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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.

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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
IMO Shortlist 2014 N2
hajimbrak   32
N a minute ago by ezpotd
Determine all pairs $(x, y)$ of positive integers such that \[\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.\]
Proposed by Titu Andreescu, USA
32 replies
hajimbrak
Jul 11, 2015
ezpotd
a minute ago
An easy geometry problem in NEHS Mock APMO
chengbilly   2
N 24 minutes ago by MathLuis
Source: own
Let $ABC$ be a triangle with circumcenter $O$ and orthocenter $H$. $AD,BE,CF$ the altitudes of $\triangle ABC$. A point $T$ lies on line $EF$ such that $DT \perp EF$. A point $X$ lies on the circumcircle of $\triangle ABC$ such that $AX,EF,DO$ are concurrent. $DT$ meets $AX$ at $R$. Prove that $H,T,R,X$ are concyclic.
2 replies
chengbilly
May 23, 2021
MathLuis
24 minutes ago
The reflection of AD intersect (ABC) lies on (AEF)
alifenix-   62
N 37 minutes ago by Rayvhs
Source: USA TST for EGMO 2020, Problem 4
Let $ABC$ be a triangle. Distinct points $D$, $E$, $F$ lie on sides $BC$, $AC$, and $AB$, respectively, such that quadrilaterals $ABDE$ and $ACDF$ are cyclic. Line $AD$ meets the circumcircle of $\triangle ABC$ again at $P$. Let $Q$ denote the reflection of $P$ across $BC$. Show that $Q$ lies on the circumcircle of $\triangle AEF$.

Proposed by Ankan Bhattacharya
62 replies
alifenix-
Jan 27, 2020
Rayvhs
37 minutes ago
PAMO 2022 Problem 1 - Line Tangent to Circle Through Orthocenter
DylanN   5
N an hour ago by Y77
Source: 2022 Pan-African Mathematics Olympiad Problem 1
Let $ABC$ be a triangle with $\angle ABC \neq 90^\circ$, and $AB$ its shortest side. Let $H$ be the orthocenter of $ABC$. Let $\Gamma$ be the circle with center $B$ and radius $BA$. Let $D$ be the second point where the line $CA$ meets $\Gamma$. Let $E$ be the second point where $\Gamma$ meets the circumcircle of the triangle $BCD$. Let $F$ be the intersection point of the lines $DE$ and $BH$.

Prove that the line $BD$ is tangent to the circumcircle of the triangle $DFH$.
5 replies
DylanN
Jun 25, 2022
Y77
an hour ago
Conditional geo with centroid
a_507_bc   6
N 2 hours ago by LeYohan
Source: Singapore Open MO Round 2 2023 P1
In a scalene triangle $ABC$ with centroid $G$ and circumcircle $\omega$ centred at $O$, the extension of $AG$ meets $\omega$ at $M$; lines $AB$ and $CM$ intersect at $P$; and lines $AC$ and $BM$ intersect at $Q$. Suppose the circumcentre $S$ of the triangle $APQ$ lies on $\omega$ and $A, O, S$ are collinear. Prove that $\angle AGO = 90^{o}$.
6 replies
a_507_bc
Jul 1, 2023
LeYohan
2 hours ago
Channel name changed
Plane_geometry_youtuber   0
2 hours ago
Hi,

Due to the search handle issue in youtube. My channel is renamed to Olympiad Geometry Club. And the new link is as following:

https://www.youtube.com/@OlympiadGeometryClub

Recently I introduced the concept of harmonic bundle. I will move on to the conjugate median soon. In the future, I will discuss more than a thousand theorems on plane geometry and hopefully it can help to the students preparing for the Olympiad competition.

Please share this to the people may need it.

