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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Nine point circle + Perpendicularities
YaoAOPS   18
N 5 minutes ago by AndreiVila
Source: 2025 CTST P2
Suppose $\triangle ABC$ has $D$ as the midpoint of $BC$ and orthocenter $H$. Let $P$ be an arbitrary point on the nine point circle of $ABC$. The line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$. The line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$. If $M$ is the midpoint of $AQ$, show that $HX \perp DM$.
18 replies
YaoAOPS
Mar 5, 2025
AndreiVila
5 minutes ago
Inequality conjecture
RainbowNeos   0
22 minutes ago
Show (or deny) that there exists an absolute constant $C>0$ that, for all $n$ and $n$ positive real numbers $x_i ,1\leq i \leq n$, there is
\[\sum_{i=1}^n \frac{x_i^2}{\sum_{j=1}^i x_j}\geq C \ln n\left(\prod_{i=1}^n x_i\right)^{\frac{1}{n}}\]
0 replies
RainbowNeos
22 minutes ago
0 replies
inequality 2905
pennypc123456789   0
23 minutes ago
Consider positive real numbers \( x, y, z \) that satisfy the condition
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3.
\]Find the maximum value of the expression
\[
P = \dfrac{yz}{\sqrt[3]{3y^2z^2+ 3x^2y^2z^2+ x^2z^2 + x^2y^2}}
+ \frac{xz}{\sqrt[3]{3x^2z^2 + 3x^2y^2z^2 + x^2y^2 + y^2z^2}}
+ \frac{xy}{\sqrt[3]{3x^2y^2 + 3x^2y^2z^2 +y^2z^2 + x^2z^2}}.
\]
0 replies
pennypc123456789
23 minutes ago
0 replies
Inspired by m4thbl3nd3r
sqing   3
N 27 minutes ago by sqing
Source: Own
Let $  a, b,c>0,b+c>a$. Prove that$$\sqrt{\frac{a}{b+c-a}}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)}\geq 1$$$$\frac{a}{b+c-a}-\frac{2a^2-b^2-c^2}{(a+b)(a+c)} \geq  \frac{4\sqrt 2}{3}-1$$
3 replies
1 viewing
sqing
Today at 3:43 AM
sqing
27 minutes ago
Inspired by qrxz17
sqing   7
N 33 minutes ago by sqing
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
7 replies
sqing
5 hours ago
sqing
33 minutes ago
Geometry problem
Whatisthepurposeoflife   2
N an hour ago by Whatisthepurposeoflife
Source: Derived from MEMO 2024 I3
Triangle ∆ABC is scalene the circle w that goes through the points A and B intersects AC at E BC at D let the Lines BE and AD intersect at point F. And let the tangents A and B of circle w Intersect at point G.
Prove that C F and G are collinear
2 replies
Whatisthepurposeoflife
Yesterday at 1:45 PM
Whatisthepurposeoflife
an hour ago
A Sequence of +1's and -1's
ike.chen   36
N an hour ago by maromex
Source: ISL 2022/C1
A $\pm 1$-sequence is a sequence of $2022$ numbers $a_1, \ldots, a_{2022},$ each equal to either $+1$ or $-1$. Determine the largest $C$ so that, for any $\pm 1$-sequence, there exists an integer $k$ and indices $1 \le t_1 < \ldots < t_k \le 2022$ so that $t_{i+1} - t_i \le 2$ for all $i$, and $$\left| \sum_{i = 1}^{k} a_{t_i} \right| \ge C.$$
