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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Trigo relation in a right angled. ISIBS2011P10
Sayan   11
N a few seconds ago by Project_Donkey_into_M4
Show that the triangle whose angles satisfy the equality
\[\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2\]
is right angled.
11 replies
+1 w
Sayan
Mar 31, 2013
Project_Donkey_into_M4
a few seconds ago
Expressing polynomial as product of two polynomials
Sadigly   3
N 5 minutes ago by Sadigly
Source: Azerbaijan Senior NMO 2021
Define $P(x)=((x-a_1)(x-a_2)...(x-a_n))^2 +1$, where $a_1,a_2...,a_n\in\mathbb{Z}$ and $n\in\mathbb{N^+}$. Prove that $P(x)$ couldn't be expressed as product of two non-constant polynomials with integer coefficients.
3 replies
Sadigly
Yesterday at 9:10 PM
Sadigly
5 minutes ago
Help me this problem. Thank you
illybest   0
15 minutes ago
Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z
0 replies
illybest
15 minutes ago
0 replies
Product of consecutive terms divisible by a prime number
BR1F1SZ   2
N 22 minutes ago by bin_sherlo
Source: 2025 Francophone MO Seniors P4
Determine all sequences of strictly positive integers $a_1, a_2, a_3, \ldots$ satisfying the following two conditions:
[list]
[*]There exists an integer $M > 0$ such that, for all indices $n \geqslant 1$, $0 < a_n \leqslant M$.
[*]For any prime number $p$ and for any index $n \geqslant 1$, the number
\[
a_n a_{n+1} \cdots a_{n+p-1} - a_{n+p}
\]is a multiple of $p$.
[/list]


2 replies
BR1F1SZ
Yesterday at 12:09 AM
bin_sherlo
22 minutes ago
No more topics!
Bushy and Jumpy and the unhappy walnut reordering
popcorn1   51
N Apr 22, 2025 by Blast_S1
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 $a<k<b$.
51 replies
popcorn1
Jul 20, 2021
Blast_S1
Apr 22, 2025
Bushy and Jumpy and the unhappy walnut reordering
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G H BBookmark kLocked kLocked NReply
Source: IMO 2021 P5
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popcorn1
1098 posts
#1 • 17 Y
Y by Functional_equation, centslordm, InternetPerson10, Mathmick51, TheStrayCat, megarnie, Aritra12, 554183, adityaguharoy, Aopamy, yshk, deplasmanyollari, eggymath, Kingsbane2139, Rounak_iitr, Funcshun840, cubres
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 $a<k<b$.
This post has been edited 2 times. Last edited by popcorn1, Jul 20, 2021, 8:40 PM
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ComiCabE
25 posts
#2 • 36 Y
Y by centslordm, Modesti, lahmacun, Mathmick51, InternetPerson10, TheStrayCat, megarnie, edfearay123, NumberX, Pitagar, oolite, guptaamitu1, VicKmath7, richrow12, CahitArf, Timmy456, Bobcats, PianoPlayer111, GioOrnikapa, WinterSecret, rayfish, FragileBonds, Aopamy, SPHS1234, jeteagle, Mehrshad, Dukejukem, Rexaria112, mathmax12, levifb, TheHimMan, Stuffybear, eggymath, Funcshun840, Kingsbane2139, cubres
Let us assume that the statement is false. At each $k$'th iteration we will color the $k$'th walnut in pink. Observe that if the inequality $a<k<b$ doesn't hold, then we must have swapped walnuts of the same sort (either both uncolored or both pink). This implies that what happens to the colors is that at each iteration Jumpy just paints a certain uncolored walnut in pink.
Now let's keep track of the number of pairs of adjacent walnuts which are both pink. At the start of the process it's $0$, and at the end it's $2021$. But observe that the parity of this quantity can change only when Jumpy colors a nut that has one pink and one uncolored neighbor. A contradiction! Q.E.D.
This post has been edited 2 times. Last edited by ComiCabE, Jul 20, 2021, 9:44 PM
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InternetPerson10
450 posts
#3 • 8 Y
Y by TheStrayCat, sarjinius, centslordm, skyguy88, ike.chen, eggymath, Rounak_iitr, cubres
Dude awesome problem.

