It's February and we'd love to help you find the right course plan!

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k a February Highlights and 2025 AoPS Online Class Information
jlacosta   0
Feb 2, 2025
We love to share what you can look forward to this month! The AIME I and AIME II competitions are happening on February 6th and 12th, respectively. Join our Math Jams the day after each competition where we will go over all the problems and the useful strategies to solve them!

2025 AIME I Math Jam: Difficulty Level: 8* (Advanced math)
February 7th (Friday), 4:30pm PT/7:30 pm ET

2025 AIME II Math Jam: Difficulty Level: 8* (Advanced math)
February 13th (Thursday), 4:30pm PT/7:30 pm ET

The F=ma exam will be held on February 12th. Check out our F=ma Problem Series course that begins February 19th if you are interested in participating next year! The course will prepare you to take the F=ma exam, the first test in a series of contests that determines the members of the US team for the International Physics Olympiad. You'll learn the classical mechanics needed for the F=ma exam as well as how to solve problems taken from past exams, strategies to succeed, and you’ll take a practice F=ma test of brand-new problems.

Mark your calendars for all our upcoming events:
[list][*]Feb 7, 4:30 pm PT/7:30pm ET, 2025 AIME I Math Jam
[*]Feb 12, 4pm PT/7pm ET, Mastering Language Arts Through Problem-Solving: The AoPS Method
[*]Feb 13, 4:30 pm PT/7:30pm ET, 2025 AIME II Math Jam
[*]Feb 20, 4pm PT/7pm ET, The Virtual Campus Spring Experience[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Feb 2, 2025
0 replies
I need this. Pls help me
Giahuytls2326   12
N 6 minutes ago by Giahuytls2326
Source: Own
Given $\triangle ABC$ , orthocenter $H$, $E$ is the center of Euler's circle of $\triangle ABC$. $X,Y,Z$ is the midpoint of $AH,BH,CH$. From $X,Y,Z$ draw the tangent of $(E)$, cut a line through $E$ perpendicular to sides $BC, CA,AB$ at $M,N,P$ .Prove that $AM,BN,CP$ concurrent
12 replies
Giahuytls2326
Thursday at 4:59 PM
Giahuytls2326
6 minutes ago
Thanks u!
Ruji2018252   4
N 7 minutes ago by sqing
Let $a,b,c>0$ and $a+b+c=abc$
Find minimum (and prove)
\[P=(ab-1)(bc+1)^6(ca-1)\]
4 replies
Ruji2018252
Yesterday at 2:38 PM
sqing
7 minutes ago
Functional Equation
AnhQuang_67   4
N 7 minutes ago by gordian.knot
Find all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying: $$x^2f(x)+f(1-x)=2x-x^4, \forall x \in \mathbb{R} $$
4 replies
AnhQuang_67
Yesterday at 4:02 PM
gordian.knot
7 minutes ago
thanks u!
Ruji2018252   1
N 27 minutes ago by sqing
Let $a,b,c\in \mathbb{R},a,b,c\ne 0$ and $a+b+c=0.$ Find minimum (and prove)
\[C=\dfrac{b^2+c^2-a^2}{b^2+c^2}+\dfrac{c^2+a^2-b^2}{c^2+a^2}+\dfrac{a^2+b^2-c^2}{a^2+b^2}\]
1 reply
1 viewing
Ruji2018252
3 hours ago
sqing
27 minutes ago
No more topics!
Classic - sum of rows and columns nonnegative
randomusername   10
N Dec 29, 2024 by Eka01
Source: 1961 All-Soviet Union Olympiad
Consider a table with one real number in each cell. In one step, one may switch the sign of the numbers in one row or one column simultaneously. Prove that one can obtain a table with non-negative sums in each row and each column.
10 replies
randomusername
Aug 4, 2015
Eka01
Dec 29, 2024
Classic - sum of rows and columns nonnegative
G H J
Source: 1961 All-Soviet Union Olympiad
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randomusername
1059 posts
#1 • 2 Y
Y by Adventure10, Mango247
Consider a table with one real number in each cell. In one step, one may switch the sign of the numbers in one row or one column simultaneously. Prove that one can obtain a table with non-negative sums in each row and each column.
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huricane
670 posts
#2 • 12 Y
Y by mathcool2009, Gh324, Eggnog82, Mathdragon245, rmtf1111, futurestar, IceWolf10, CrazyMathMan, Mathlover_1, Adventure10, Mango247, Funcshun840
Let $S$ be the sum of all numbers in the cells of the table.From all the possible configurations(a configuration is the table after some steps) we will choose the configuration for which $S$ is maximal.Let's suppose that there is a column of that configuration with the propriety that the sum of the numbers inscribed in the column's cells is negative.Let $a_1,a_2,...,a_n$ be the numbers from that column.Then $a_1+a_2+...+a_n<0$.In one step we change the signs of the numbers of that column.Then the new sum $S'$ of all the numbers of that configuration is equal to $S-2(a_1+a_2+...+a_n)>S$,contradicting the maximality of $S$!Thus there is no column such that the sum of the number inscribed in the cells of that column is negative.In the same manner we can prove that there will be no row such that the sum of the numbers inscribed in the cells of that row is negative.
This post has been edited 1 time. Last edited by huricane, Aug 4, 2015, 10:12 AM
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leonarda
6 posts
#3 • 2 Y
Y by Adventure10, Mango247
huricane wrote:
Let $S$ be the sum of all numbers in the cells of the table.From all the possible configurations(a configuration is the table after some steps) we will choose the configuration for which $S$ is maximal.Let's suppose that there is a column of that configuration with the propriety that the sum of the numbers inscribed in the column's cells is negative.Let $a_1,a_2,...,a_n$ be the numbers from that column.Then $a_1+a_2+...+a_n<0$.In one step we change the signs of the numbers of that column.Then the new sum $S'$ of all the numbers of that configuration is equal to $S-2(a_1+a_2+...+a_n)>S$,contradicting the maximality of $S$!Thus there is no column such that the sum of the number inscribed in the cells of that column is negative.In the same manner we can prove that there will be no row such that the sum of the numbers inscribed in the cells of that row is negative.