36 replies
ike.chen
Jul 9, 2023
maromex
an hour ago
Basic ideas in junior diophantine equations
Maths_VC   1
N an hour ago by grupyorum
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
1 reply
Maths_VC
Tuesday at 7:54 PM
grupyorum
an hour ago
Inspired by qrxz17
sqing   3
N an hour ago by sqing
Source: Own
Let $ a,b,c $ be reals such that $ (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 = 28 $ and $  (a^2+b^2+c^2)^2 =16. $ Find the value of $ a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1).$
3 replies
sqing
5 hours ago
sqing
an hour ago
Hardest in ARO 2008
discredit   30
N an hour ago by Phat_23000245
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
30 replies
discredit
Jun 11, 2008
Phat_23000245
an hour ago
Find the value
sqing   12
N an hour ago by Phat_23000245
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
12 replies
sqing
Jun 22, 2024
Phat_23000245
an hour ago
Midpoints of arcs form a similar triangle
v_Enhance   19
N an hour ago by Adywastaken
Source: USA TSTST 2013, Problem 1
Let $ABC$ be a triangle and $D$, $E$, $F$ be the midpoints of arcs $BC$, $CA$, $AB$ on the circumcircle. Line $\ell_a$ passes through the feet of the perpendiculars from $A$ to $DB$ and $DC$. Line $m_a$ passes through the feet of the perpendiculars from $D$ to $AB$ and $AC$. Let $A_1$ denote the intersection of lines $\ell_a$ and $m_a$. Define points $B_1$ and $C_1$ similarly. Prove that triangle $DEF$ and $A_1B_1C_1$ are similar to each other.
19 replies
v_Enhance
Aug 13, 2013
Adywastaken
an hour ago
find question
mathematical-forest   4
N an hour ago by GreekIdiot
Are there any contest questions that seem simple but are actually difficult? :-D
4 replies
mathematical-forest
3 hours ago
GreekIdiot
an hour ago
4 var inequality
SunnyEvan   1
N an hour ago by SunnyEvan
Source: Own
Let $ x,y,z,t \in R^+ ,$ such that : $ (x+y+z+t)^2 = x+y+z+t + (x+z)(y+t) $ and $ x \geq y \geq z \geq t .$
Try to prove or disprove : $$ \frac{2 \sqrt{x+y+z+t +(x+t)(y+z)}}{x^2+y^2+z^2+t^2 +3xz+3yt+xt+yz} \geq \frac{11(x+y)(z+t)-(x+y+z+t)}{x+y+z+t +(x+z)(y+t)} $$
1 reply
SunnyEvan
Today at 7:07 AM
SunnyEvan
an hour ago
Bushy and Jumpy and the unhappy walnut reordering
popcorn1   53
N May 18, 2025 by lksb
Source: 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 $a<k<b$.
53 replies
popcorn1
Jul 20, 2021
lksb
May 18, 2025
Bushy and Jumpy and the unhappy walnut reordering
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2021 P5
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ihatemath123
3449 posts
#43 • 1 Y
Y by cubres
Call a walnut that has been the center of a move a "changed" walnut.