Thank you!
0 replies
Plane_geometry_youtuber
2 hours ago
0 replies
IMO Shortlist 2010 - Problem G1
Amir Hossein   134
N 2 hours ago by happypi31415
Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom
134 replies
Amir Hossein
Jul 17, 2011
happypi31415
2 hours ago
Divisors on number
RagvaloD   34
N 3 hours ago by cubres
Source: All Russian Olympiad 2017,Day1,grade 10,P5
$n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all divisors for some $m$ (except $1,m$). Find all such $n$.
34 replies
RagvaloD
May 3, 2017
cubres
3 hours ago
IMO ShortList 2002, number theory problem 2
orl   59
N 3 hours ago by cubres
Source: IMO ShortList 2002, number theory problem 2
Let $n\geq2$ be a positive integer, with divisors $1=d_1<d_2<\,\ldots<d_k=n$. Prove that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is always less than $n^2$, and determine when it is a divisor of $n^2$.
59 replies
orl
Sep 28, 2004
cubres
3 hours ago
None of the circles contains the pentagon - ILL 1970, P34
Amir Hossein   1
N 3 hours ago by legogubbe
In connection with a convex pentagon $ABCDE$ we consider the set of ten circles, each of which contains three of the vertices of the pentagon on its circumference. Is it possible that none of these circles contains the pentagon? Prove your answer.
1 reply
Amir Hossein
May 21, 2011
legogubbe
3 hours ago
interesting incenter/tangent circle config
LeYohan   0
3 hours ago
Source: 2022 St. Mary's Canossian College F4 Final Exam Mathematics Paper 1, Q 18d of 18 (modified)
$BC$ is tangent to the circle $AFDE$ at $D$. $AB$ and $AC$ cut the circle at $F$ and $E$ respectively. $I$ is the in-centre of $\triangle ABC$, and $D$ is on the line $AI$. $CI$ and $DE$ intersect at $G$, while $BI$ and $FD$ intersect at $P$. Prove that the points $P, F, G, E$ lie on a circle.
0 replies
LeYohan
3 hours ago
0 replies
interesting geo config (2/3)
Royal_mhyasd   5
N 3 hours ago by Royal_mhyasd
Source: own
Let $\triangle ABC$ be an acute triangle and $H$ its orthocenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = |\angle ABC-\angle ACB|$. Define $Q$ and $R$ as points on the parallels through $B$ to $AC$ and through $C$ to $AB$ similarly. If $P,Q,R$ are positioned around the sides of $\triangle ABC$ as in the given configuration, prove that $P,Q,R$ are collinear.
5 replies
Royal_mhyasd
Yesterday at 11:36 PM
Royal_mhyasd
3 hours ago
interesting geometry config (3/3)
Royal_mhyasd   2
N 3 hours ago by Royal_mhyasd
Let $\triangle ABC$ be an acute triangle, $H$ its orthocenter and $E$ the center of its nine point circle. Let $P$ be a point on the parallel through $C$ to $AB$ such that $\angle CPH = |\angle BAC-\angle ABC|$ and $P$ and $A$ are on different sides of $BC$ and $Q$ a point on the parallel through $B$ to $AC$ such that $\angle BQH = |\angle BAC - \angle ACB|$ and $C$ and $Q$ are on different sides of $AB$. If $B'$ and $C'$ are the reflections of $H$ over $AC$ and $AB$ respectively, $S$ and $T$ are the intersections of $B'Q$ and $C'P$ respectively with the circumcircle of $\triangle ABC$, prove that the intersection of lines $CT$ and $BS$ lies on $HE$.

final problem for this "points on parallels forming strange angles with the orthocenter" config, for now. personally i think its pretty cool :D
2 replies
Royal_mhyasd
Today at 7:06 AM
Royal_mhyasd
3 hours ago
Convex Quadrilateral with Bisector Diagonal
matinyousefi   8
N 4 hours ago by lpieleanu
Source: Germany TST 2017
In a convex quadrilateral $ABCD$, $BD$ is the angle bisector of $\angle{ABC}$. The circumcircle of $ABC$ intersects $CD,AD$ in $P,Q$ respectively and the line through $D$ parallel to $AC$ cuts $AB,AC$ in $R,S$ respectively. Prove that point $P,Q,R,S$ lie on a circle.
8 replies
matinyousefi
Apr 11, 2020
lpieleanu
4 hours ago
EGMO magic square
Lukaluce   16
N Apr 23, 2025 by zRevenant
Source: EGMO 2025 P6
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 $\frac{R}{C}$?

