference of $2$. So far we showed that for every even $n$ we can make an $2*{n/2}$ rectangle that satisfies this problem. Know we will show when $n$ is odd. When $n\equiv1(mod4)$ we can make the $4$ vertices as $n, (n-10)/2, n-1, (n-8)/2$ and make the other 2 sides as a rainbow(after 1) and put the 1 at the side that is 1 less than the other. Now let's look at the case when $n\equiv3(mod4)$. First find the first attempt when $n$ is $7$ that is $3, 1, 6, 2, 7, 4, 5$. And we will add the next 4 numbers at each step like the rainbow thing I said previously. Then we are done! For all $n$ we have shown at every case we can make this as a rectangle shape at the plane.

Presented a solution here to help visualize! https://youtu.be/xkIm0k1FE-8

Who knows which country will be coordinating at the APMO 2024?

OG! Claim: There exists a polygon of sides $\sqrt(2), 2\sqrt(2) \cdots n\sqrt(2)$ such that none of the interior or exterior angles is $90$ degrees for $n \geq 4$. Proof: First observe that existence of such a polygon is guaranteed since for $n \geq 4$, \sqrt(2)+ 2\sqrt(2) +\cdots +(n-1)\sqrt 2 > n\sqrt(2)$ Now assume the polygon has a right angle (Will continue later)

The author is Michelle Chen, Australia.