For the sake of contradiction, assume that on every move, Jumpy swaps two walnuts that are either both less than the center one, or both greater. Then, Jumpy is either swapping two changed walnuts, or two yet-to-be-changed walnuts. Either way, one walnut becomes changed at each move, and the positions of the others remain constant.

Number each hole in clockwise order: $1, 2, \dots, 2021$. Because of what we established before, once a hole contains a changed walnut, it will always contain a changed walnut. Therefore, we may refer to each hole as "changed" or "unchanged" - each move is equivalent to picking a hole and changing it. Set, WLOG, walnut $1$ in hole $1$. Then, at some point, we must change hole $2$ - before we do that though, hole $3$ has to be changed. When we change hole $3$, holes $2$ and $4$ must both be unchanged, so we also must change hole $3$ before hole $4$. So when we change hole $4$, both holes $3$ and holes $5$ must be changed. This pattern continues, until we have that hole $2022$ must be changed after hole $2021$, which is a contradiction, since hole $2022$ is hole $1$, which we assumed was changed first.

Therefore, Jumpy must swap two walnuts $a$ and $b$ with $a < k < b$.
Z K Y
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huashiliao2020
1292 posts
#44 • 1 Y
Y by cubres
Assign a walnut a signature if it's been the center of a reordering the number 1, and 0 otherwise. AFTSOC Jumpy is swapping two walnuts that are both less or greater than the indice of the center walnut; in particular, Jumpy swaps two walnuts with the same signature, hence a hole never changes it's signature. WLOG set walnut 1 in hole 1 by shifting hole \#s; when we change hole 2, we must have that hole 3 had a signature of 1, since hole 1 has a signature of 1, whence hole 4 has a signature of 0 before changing hole 3, etc. We continue this way until we get that hole 2022=1 must have a signature of 0 before changing hole 2021, contradiction since we gave hole 1 a signature of 1 first. We conclude. :surf:

Remark. Signature seems like such a common word to assign now, I've seen it in several problems lol
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pi271828
3371 posts
#45 • 1 Y
Y by cubres
We assume the opposite for the sake of contradiction. Label a hole $k$ black on the kth move when it is chosen, and keep it black for the remaining of the process. Keep a hole white if it hasn't been chosen yet. Notice that on each move we should be swapping holes of the same color, and obviously we will turn the middle hole from white to black. Also note that once a slot holds a black hole, it must always hold a black hole otherwise the problem is proved. Now, simply consider an arc of the circle that is $B\underbrace{WW \dots WW}_{x}B$. The color of the endpoints of this arc can never be changed otherwise the problem is proved, so we shift our attention to the inner part. We prove by strong induction that if $2 \mid x$, then there must be a move that swaps two holes of opposite color. The base case of $x = 2$ is clear. Now assume that for an even $n$, that all evens less than $n$ follow the hypothesis. One of the inner $W$'s will be chosen and will be turned to a $B$. Now, we have $B\underbrace{W \dots W}_{a}B\underbrace{W \dots W}_{b}B$, where $a+b$ is odd, implying that one of $a$ and $b$ is even, so we have completed induction. Now, notice that after the first two moves, there must be an arc that will contain an even number of $W$'s that connects the two black holes because $2021$ is odd, so we are done.
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Shreyasharma
684 posts
#46 • 2 Y
Y by eggymath, cubres
Assume for the sake of contradiction otherwise. Rephrase the problem in terms of the color of a walnut. Color the $k$-th walnut red on the $k$-th turn. Note that as we swap two walnuts of the same color on the $k$-th term by assumption, the coloring remains the same.

Clearly we begin with $0$ pairs of adjacent red walnuts and end with $2021$ pairs of adjacent red walnuts. Now on any given turn a walnut that changes color has two neighbors of the same color, so the pairs of adjacent red walnuts either goes up by $2$ or by $0$, a contradiction.
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RGB
19 posts
#47 • 2 Y
Y by eggymath, cubres
Consider the process that happens after Jumpy rearranges the walnuts.

Initially, each walnut has a unique position around the circular pattern of holes. After Jumpy's sequence of moves, some walnuts might return to their original positions, while others might have moved around.

Now, let's imagine that none of the swaps involve a pair of walnuts where one walnut has a number smaller than $k$ and the other walnut has a number larger than $k$. This means that for every swap that occurs during Jumpy's sequence of moves, either both walnuts have numbers smaller than $k$, or both walnuts have numbers larger than $k$.

Let's consider the total displacement of all the walnuts that have numbers smaller than $k$ after Jumpy's moves. Similarly, let's consider the total displacement of all the walnuts that have numbers larger than $k$ after Jumpy's moves.

The sum of displacements of walnuts smaller than $k$ and walnuts larger than $k$ around the circle must be equal in magnitude but opposite in direction (due to the circular nature of the arrangement). This is because each swap results in a net displacement of $0$ for the overall circle, as each walnut returns to its original position after a full circle.

If there were no swap that involves a pair of walnuts where one walnut has a number smaller than $k$ and the other walnut has a number larger than $k$, then the sum of displacements of walnuts smaller than $k$ and the sum of displacements of walnuts larger than $k$ would have to be different, leading to a contradiction.