Proposed by Paulius Aleknavičius, Lithuania, and Anghel David Andrei, Romania
16 replies
Lukaluce
Apr 14, 2025
zRevenant
Apr 23, 2025
EGMO magic square
G H J
G H BBookmark kLocked kLocked NReply
Source: EGMO 2025 P6
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Lukaluce
274 posts
#1 • 4 Y
Y by RainbowJessa, radian_51, farhad.fritl, dangerousliri
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 $\frac{R}{C}$?

Proposed by Paulius Aleknavičius, Lithuania, and Anghel David Andrei, Romania
This post has been edited 2 times. Last edited by Lukaluce, Apr 29, 2025, 10:37 AM
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aaaa_27
4 posts
#2 • 1 Y
Y by RainbowJessa
2 combi in a day?
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R8kt
303 posts
#3 • 29 Y
Y by BR1F1SZ, oVlad, alexanderhamilton124, GuvercinciHoca, Ciobi_, Assassino9931, megarnie, Triangle_Center, Kimchiks926, chirita.andrei, qwedsazxc, EpicBird08, EeEeRUT, CerealCipher, RainbowJessa, Sedro, khina, ihatemath123, Miquel-point, Yiyj1, aidan0626, farhad.fritl, Rox_, RaduAndreiLecoiu, MuhammadAmmar, ZVFrozel, paintingredflagsgreen3761, math_comb01, qlip
This problem was created by me (Paulius Aleknavičius, Lithuania) and a fitting official solution was provided by the one and only David-Andrei Anghel (Romania). Hard to believe two of my problems made EGMO this year (P1 and P6)! I hope you guys enjoyed them.
This post has been edited 1 time. Last edited by R8kt, Apr 14, 2025, 12:02 PM
Reason: .
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DottedCaculator
7357 posts
#4 • 5 Y
Y by EeEeRUT, khina, qwedsazxc, radian_51, aidan0626
The largest possible value of $\frac RC$ is $\frac{2025}{89}$. For the construction, partition the left $45$ columns into $45\times45$ squares, and put $\frac1{45}$ in each of the main diagonals of the squares and $0$s elsewhere in those squares. Fill the rest of the board with $\frac1{2025}$. Then, $R=45$ and $C=1+\frac{1980}{2025}=\frac{89}{45}$, so $\frac RC=\frac{2025}{89}$.

Now, we show $\frac RC\leq\frac{2025}{89}$. For each row, circle the largest number in the row. Then, let $a_i$ be the number of circles in column $i$ and let $s_i$ be the sum of the circles in column $i$. Then, we can pick $C\geq\sum_{a_i>0}\left(\frac{s_i}{a_i}-\frac1{2025}\right)+1$. By AM-GM, $\frac{s_i}{a_i}\geq-\frac1{2025}a_i+\frac2{45}\sqrt{s_i}$, so
\begin{align*}
C&\geq\sum_{a_i>0}\left(-\frac1{2025}a_i+\frac2{45}\sqrt{s_i}-\frac1{2025}\right)+1\\
&=\sum_{a_i>0}\left(\frac2{45}\sqrt{s_i}-\frac1{2025}\right)\\
&\geq\sum_{a_i>0}\frac{89}{2025}s_i\\
&=\frac{89}{2025}R,
\end{align*}as $\frac2{45}\sqrt{s_i}-\frac1{2025}\geq\frac{89}{2025}s_i$ is equivalent to $(\sqrt{s_i}-1)(89\sqrt{s_i}-1)\leq0$ since each circled number must be at least $\frac1{2025}$. Therefore, $\frac RC\leq\frac{2025}{89}$.
This post has been edited 2 times. Last edited by DottedCaculator, Apr 14, 2025, 2:16 PM
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Polyquadratus
4 posts
#5 • 5 Y
Y by Triangle_Center, R8kt, oVlad, Yiyj1, aidan0626
R8kt wrote:
This problem was created by me (Paulius Aleknavičius, Lithuania) and a fitting official solution was provided by the one and only David-Andrei Anghel (Romania). Hard to believe two of my problems made EGMO this year (P1 and P6)! I hope you guys enjoyed them.

