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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
Product of f(m) multiple of odd integers
buzzychaoz   24
N 33 minutes ago by cursed_tangent1434
Source: China Team Selection Test 2016 Test 2 Day 2 Q4
Set positive integer $m=2^k\cdot t$, where $k$ is a non-negative integer, $t$ is an odd number, and let $f(m)=t^{1-k}$. Prove that for any positive integer $n$ and for any positive odd number $a\le n$, $\prod_{m=1}^n f(m)$ is a multiple of $a$.
24 replies
buzzychaoz
Mar 21, 2016
cursed_tangent1434
33 minutes ago
Domain of (a, b) satisfying inequality with fraction
Kunihiko_Chikaya   1
N 34 minutes ago by Mathzeus1024
Source: 2014 Kyoto University entrance exam/Science, Problem 4
For real constants $a,\ b$, define a function $f(x)=\frac{ax+b}{x^2+x+1}.$

Draw the domain of the points $(a,\ b)$ such that the inequality :

\[f(x) \leq f(x)^3-2f(x)^2+2\]

holds for all real numbers $x$.
1 reply
Kunihiko_Chikaya
Feb 26, 2014
Mathzeus1024
34 minutes ago
Generic Real-valued FE
lucas3617   0
34 minutes ago
$f: \mathbb{R} -> \mathbb{R}$, find all functions where $f(2x+f(2y-x))+f(-x)+f(y)=2f(x)+f(y-2x)+f(2y)$ for all $x$,$y \in \mathbb{R}$
0 replies
lucas3617
34 minutes ago
0 replies
Solve All 6 IMO 2024 Problems (42/42), New Framework Looking for Feedback
Blackhole.LightKing   4
N an hour ago by Blackhole.LightKing
Hi everyone,

I’ve been experimenting with a different way of approaching mathematical problem solving — a framework that emphasizes recursive structures and symbolic alignment rather than conventional step-by-step strategies.

Using this method, I recently attempted all six problems from IMO 2024 and was able to arrive at what I believe are valid full-mark solutions across the board (42/42 total score, by standard grading).

However, I don’t come from a formal competition background, so I’m sure there are gaps in clarity, communication, or even logic that I’m not fully aware of.

If anyone here is willing to take a look and provide feedback, I’d appreciate it — especially regarding:

The correctness and completeness of the proofs

Suggestions on how to make the ideas clearer or more elegant

Whether this approach has any broader potential or known parallels

I'm here to learn more and improve the presentation and thinking behind the work.

You can download the Solution here.

https://agi-origin.com/assets/pdf/AGI-Origin_IMO_2024_Solution.pdf


Thanks in advance,
— BlackholeLight0


4 replies
Blackhole.LightKing
Yesterday at 12:14 PM
Blackhole.LightKing
an hour ago
Inequalities from SXTX
sqing   20
N 2 hours ago by sqing
T702. Let $ a,b,c>0 $ and $ a+2b+3c=\sqrt{13}. $ Prove that $$ \sqrt{a^2+1} +2\sqrt{b^2+1} +3\sqrt{c^2+1} \geq 7$$S
T703. Let $ a,b $ be real numbers such that $ a+b\neq 0. $. Find the minimum of $ a^2+b^2+(\frac{1-ab}{a+b} )^2.$
T704. Let $ a,b,c>0 $ and $ a+b+c=3. $ Prove that $$ \frac{a^2+7}{(c+a)(a+b)} + \frac{b^2+7}{(a+b)(b+c)} +\frac{c^2+7}{(b+c)(c+a)}  \geq 6$$S
20 replies
sqing
Feb 18, 2025
sqing
2 hours ago
Theory of Equations
P162008   3
N 3 hours ago by P162008
Let $a,b,c,d$ and $e\in [-2,2]$ such that $\sum_{cyc} a = 0, \sum_{cyc} a^3 = 0, \sum_{cyc} a^5 = 10.$ Find the value of $\sum_{cyc} a^2.$
3 replies
P162008
Wednesday at 11:27 AM
P162008
3 hours ago
Fun & Simple puzzle
Kscv   7
N 3 hours ago by vanstraelen
$\angle DCA=45^{\circ},$ $\angle BDC=15^{\circ},$ $\overline{AC}=\overline{CB}$

$\angle ADC=?$
7 replies
Kscv
Apr 13, 2025
vanstraelen
3 hours ago
A problem involving modulus from JEE coaching
AshAuktober   7
N 5 hours ago by Jhonyboy
Solve over $\mathbb{R}$:
$$|x-1|+|x+2| = 3x.$$
(There are two ways to do this, one being bashing out cases. Try to find the other.)
7 replies
1 viewing
AshAuktober
Apr 21, 2025
Jhonyboy
5 hours ago
FB = BK , circumcircle and altitude related (In the World of Mathematics 516)
parmenides51   5
N Today at 4:05 AM by jasperE3
Let $BT$ be the altitude and $H$ be the intersection point of the altitudes of triangle $ABC$. Point $N$ is symmetric to $H$ with respect to $BC$. The circumcircle of triangle $ATN$ intersects $BC$ at points $F$ and $K$. Prove that $FB = BK$.