In the $k^{\text{th}}$ move, color walnut $k$ green. Then it suffices to show a green walnut and a brown walnut are swapped at one point. For the sake of contradiction assume otherwise, that is, every swap swaps two walnuts of the same color.

Then the state of the colors does not change after every swap, so we may reformulate the problem to the following:
Reformulation wrote:
There are $2021$ brown walnuts in a circle. In each move, Jumpy may choose one walnut whose neighbors are of the same color, and color it green. Show that Jumpy cannot color all walnuts green.
First, consider a gap of $2n$ brown walnuts between two green walnuts on either side. It's impossible to color all these walnuts green. To see this, note that if $n = 1$, neither walnut can be colored (as both neighbors are of different colors). Otherwise, make one move (it cannot be on the first or last brown walnut). This move splits the brown walnuts into two groups, with $2n-1$ brown walnuts in total. Because this is odd, the groups are of different parities, so one group must have an even number of walnuts. Then keep splitting this group until you get a gap of two brown walnuts, at which point you can't color it.

Now make two arbitrary moves; these divide the $2019$ brown walnuts into two groups. As $2019$ is odd, one of these groups must have an even number of walnuts, which you can't color all green. Yay! :DD
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square_root_of_3
78 posts
#4 • 5 Y
Y by centslordm, megarnie, PianoPlayer111, eggymath, cubres
Call a walnut obedient if one of its neighbours is larger than it, and the other one is smaller. We call a process bad if the problem's condition is not satisfied. We want to prove that there are no bad processes.

Lemma 1. The number of obedient walnuts is of the same parity as the number of walnuts.
Proof
Lemma 2. In a bad process, for any $j \in \{0,\ldots, 2021\}$, after $j$ moves, the number of obedient walnuts with labels greater than $j$ is odd.
Proof
Now, if a process is bad, after $2021$ moves there is an odd number of obedient walnuts greater than $2021$, which is a contradiction. Therefore, bad processes don't exist.
This post has been edited 3 times. Last edited by square_root_of_3, Jul 23, 2021, 12:02 PM
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TheStrayCat
161 posts
#5 • 4 Y
Y by centslordm, megarnie, eggymath, cubres
(Found a mistake in my solution, please disregard)
This post has been edited 3 times. Last edited by TheStrayCat, Jul 21, 2021, 4:21 AM
Reason: Found a mistake
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SnowPanda
186 posts
#6 • 4 Y
Y by centslordm, megarnie, eggymath, cubres
Solution
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VulcanForge
626 posts
#7 • 6 Y
Y by Inshaallahgoldmedal, centslordm, megarnie, eggymath, i3435, cubres
Solved with Isaac Zhu, Jeffery Chen, Kevin Wu, Albert Wang, Nacho Cho, Linus Hamilton, Sam Zhang.

Assume for contradiction otherwise. On the $k$-th move, paint the $k$-th acorn red. The key observation due to the assumption at the beginning is that if we are swapping around acorn $k$, then both it's neighbors must be the same color: otherwise the problem would be finished.

Now this gives a contradiction: if we draw an edge between every pair of adjacent red acorns, at each step we gain an even number of edges. But we're supposed to have $2021$ edges at the end, contradiction.
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AnonymousBunny
339 posts
#8 • 3 Y
Y by centslordm, eggymath, cubres
Write the number 0 on each walnut initially. Suppose we change the number on walnut k to 1 on turn k. Assume the problem statement is false, then the neighbors of walnut k at turn k have the same number, so swapping them has no effect on the number sequence.

So we are changing a 0 to 1 at each step and we want to show at some point, we change a sequence like 100 to 110 or 001 to 011.

At any given time, consider the maximal contiguous runs of 0. After the first move, there is one contiguous run of length 2020. I claim that given an initial configuration of 2k consecutive zeroes surroubded by 1s at the endpoints (i.e., $1 0^{2k} 1$) (*), if we change the 0's to 1's one by one, we must hit a forbidden configuration as mentioned above.

For this we induct. Base case k=1 is trivial. For induction step note that if we change a 0 to 1 in a block of 2k consecutive 0s, we break this block into two blocks of size p,q where p+q=2k-1 is odd, so one of them must be even, say p. Induct on the block of length p.