Therefore, there must be a swap involving a pair of walnuts $a$ and $b$ such that $a<k<b$ during Jumpy's sequence of moves.
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cj13609517288
1924 posts
#48 • 2 Y
Y by eggymath, cubres
Suppose that after the $m$-th move, walnut $m$'s value becomes $0$. If we can find a value of $k$ that works here, it must also work in the original problem as if $m<k$, then $0<k$, and if $m>k$, then $m$ hasn't turned into $0$ yet. Therefore, we now want to show that $0$ and something $\ne 0$ swapped during the process(call this a "forbidden move").

Suppose not. After the first move, there is a "contiguous chain" of $2020$ nonzero numbers. We can never have an end of a chain perform their swap unless the chain is of length $1$.

Claim. An even chain will always cause a forbidden move.
Proof. Let the chain be of length $2n$. We will strong induct on $n$. Base case $n=1$ is trivial, so assume that $1,2,\dots,n-1$ work. Then on the first move of that chain, the chain will be split into two parts of length $a$ and $2n-1-a$, one of which will have a smaller even length, which will cause a forbidden move.

Thus our chain of $2020$ will cause a forbidden move, contradiction. $\blacksquare$
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lnzhonglp
120 posts
#49 • 2 Y
Y by eggymath, cubres
Suppose otherwise. On the $k$th turn, color the $k$th walnut red. Each walnut we color red must be between two walnuts of the same color. The number of pairs of adjacent red walnuts changes by $2$ or $0$, but we must go from $0$ to $2021$. Therefore, this is impossible.
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OronSH
1748 posts
#50 • 4 Y
Y by megarnie, eggymath, ihatemath123, cubres
Consider the same operation, but after a swap you add $2021$ to the label of the walnut just operated on. Every swap now satisfies $k<a,b.$ We want to then show that there is some swap that swaps a number $\le 2021$ and a number $>2021.$ This is doable iff at every step, between every two consecutive $>2021$ walnuts there is an odd number of $\le 2021$ walnuts (by induction), but this is false after the first step, since there are $2020$ walnuts between the one with label $2022$ and itself.
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dolphinday
1329 posts
#51 • 2 Y
Y by eggymath, cubres
At the start, paint all walnuts white and then paint walnut $k$ on the $k$th turn black.
Then notice that the condition $a < k < b$ is equivalent to swapping two walnuts of differing colors. So then FTSOC assume that throughout, we always swap two walnuts of the same color.
However notice that the number of adjacent pairs of black walnuts is always $0 \pmod 2$ as we can only add a black walnut in between two black walnuts or in between two white walnuts. This is a contradiction as our final state with $2021$ black walnuts has an odd number, so we are done.
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HamstPan38825
8868 posts
#52 • 1 Y
Y by cubres
I found this very straightforward for IMO5? This is one of those ``quasi-invariant" problems where simplifying the structure inexplicably yields a neat solution.

Call an acorn large if we have already performed a move on it and small otherwise. Assume for the sake of contradiction that there exists an ordering for which the condition is not true. Let the dynamic variable $k$ denote the number of groups between two consecutive large acorns that have an even number of small acorns between them.

Claim. The parity of $k$ is constant.

Proof. In any move that does not satisfy the condition, one small acorn surrounded by either two large or two small acorns is converted into a large acorn. If the acorn were originally part of an odd group, the first case eliminates this odd group, and the second case splits the odd group into two even or two odd groups. If the acorn were originally part of an even group, the first case is impossible, and the second case splits the even group into one even and one odd group. In both cases, the parity of $k$ does not change. $\blacksquare$