I <3 David Andrei Anghel
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oVlad
1746 posts
#6 • 11 Y
Y by khina, Triangle_Center, chirita.andrei, Yiyj1, aidan0626, Ciobi_, farhad.fritl, SomeonesPenguin, paintingredflagsgreen3761, ravengsd, Double07
Polyquadratus wrote:
R8kt wrote:
This problem was created by me (Paulius Aleknavičius, Lithuania) and a fitting official solution was provided by the one and only David-Andrei Anghel (Romania). Hard to believe two of my problems made EGMO this year (P1 and P6)! I hope you guys enjoyed them.

I <3 David Andrei Anghel

Back off he's mine.
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Euclid9876
5 posts
#7
Y by
Did anyone solve this in the actual competition?
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YaoAOPS
1541 posts
#8 • 1 Y
Y by radian_51
We claim the answer is $\frac{R}{C} = \frac{2025}{89}$.

Construction: Divide the board into a $2025 \times 45$ rectangle with $45$ columns and a $2025 \times 2020$ rectangle. Fill the $2025 \times 2020$ rectangle with all $\frac{1}{2025}$, and fill the $2025 \times 45$ with $2025$ occurences of $\frac{1}{45}$ with $45$ occurences per column and one per row. Then
\[
	\frac{R}{C} = \frac{2025 \cdot \frac{1}{45}}{45 \cdot \frac{1}{45} + \frac{2025 - 45}{45}} = \frac{2025}{89}
\]
Bound: Color the largest cell in each row red, breaking ties arbitrarily.

Claim: We may assume that two red cells in the same column have the same value.
Proof: Replace both rows with their average, $R$ remains the same and $C$ remains the same or decreases. $\blacksquare$

Now, WLOG let the red cells be in columns $1$ through $i$ with $a_i$ in the $i$th column. We then have that $a_1 + a_2 + \dots + a_i = 2025$. Note that the red cells in the $i$th column have value at most $\frac{1}{2025} \le r_i \le \frac{1}{a_i}$. Let $N = \frac{2025}{89}$. We then want to show that
\[
	a_1r_1 + a_2r_2 + \dots + a_ir_i \le N \cdot \left(r_1 + r_2 + r_3 + \dots + \frac{2025 - i}{2025}\right)
\]This is tightest when the $a_i < N$ have $r_i = \frac{1}{2025}$ and the $a_i > N$ have $r_i = \frac{1}{a_i}$. We can thus rewrite this as
\[
	j + \frac{t}{2025} \le N \cdot \left(\frac{1}{a_1} + \dots + \frac{1}{a_j} + \frac{i-j}{2025} + \frac{2025 - i}{2025}\right)
\]where $a_1 + \dots + a_j = 2025 - t$. Since $t \ge i-j \ge \frac{t}{N}$, this becomes
\[
	\frac{j}{N} \le \frac{1}{a_1} + \dots + \frac{1}{a_j} + \frac{2025 - i}{2025}
\]which with AM-HM and applying $t \ge i-j$ becomes
\[
	\frac{j}{N} \le \frac{j^2}{a+j} + \frac{a}{2025}
\]with $a \le 2025-j$. The RHS is minimized at $a = 44j$ as the only root. If $j \le 45$, this is tightest at $a = 44j$ and becomes
\[
	\frac{j}{N} \le \frac{j^2}{45j} + \frac{44j}{2025} \iff
	\frac{89j}{2025} \le \frac{45j}{2025j} + \frac{44j}{2025}
\]which holds. If $j \ge 45$, this becomes
\[
	\frac{j}{N} \le \frac{j^2}{2025} + \frac{2025-j}{2025} \iff 
    89j \le j^2 + 2025 - j \iff (j-45)^2 \ge 0 
\]which holds.
This post has been edited 3 times. Last edited by YaoAOPS, Apr 14, 2025, 8:03 PM
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Yiyj1
1271 posts
#9
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Polyquadratus wrote:
R8kt wrote:
This problem was created by me (Paulius Aleknavičius, Lithuania) and a fitting official solution was provided by the one and only David-Andrei Anghel (Romania). Hard to believe two of my problems made EGMO this year (P1 and P6)! I hope you guys enjoyed them.