(V. Starodub, Kyiv)
5 replies
parmenides51
Apr 19, 2020
jasperE3
Today at 4:05 AM
Double Sum
P162008   0
Today at 3:22 AM
Let $\xi = \lim_{m \to\infty} \sum_{n=4}^{m} \sum_{k=2}^{n-2} \frac{1}{\binom{n}{k}}.$ If the value of $\xi$ can be written as $\frac{m}{n}$ where $m$ and $n$ are co-prime positive integers then compute the value of $m^3 + n^3.$
0 replies
P162008
Today at 3:22 AM
0 replies
Polynomial Limit
P162008   0
Today at 1:47 AM
Let $p = \lim_{y\to\infty} \left(\frac{2}{y^2} \left(\lim_{z\to\infty} \frac{1}{z^4} \left(\lim_{x\to\infty} \frac{((y^2 + y + 1)x^k + 1)^{z^2 + z + 1} - ((z^2 + z + 1)x^k + 1)^{y^2 + y + 1}}{x^{2k}}\right)\right)\right)^y$ where $k \in N$ and $q = \lim_{n\to\infty} \left(\frac{\binom{2n}{n}. n!}{n^n}\right)^{1/n}$ where $n \in N$. Find the value of $p.q.$
0 replies
P162008
Today at 1:47 AM
0 replies
Octagon Problem
Shiyul   4
N Today at 1:43 AM by Sid-darth-vater
The vertices of octagon $ABCDEFGH$ lie on the same circle. If $AB = BC = CD = DE = 11$ and $EF = FG = GH = HA = sqrt2$, what is the area of octagon $ABCDEFGH$?

I approached this problem by noticing that the area of the octagon is the area of the eight isoceles triangles with lengths $r$, $r$, and $sqrt2$ or 11. However, I didn't know how to find the radius. Can anyone give me a hint?
4 replies
Shiyul
Yesterday at 11:41 PM
Sid-darth-vater
Today at 1:43 AM
Polynomial Limit
P162008   2
N Today at 1:30 AM by P162008
If $P_{n}(x) = \prod_{k=1}^{n} \left(x + \frac{1}{2^k}\right) = \sum_{k=0}^{n} a_{k} x^k$ then find the value of $\lim_{n \to \infty} \frac{a_{n - 2}}{a_{n - 4}}.$
2 replies
P162008
Wednesday at 11:55 AM
P162008
Today at 1:30 AM
Sequence problem I never used
Sedro   1
N Yesterday at 9:01 PM by mathprodigy2011
Let $\{a_n\}_{n\ge 1}$ be a sequence of reals such that $a_1=1$ and $a_{n+1}a_n = 3a_n+2$ for all positive integers $n$. As $n$ grows large, the value of $a_{n+2}a_{n+1}a_n$ approaches the real number $M$. What is the greatest integer less than $M$?
1 reply
Sedro
Yesterday at 4:19 PM
mathprodigy2011
Yesterday at 9:01 PM
Adjacency matrices filled with free variables
darij grinberg   2
N Sep 12, 2013 by MatthewOrtega
Source: an idea looking for applications
Some time ago I have tried to prove Hall's matching/marriage theorem on my own. As a result, I obtained a rather tedious linear-algebraic proof which I don't deem particularly interesting on its own, but it contains an idea that I would love to see applied at some more serious questions.

The proof can be found in a note called "An algebraic approach to Hall's matching theorem" on my website. The note itself is a pain to read and is just there to make sure that the proof has no flaws. If you are interested in the proof, try out the shortened version of the note, which is still 7 pages long (the first 2 of them being prerequisites, however). Given the fact that the Hall theorem can be shown on half a page (though I don't find the proofs quite natural, honestly), this is not quite a sensation.

Anyway, as I said, what matters is the idea, which is basically to fill an adjacency matrix or another kind of characteristic matrix not with $ 0$'s and $ 1$'s but with $ 0$'s and free variables. In the case of the Hall theorem, we have a bipartite graph with $ n$ blue vertices $ B_1$, $ B_2$, ..., $ B_n$ and $ n$ green vertices $ G_1$, $ G_2$, ..., $ G_n$. (Have you guessed that B and G stand for blue and green?) We take an arbitrary field $ K$, for instance $ K = \mathbb{Q}$, and denote by $ L$ the field of all rational functions of $ n^2$ indeterminates $ X_{1,1}$, $ X_{1,2}$, ..., $ X_{n,n}$ (one indeterminate $ X_{i,j}$ for each pair $ \left(i,j\right)\in\left\{1,2,...,n\right\}^2$) over $ K$. Then, we consider the matrix $ S\in\mathrm{M}_n\left(L\right)$ which has, in its $ i$-th row and $ j$-th column, the entry $ \left\{\begin{array}{c} X_{i,j},\text{ if }G_{j}\text{ is connected with }B_{i} ; \\
0,\text{ if }G_{j}\text{ is not connected with }B_{i} \end{array}\right.$. This matrix "represents" the bipartite graph and has the nice property that it is invertible if and only if the bipartite graph has a perfect matching. My proof of Hall's theorem exploits the properties of this matrix (alas, in a rather ugly way).