(*) We are "unwrapping" the circle here after the first move.
This post has been edited 1 time. Last edited by AnonymousBunny, Jul 21, 2021, 3:48 AM
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TheUltimate123
1740 posts
#9 • 8 Y
Y by centslordm, HamstPan38825, sabkx, math_comb01, eggymath, Dissonant, Rounak_iitr, cubres
Call a walnut \(k\) conservative if on the \(k\)th move Jumpy swaps walnuts \(a\) and \(b\) with \(a,b<k\), and call a walnut \(k\) liberal if on the \(k\)th move Jumpy swaps walnuts \(a\) and \(b\) with \(a,b>k\). Assume for contradiction every walnut is either conservative or liberal.

Claim: If walnut \(k\) is conservative, then it does not move after the \(k\)th move.

Proof. Indeed, I contend the neighbors of \(k\) will always be smaller than the move number. After the \(k\)th move, the neighbors of \(k\) are \(a,b<k\) by design. If on move \(t\), a neighbor of \(k\) is swapped out, since the original neighbor was less than \(t\), the new neighbor is also less than \(t\). \(\blacksquare\)

Claim: If \(k\) is conservative, then its neighbors in the final position are not conservative.

Proof. If the \(k\)th move swaps \(a\) and \(b\), with \(a,b<k\), then \(a\) and \(b\) are not conservative since they move on the \(k\)th move. Whenever either neighbor of \(k\) is replaced, each resulting neighbor \(i\) is not conservative, since it is swapped after the \(i\)th move. \(\blacksquare\)

It follows that there are at most 1010 conservative walnuts. Analogously there are at most 1010 liberal walnuts, contradicting that all walnuts are either conservative or liberal.
This post has been edited 1 time. Last edited by TheUltimate123, Jul 22, 2021, 4:41 PM
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MarkBcc168
1595 posts
#10 • 7 Y
Y by L567, centslordm, PIartist, sabkx, eggymath, Funcshun840, cubres
Assume for a contradiction that Jumpy never swaps two walnuts $a,b$ with $a<k<b$ in the $k$-th move.

Before Jumpy can swap walnuts, Bushy woke up and observed Jumpy from afar. Bushy is angry upon discovering that Jumpy randomly swaps his beautifully arranged walnuts, so after the $k$-th move, Bushy steps in and removes the walnut $k$ from the hole. Observe that from what we assumed, a vacant hole cannot become occupied with walnut and vice versa.

After Bushy removes the walnut $1$, the remaining walnuts become a single contiguous segment of $2020$ walnuts, which is even. Therefore, when a walnut is removed from this segment, it will be split into an odd and an even segment; we keep our eyes on this even segment. Considering this even segment further, we see that it will eventually be divided to a smaller even segment. Thus, by repeating this argument, eventually, we will get a segment of length $2$ that Bushy cannot take away any either walnut, a contradiction.
This post has been edited 2 times. Last edited by MarkBcc168, Jul 21, 2021, 10:42 PM
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Richangles
139 posts
#11 • 2 Y
Y by centslordm, cubres
I expected this problem to be NT, since P1 was either Algebra or combo. So has the PSC dropped the method of P1, P2, P4 & P5 being a permutation of A, C, G & N?
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jasperE3
11325 posts
#12 • 3 Y
Y by centslordm, eggymath, cubres
No, P1 was NT.
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megarnie
5607 posts
#13 • 4 Y
Y by centslordm, Han-htr, eggymath, cubres
Solution
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554183
484 posts
#14 • 3 Y
Y by centslordm, eggymath, cubres
Wu
On the $k$th move, color walnut $k$ black, and let the other walnuts remain brown. We count the number of pairs of adjacent black walnuts. It’s easy to see that if the opposite is assumed, then the change in the quantity is even. But we require the quantity to become $2021$ which is odd at the end.
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starchan
1609 posts
#15 • 2 Y
Y by centslordm, cubres
Solution?
This post has been edited 1 time. Last edited by starchan, Jul 28, 2021, 9:52 AM
Reason: Fixed an error in my previous solution.
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