I <3 David Andrei Anghel

who doesnt
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cj13609517288
1926 posts
#10
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This is not a "complete" solution (I used WolframAlpha because I wasn't sure if my algebra would finish), but I will post it for storage.

https://manifold.markets/JiaheLiu/will-a-problem-on-the-2025-imo-nont#r0bnhtwdgy moment

Replace $2025$ with $n^2$. The answer is $\boxed{\frac{n^2}{2n-1}}$, achieved by generalizing the following example for $n=2$ (everything should be divided by $4$):

2011
2011
0211
0211

Now for the proof, consider the scorer for each row, and rearrange the rows and columns so that the scorers are "sorted", like so:

X???
X???
?X??
??X?

Say there are $\ell$ columns that contains scorers. Note that the non-scorers in a column with more than $\frac{n^2}{2n-1}$ scorers should be distributed to the scorers to maximize the ratio, the converse is true (distribute among non-scorers if there aren't enough scorers).

After a lot of algebra, eventually it turns out that we want to prove
\[\frac{n^2a+n^2-s}{n^2-a+\frac{n^2a^2}{s}}\le\frac{n^2}{2n-1}.\]After more bashing, we realize that this is optimized when $a=\frac{n}{s}$, and eventually we get that this inequality is true. $\blacksquare$
This post has been edited 1 time. Last edited by cj13609517288, Apr 14, 2025, 8:44 PM
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MathLuis
1559 posts
#11 • 1 Y
Y by MS_asdfgzxcvb
Give all cells that contain some *uniquely chosen* value from the $r_i$'s the yellow color, now on each column $i$ consider $s_i$ the sum of values of yellow cells and let $y_i$ the number of yellow cells it has.
Notice from the maximun condition we must have that each $c_i$ is greater than $\frac{1}{2025}$ but obviously this won't always be sharp so in order to make the relevant info above more useful we will take in consideration the existence of the yellow cells and the thing mentioned above to get that $C \ge \sum_{y_i>0} \left(\frac{s_i}{y_i}-\frac{1}{2025} \right)+1$ and all we want is for this to be in terms of $s_i$ only as $R=\sum_{y_i>0} s_i$ is trivially true by double counting so now we will split sum of $y_i$'s and $s_i$'s by using AM-GM as it is true that $\frac{s_i}{y_i}+\frac{y_i}{2025} \ge \frac{2}{45} \cdot \sqrt{s_i}$ and therefore $C \ge \sum_{y_i>0} \left(\frac{2}{45} \cdot \sqrt{s_i}-\frac{y_i}{2025}-\frac{1}{2025} \right)+1=\frac{2}{45} \cdot \left( \sum_{y_i>0} \sqrt{s_i}-\frac{1}{90} \right)$ And now here we need $s_i$ to show up again in a way we can get rid of the square roots, notice that each $s_i \ge \frac{1}{2025}$ and thus $\sqrt{s_i} \ge \frac{1}{45}$ but also trivially $1 \ge \sqrt{s_i}$ which should lead to picking $(\sqrt{s_i}-1)(k\sqrt{s_i}-1) \le 0$ for some $k \ge 45$, now expanding this gives $ks_i-(k+1)\sqrt{s_i}+1 \le 0$ and therefore $ks_i+1 \le (k+1)\sqrt{s_i}$ so our pick here will be $k=89$ to get rid of each $\frac{1}{90}$ in which case we get that $\sqrt{s_i}-\frac{1}{90} \ge \frac{89s_i}{90}$ this so that we don't have the need to summon each $y_i$ again...
And of course this gives $\frac{R}{C} \le \frac{2025}{89}$ so tracing back the equality this gives that in each of the columns where $y_i>0$ we have that the only nonzero cells are the ones in yellow, also from the AM-GM we have that $y_i=45$ if $y_i>0$ which means we must have exactly $45$ such rows, so as a construction consider the first 45 columns, and split the board in $45 \times 45$'s then on the first column of these place the $\frac{1}{45}$ on the same main diagonal with the same direction for each such $45 \times 45$ and the rest of cells must be zero, also from the first bound of $C$ we have followed acordingly all $s_i$'s to be equal which is why we picked $\frac{1}{45}$ but also notice equality happens when every other cell is $\frac{1}{2025}$ and thus that will be our construction, thus we are done :cool:.
This post has been edited 2 times. Last edited by MathLuis, Apr 15, 2025, 4:41 AM
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AbbyWong
171 posts
#13
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@R9182 ChatGPT gave me 2025 lol.
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YaoAOPS
1541 posts
#14 • 1 Y
Y by R9182
Unfortunately, the only charitable thing you can give the AI is that it found the construction.
R9182 wrote:
We now explain briefly why no doubly stochastic matrix can yield a larger value of \(\frac{R}{C}\). At the heart of the matter is the trade–off already discussed. If one tries to increase some row’s maximum beyond \(\frac{1}{45}\), then (since the sum of its row is 1) one must concentrate even more mass in a single entry. But then the column that “carries” that mass will have its maximum raised. In our construction the choice of “heavy” columns has been optimized: by “packing” exactly 45 large entries of size \(1/45\) in each such column, we guarantee that each heavy column carries the minimal possible maximum to serve 45 rows. (A short optimization shows that if one designates \(k\) columns to serve as the targets for high entries and distributes the mass evenly among them so that every heavy row gets exactly \(1/r\) in the column it uses, then one may obtain a ratio of
\[ \frac{n}{\,k+r-1}\,, \]with the constraint \(k\cdot r=n\). This expression is maximized when \(k=r=\sqrt{n}\); in our case \(\sqrt{2025}=45\), and the maximum value is exactly
\[ \frac{2025}{45+45-1}=\frac{2025}{89}\,. \])