Seems to me the provided proof has nothing to do with the fact that starting with arbitrary table we have to reach the state where sums of all individual rows/columns has to be positive, using only the two allowed operations: changing the signs of elements within a certain row/column.
It appears that solution provided merely proves if we have somehow acquire already a table where a sum of every row/column happens to be positive than the total sum of entire table cannot be lower than table where a sum of any row/column would be negative.
Again, how that proves that we can actually reach the state where all rows/columns are positive if we use only two of the above operations.
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huricane
670 posts
#4 • 3 Y
Y by CrazyMathMan, Adventure10, Mango247
Note that there is a finite number of ways to switch the rows\columns.
So my solution proves that if we try all possibilities, one must be the good one.

Anyways, if you want an algorithmic solution,here is one:
Algorithm
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PCChess
548 posts
#5
Y by
First, observe that there are a finite number of arrangements of numbers since there are only 2 possible numbers for each cell. Let $S$ be the sum of all the numbers in the table. For each move, assume that there is one column or row that is negative (if not then we are done), and switch the signs of all the numbers in that row/column. Since the sum of the numbers in that row/column were originally negative, the sum now becomes positive, and $S$ increases. Hence, for each move, $S$ is strictly increasing. Furthermore, since there are a finite number of arrangements, after a finite number of moves, we will either reach a position where all the numbers in the table are nonnegative or the sums of all the rows/columns are nonnegative.
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Sprites
478 posts
#6
Y by
Consider the obvious Matrix representation of the problem.$$\begin{bmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} & . & . & . & a_{1,n}\\
. & . & . & . & . & . & . & .\\
. & . & . & . & . & . & . & .\\
. & . & . & . & . & . & . & .\\
. & . & . & . & . & . & . & .\\
. & . & . & . & . & . & . & .\\
a_{m,1} & a_{m,2} & a_{m,3} & a_{m,4} & . & . & . & a_{m,n}

\end{bmatrix}$$Lable the rows and columns as $C_i$ and $R_i$
Consider two subsequences of $\mathcal{C}_i$ and $\mathcal{R}_i$;$\mathcal{C}_{k_i}$ and $\mathcal{R}_{j_i}$ such that $$\sum_{j \in \mathcal{C}_{k_i} \text{ or } \mathcal{R}_{j_i}}<0$$Algorithm for positive sum:-
$\;\;\;\; \bullet$ Change signs of both $\mathcal{R}_{j_i}$ and $\mathcal{C}_{k_i}$.
$\;\;\;\; \bullet$ Consider $\mathcal{C}_{k_i} \cup\mathcal{R}_{j_i}$;it is an invariant i.e after we change signs of both the sets it doesn't change.
$\;\;\;\; \bullet$ FTSOC,assume that the algorithm get's stuck.WLOG,it get's stuck at $\mathcal{C}_1$ and $\mathcal{R}_1$ then we must have $$\sum_{j \in \mathcal{C}_{k_i} \text{ and } \mathcal{R}_{j_i}} <2 \cdot \left[\mathcal{C}_{k_i} \cup\mathcal{R}_{j_i} \right]$$Then simply change all $-$ve numbers in $\mathcal{R}_1$ and $\mathcal{C}_1$ to positive and re-apply the algorithm as long as all sums are positive.$\blacksquare$
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Saucepan_man02
1288 posts
#7 • 1 Y
Y by Sreepranad
Here is a quick solution:

Let $S$ denote the sum of all the numbers in every row and column. Then each square is counted twice in $S$.

Since the number of rows and columns are finite, we take the row/column $L$ with minimum sum $T$. If $T < 0$, then we switch the signs in $L$ and this step increases $S$. If $T \ge 0$, we stop this process.

We will prove that this algorithm will terminate after finite operations. Notice that $S$ is mono-variant during this algorithm. Since $S$ have take only finite number of values, this algorithm terminates after finite operations.
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mannshah1211
651 posts
#8
Y by
We'll use the following algorithm: While there is some row/column with negative sum in the grid, apply the flipping operation on that row or column. Observe that since there is only a finite number of possibilities for each number of the table (either $x$, or $-x$), thus, there is only a finite number of possible sums of the grid, and the sum only monotonically increases in one iteration of the algorithm the algorithm can't go on indefinitely, and clearly the end result of the algorithm will be a valid grid, since the algorithm stops at that point, so we are done.
This post has been edited 1 time. Last edited by mannshah1211, Jan 4, 2024, 12:08 PM
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AshAuktober
868 posts
#9
Y by
Consider, over all possible configurations reachable, the one with maximum sum of numbers. If it has a negative-summing column or row, flipping the signs in it increases the sum, contradiction. Thus it must satisfy the condition. $\square$
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isaacren
110 posts
#10
Y by
oh my you guys are good at this (both latex and math) :thumbup:
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Eka01
204 posts
#11
Y by
If a row or a column has a negative sum, flip the sign of its numbers. This makes the sum of that row/column positive. This also increases the overall sum of numbers on the board as all the rows except the one we disturbed are left untouched. Hence the sum of numbers will always increase, as long as we can make a move but the sum cannot increase indefinitely as it is bounded by the condition when all the numbers are positive, so we eventually become unable to make a move, but this happens precisely when all the sums are non negative, hence we are done.
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