Note that $ S$ is not an adjacency matrix, but we could easily define a similar variation on the concept of adjacency matrices, or incidence matrices. It is less clear to me, however, under which circumstances such variations can be of use. Therefore two questions:

- Is the idea/trick of filling characteristic matrices with free variables new?
- Can anyone find a less simple application of the trick?

Darij
2 replies
darij grinberg
Oct 3, 2007
MatthewOrtega
Sep 12, 2013
Adjacency matrices filled with free variables
G H J
Source: an idea looking for applications
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darij grinberg
6555 posts
#1 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Some time ago I have tried to prove Hall's matching/marriage theorem on my own. As a result, I obtained a rather tedious linear-algebraic proof which I don't deem particularly interesting on its own, but it contains an idea that I would love to see applied at some more serious questions.

The proof can be found in a note called "An algebraic approach to Hall's matching theorem" on my website. The note itself is a pain to read and is just there to make sure that the proof has no flaws. If you are interested in the proof, try out the shortened version of the note, which is still 7 pages long (the first 2 of them being prerequisites, however). Given the fact that the Hall theorem can be shown on half a page (though I don't find the proofs quite natural, honestly), this is not quite a sensation.

Anyway, as I said, what matters is the idea, which is basically to fill an adjacency matrix or another kind of characteristic matrix not with $ 0$'s and $ 1$'s but with $ 0$'s and free variables. In the case of the Hall theorem, we have a bipartite graph with $ n$ blue vertices $ B_1$, $ B_2$, ..., $ B_n$ and $ n$ green vertices $ G_1$, $ G_2$, ..., $ G_n$. (Have you guessed that B and G stand for blue and green?) We take an arbitrary field $ K$, for instance $ K = \mathbb{Q}$, and denote by $ L$ the field of all rational functions of $ n^2$ indeterminates $ X_{1,1}$, $ X_{1,2}$, ..., $ X_{n,n}$ (one indeterminate $ X_{i,j}$ for each pair $ \left(i,j\right)\in\left\{1,2,...,n\right\}^2$) over $ K$. Then, we consider the matrix $ S\in\mathrm{M}_n\left(L\right)$ which has, in its $ i$-th row and $ j$-th column, the entry $ \left\{\begin{array}{c} X_{i,j},\text{ if }G_{j}\text{ is connected with }B_{i} ; \\
0,\text{ if }G_{j}\text{ is not connected with }B_{i} \end{array}\right.$. This matrix "represents" the bipartite graph and has the nice property that it is invertible if and only if the bipartite graph has a perfect matching. My proof of Hall's theorem exploits the properties of this matrix (alas, in a rather ugly way).

Note that $ S$ is not an adjacency matrix, but we could easily define a similar variation on the concept of adjacency matrices, or incidence matrices. It is less clear to me, however, under which circumstances such variations can be of use. Therefore two questions:

- Is the idea/trick of filling characteristic matrices with free variables new?
- Can anyone find a less simple application of the trick?

Darij
This post has been edited 1 time. Last edited by darij grinberg, Nov 18, 2008, 12:56 PM
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Ravi B
3083 posts
#2 • 4 Y
Y by Adventure10, Mango247, and 2 other users
The idea of free variables for matching has been used before. For example, Tutte (1947) showed an analogous if-and-only-if result for non-bipartite graphs, in which the matrix is a certain skew-symmetric matrix.

The idea became popular in computer science about 20 years ago, because it led to randomized parallel algorithms for matching. The trick is to assign random values to the free variables. With high probability, if the graph has a matching, then the numerical determinant will be non-zero. See Mulmuley, Vazirani, and Vazirani (1987), "Matching is as easy as matrix inversion", Combinatorica.
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MatthewOrtega
1 post
#3 • 2 Y
Y by Adventure10, Mango247
Ravi B wrote:
The idea of free variables for matching has been used before. For example, Tutte (1947) showed an analogous if-and-only-if result for non-bipartite graphs, in which the matrix is a certain skew-symmetric matrix.

The idea became popular in computer science about 20 years ago, because it led to randomized parallel algorithms for matching. The trick is to assign random values to the free variables. With high probability, if the graph has a matching, then the numerical determinant will be non-zero. See Mulmuley, Vazirani, and Vazirani (1987), "Matching is as easy as matrix inversion", Combinatorica.
hello friend I am working on similar concept so can you tell about the key points which I must take care of?Waiting for reply thanks in advance:)
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