A more “dual” formulation of the problem (via linear programming duality) leads to the same optimum. We omit the details for brevity.

This is nonsensical and does not go into actually proving the bound.
R9182 wrote:
A careful (and somewhat lengthy) verification shows that one may indeed “patch” the matrix so that all the doubly stochastic constraints are met. (There are several ways to complete the construction; one may view the process as “gluing” together an extreme point of the Birkhoff–von Neumann polytope on a \(45\times45\) block with the “flat” matrix on the remaining cells.)

Throwing the words Birkhoff–von Neumann around in a doubly stochastic matrix problem does not actually prove anything other than looking impressive.
R9182 wrote:
- If we partition the \(2025\) rows into \(k\) groups (with equal size, say) and have each group’s rows use a distinct column for their row–maximum, then in any column which is so chosen the column sum constraint forces the common maximum to be equal to the reciprocal of the group’s size.
- At the same time, by “filling in” the remaining cells very uniformly the maximum of every other column may be made as low as possible.

This is a highly nontrivial step and most of the problem's difficulty lies in proving this. Showing that these maximums are equal is also nontrivial.
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R9182
8 posts
#15
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YaoAOPS wrote:
Unfortunately, the only charitable thing you can give the AI is that it found the construction.
R9182 wrote:
We now explain briefly why no doubly stochastic matrix can yield a larger value of \(\frac{R}{C}\). At the heart of the matter is the trade–off already discussed. If one tries to increase some row’s maximum beyond \(\frac{1}{45}\), then (since the sum of its row is 1) one must concentrate even more mass in a single entry. But then the column that “carries” that mass will have its maximum raised. In our construction the choice of “heavy” columns has been optimized: by “packing” exactly 45 large entries of size \(1/45\) in each such column, we guarantee that each heavy column carries the minimal possible maximum to serve 45 rows. (A short optimization shows that if one designates \(k\) columns to serve as the targets for high entries and distributes the mass evenly among them so that every heavy row gets exactly \(1/r\) in the column it uses, then one may obtain a ratio of
\[ \frac{n}{\,k+r-1}\,, \]with the constraint \(k\cdot r=n\). This expression is maximized when \(k=r=\sqrt{n}\); in our case \(\sqrt{2025}=45\), and the maximum value is exactly
\[ \frac{2025}{45+45-1}=\frac{2025}{89}\,. \])

A more “dual” formulation of the problem (via linear programming duality) leads to the same optimum. We omit the details for brevity.