But after the first move, $k = 1$ is odd, and after all moves are complete, all acorns are large, so $k=0$. This yields a contradiction.
This post has been edited 1 time. Last edited by HamstPan38825, Jun 15, 2024, 12:18 AM
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Martin2001
166 posts
#53 • 1 Y
Y by cubres
Rephrase the problem as so: Let there be $2021$ acorns arranged in a circle, and for each move we can color exactly one acorn red. We show that it is impossible to completely color all acorns red without coloring an acorn red where the immediately adjacent acorns are different colors.
We show that it is impossible for a string with an even number of acorns, where both endpoints of the string start as red. Note that this is an easy induction starting from where there are $4$ acorns. Because the original circle of $2021$ acorns can be arranged as a string with $2022$ acorns with both endpoints starting as red, we're done$.\blacksquare$
This post has been edited 1 time. Last edited by Martin2001, Aug 26, 2024, 1:04 AM
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Mathandski
773 posts
#54 • 1 Y
Y by cubres
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Blast_S1
363 posts
#55 • 1 Y
Y by cubres
Let $f(n)$ be $1$ if both neighbors of $n$ are greater, $-1$ if both are smaller, and $0$ otherwise. Since the sum of $f(n)$ over all $n$ is $0$, and since $2021$ is odd, it follows that there exist an odd number of $n$ such that $f(n) = 0$.

Consider the $k^\text{th}$ move. If $f(k) = 0$, then we're done. If $f(k) = 1$, then the move only affects $f(n)$ for $n > k$. If $f(k) = -1$, then the move only affects $f(n)$ for $n < k$. The parity of the number of $n > k$ for which $f(n) = 0$ is invariant in the latter two cases. The end.
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Bonime
38 posts
#56
Y by
The hardest part is to consider a coloring argument. After the $k$-th move, color walnut $k$ black.

Once we do this, the actual numbers on the walnuts are no longer important. We only need to prove that at some point, two walnuts of different colors are swapped. Suppose, for contradiction, that this never happens: that is, at every move, the two walnuts being swapped are either both black or both uncolored (white).

Let us define a $\textit{block}$ of black walnuts as a maximal chain of consecutive black walnuts (in the circular arrangement).
$\textbf{Claim.}$ At every instant $t$, each block of black walnuts has odd size.

$\textit{Proof of Claim.}$ At each move, exactly one new walnut is colored black. Initially, there are no black walnuts. The first black walnut forms a block of size 1, which is odd.
Now observe how the blocks evolve. When a black walnut is added adjacent to an existing block, that block increases its size by $1$, changing its parity. But in order for a black walnut to be added in a way that merges two existing blocks, it must lie between them. Merging two blocks with a single walnut between them results in a new block of a parity different from the sum of the parity of these two.
Since we always begin with blocks of size 1 (odd), and we only merge blocks in a way that preserves oddness, all blocks must have odd size at any point in time. $\blacksquare$

Now consider time $t = 2020$, just before the final move. At this moment, exactly $2020$ walnuts have been colored black, and so we'll have an even size block. This contradiction shows that our assumption was false.

Therefore, at some point, Jumpy must swap two walnuts $a$ and $b$ such that $a < k < b$. $\blacksquare$
This post has been edited 1 time. Last edited by Bonime, May 18, 2025, 5:56 PM
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lksb
180 posts
#57 • 1 Y
Y by Bonime
Let a walnut $k$ be "big" if both its neighbors are smaller than $k$ and "small" if both of them are bigger than $k$. Assume FTSOC that all walnuts are either big or small.

Claim 1: If $k$ is big, then it doesn't move after the $k$-th move
Proof: As both neighbors of $k$, $a$ and $b$, are smaller than it, after the $k$-th move, if $k$ moves, then this move gotta be either the $a$-th or $b$-th move, however, they were performed. Therefore, $k$ doesn't move.

Claim 2: If $k$ is big, then its final neighbors are not big
Proof: If in the $k$-th move, the neighbors are $a$ and $b$, with $a,b<k$, then they are not big, as they moved in the $k$-th move. Whenever a neighbor of $k$ is replaced, its new neighbor $i$ is not big, as it is swapped after the $i$-th move.

With this, there are at most $1010$ big walnuts, as well as at most $1010$ small walnuts. Therefore, there is at least $2021-1010-1010=1$ walnut which is neither big nor small.
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