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Incredible combinatorics problem
A_E_R   2
N 11 minutes ago by quacksaysduck
Source: Turkmenistan Math Olympiad - 2025
For any integer n, prove that there are exactly 18 integer whose sum and the sum of the fifth powers of each are equal to the integer n.
2 replies
A_E_R
2 hours ago
quacksaysduck
11 minutes ago
What is the likelihood the last card left in the deck is black?
BEHZOD_UZ   1
N an hour ago by sami1618
Source: Yandex Uzbekistan Coding and Math Contest 2025
You have a deck of cards containing $26$ black and $13$ red cards. You pull out $2$ cards, one after another, and check their colour. If both cards are the same colour, then a black card is added to the deck. However, if the cards are of different colours, then a red card is used to replace them. Once the cards are taken out of the deck, they are not returned to the deck, and thus the number of cards keeps reducing. What is the likelihood the last card left in the deck is black?
1 reply
BEHZOD_UZ
an hour ago
sami1618
an hour ago
abc(a+b+c)=3, show that prod(a+b)>=8 [Indian RMO 2012(b) Q4]
Potla   31
N an hour ago by sqing
Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that we have
\[(a+b)(b+c)(c+a)\geq 8.\]
Also determine the case of equality.
31 replies
Potla
Dec 2, 2012
sqing
an hour ago
AGI-Origin Solves Full IMO 2020–2024 Benchmark Without Solver (30/30) beat Alpha
AGI-Origin   10
N an hour ago by TestX01
Hello IMO community,

I’m sharing here a full 30-problem solution set to all IMO problems from 2020 to 2024.

Standard AI: Prompt --> Symbolic Solver (SymPy, Geometry API, etc.)

Unlike AlphaGeometry or symbolic math tools that solve through direct symbolic computation, AGI-Origin operates via recursive symbolic cognition.

AGI-Origin:
Prompt --> Internal symbolic mapping --> Recursive contradiction/repair --> Structural reasoning --> Human-style proof

It builds human-readable logic paths by recursively tracing contradictions, repairing structure, and collapsing ambiguity — not by invoking any external symbolic solver.

These results were produced by a recursive symbolic cognition framework called AGI-Origin, designed to simulate semi-AGI through contradiction collapse, symbolic feedback, and recursion-based error repair.

These were solved without using any symbolic computation engine or solver.
Instead, the solutions were derived using a recursive symbolic framework called AGI-Origin, based on:
- Contradiction collapse
- Self-correcting recursion
- Symbolic anchoring and logical repair

Full PDF: [Upload to Dropbox/Google Drive/Notion or arXiv link when ready]

This effort surpasses AlphaGeometry’s previous 25/30 mark by covering:
- Algebra
- Combinatorics
- Geometry
- Functional Equations

Each solution follows a rigorous logical path and is written in fully human-readable format — no machine code or symbolic solvers were used.

I would greatly appreciate any feedback on the solution structure, logic clarity, or symbolic methodology.

Thank you!

— AGI-Origin Team
AGI-Origin.com
10 replies
AGI-Origin
6 hours ago
TestX01
an hour ago
FE solution too simple?
Yiyj1   6
N an hour ago by Primeniyazidayi
Source: 101 Algebra Problems from the AMSP
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$f(f(x)+y) = f(x^2-y)+4f(x)y$$holds for all pairs of real numbers $(x,y)$.

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?
6 replies
Yiyj1
Apr 9, 2025
Primeniyazidayi
an hour ago
Two very hard parallel
jayme   5
N 2 hours ago by jayme
Source: own inspired by EGMO
Dear Mathlinkers,

1. ABC a triangle
2. D, E two point on the segment BC so that BD = DE= EC
3. M, N the midpoint of ED, AE
4. H the orthocenter of the acutangle triangle ADE
5. 1, 2 the circumcircle of the triangle DHM, EHN
6. P, Q the second point of intersection of 1 and BM, 2 and CN
7. U, V the second points of intersection of 2 and MN, PQ.

Prove : UV is parallel to PM.

Sincerely
Jean-Louis
5 replies
jayme
Yesterday at 12:46 PM
jayme
2 hours ago
Number theory
XAN4   1
N 2 hours ago by NTstrucker
Source: own
Prove that there exists infinitely many positive integers $x,y,z$ such that $x,y,z\ne1$ and $x^x\cdot y^y=z^z$.
1 reply
XAN4
Apr 20, 2025
NTstrucker
2 hours ago
R+ FE with arbitrary constant
CyclicISLscelesTrapezoid   25
N 3 hours ago by DeathIsAwe
Source: APMO 2023/4
Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f \colon \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that \[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\]
25 replies
CyclicISLscelesTrapezoid
Jul 5, 2023
DeathIsAwe
3 hours ago
Combo with cyclic sums
oVlad   1
N 4 hours ago by ja.
Source: Romania EGMO TST 2017 Day 1 P4
In $p{}$ of the vertices of the regular polygon $A_0A_1\ldots A_{2016}$ we write the number $1{}$ and in the remaining ones we write the number $-1.{}$ Let $x_i{}$ be the number written on the vertex $A_i{}.$ A vertex is good if \[x_i+x_{i+1}+\cdots+x_j>0\quad\text{and}\quad x_i+x_{i-1}+\cdots+x_k>0,\]for any integers $j{}$ and $k{}$ such that $k\leqslant i\leqslant j.$ Note that the indices are taken modulo $2017.$ Determine the greatest possible value of $p{}$ such that, regardless of numbering, there always exists a good vertex.
1 reply
oVlad
Yesterday at 1:41 PM
ja.
4 hours ago
Incircle of a triangle is tangent to (ABC)
amar_04   11
N 4 hours ago by Nari_Tom
Source: XVII Sharygin Correspondence Round P18
Let $ABC$ be a scalene triangle, $AM$ be the median through $A$, and $\omega$ be the incircle. Let $\omega$ touch $BC$ at point $T$ and segment $AT$ meet $\omega$ for the second time at point $S$. Let $\delta$ be the triangle formed by lines $AM$ and $BC$ and the tangent to $\omega$ at $S$. Prove that the incircle of triangle $\delta$ is tangent to the circumcircle of triangle $ABC$.
11 replies
amar_04
Mar 2, 2021
Nari_Tom
4 hours ago
a