This is nonsensical and does not go into actually proving the bound.
R9182 wrote:
A careful (and somewhat lengthy) verification shows that one may indeed “patch” the matrix so that all the doubly stochastic constraints are met. (There are several ways to complete the construction; one may view the process as “gluing” together an extreme point of the Birkhoff–von Neumann polytope on a \(45\times45\) block with the “flat” matrix on the remaining cells.)

Throwing the words Birkhoff–von Neumann around in a doubly stochastic matrix problem does not actually prove anything other than looking impressive.
R9182 wrote:
- If we partition the \(2025\) rows into \(k\) groups (with equal size, say) and have each group’s rows use a distinct column for their row–maximum, then in any column which is so chosen the column sum constraint forces the common maximum to be equal to the reciprocal of the group’s size.
- At the same time, by “filling in” the remaining cells very uniformly the maximum of every other column may be made as low as possible.

This is a highly nontrivial step and most of the problem's difficulty lies in proving this. Showing that these maximums are equal is also nontrivial.

Oh, thanks for the detailed analysis of the AI's attempt at this problem. It seems that AI is still far from actually being able to solve problems of this kind. I see what you're saying — it skipped the highly non-trivial and important steps without providing any proof.
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Assassino9931
1383 posts
#16 • 1 Y
Y by sami1618
Both this problem and EGMO 2024 P6 are truly mathematically brilliant and have some research flavour. However, there goes once again of Problem 6 having no solution from a European country and barely any from all countries. People, please, if a competition has 6 problems, then really try to make it to have 6 problems. Day 1 was very well balanced, congrats to the PSC, and will remain a good example for aiming a balanced paper in the near and far future!
This post has been edited 3 times. Last edited by Assassino9931, Apr 16, 2025, 6:44 AM
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Iveela
117 posts
#17 • 2 Y
Y by R9182, thdnder
@above I think we should view mathematical olympiads as a celebration of the beauty of mathematics rather than a test for university admissions and whatnot. That is not to say that having a balanced paper is unimportant, however including one very difficult but amazing problem does not take away from the quality of the contest. I think these problems excel in that respect. The same logic goes for problems like IMO 2023 P6 as well.
This post has been edited 4 times. Last edited by Iveela, Apr 19, 2025, 2:16 PM
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zRevenant
14 posts
#18
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(Livesolved on YouTube: Art of Olympiad Mathematics)

Answer: $\frac{89}{2025}$.

Proof. For example, let's take rightmost $45$ columns and place $45$ fractions $\frac{1}{45}$ in $45$ squares in every column, such that no two are in the same row. In the rest of the columns we placed $\frac{1}{2025}$. This construction clearly works to give the answer $\frac{89}{2025}$.

For bound, let's mark all the cells that are biggest in a row $red$. Then, we suppose there are $k$ columns that have $red$ cells. Suppose there are exactly $x_1, x_2, ..., x_k$ numbers in those columns and the sum in each row is $a_1, a_2, ..., a_k$. Then, clearly $R=r_1+...+r_{2025} \le k$ because they are contained in $k$ columns. $C=c_1+...+c_{2025} \ge \frac{1}{2025} \cdot (2025-k) + \frac{a_1}{x_1} + ... + \frac{a_k}{x_k}$. In order to bound further, we can use Titu's lemma for when everything except for $x_i$'s is fixed, which yields that $a_i=t \cdot x_i$. Hence, we want to bound $\frac{2025t}{\frac{2025-k}{2025}+kt}$, which is greater or equal to $\frac{R}{C}$, which we continue by dividing both numerator and denominator by $t$. Now, we just have to bound $t$: $t=\frac{a_1+...+a_k}{x_1+...+x_k} \le \frac{k}{2025}$ which will get us to $\frac{R}{C} \le \frac{2025}{k+\frac{2025}{k}-1}$ which yields the needed answer.
This post has been edited 4 times. Last edited by zRevenant, Apr 23, 2025, 10:30 AM
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