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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

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
black and
red cards. You pull out
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
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
be positive real numbers such that
Prove that we have
![\[(a+b)(b+c)(c+a)\geq 8.\]](//latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
Also determine the case of equality.


![\[(a+b)(b+c)(c+a)\geq 8.\]](http://latex.artofproblemsolving.com/2/a/6/2a63f19de634fce70ff4ebc02ad13d442a20c378.png)
Also determine the case of equality.
31 replies

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
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
FE solution too simple?
Yiyj1 6
N
an hour ago
by Primeniyazidayi
Source: 101 Algebra Problems from the AMSP
Find all functions
such that the equality
holds for all pairs of real numbers
.
My solution
I feel like my solution is too simple. Is there something I did wrong or something I missed?



My solution
Clearly,
is an obvious solution. Now, let
. Then, we have
or
. Therefore, the solutions are
.





I feel like my solution is too simple. Is there something I did wrong or something I missed?
6 replies
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
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
Number theory
XAN4 1
N
2 hours ago
by NTstrucker
Source: own
Prove that there exists infinitely many positive integers
such that
and
.



1 reply

R+ FE with arbitrary constant
CyclicISLscelesTrapezoid 25
N
3 hours ago
by DeathIsAwe
Source: APMO 2023/4
Let
be a given positive real and
be the set of all positive reals. Find all functions
such that



![\[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\]](http://latex.artofproblemsolving.com/b/0/6/b069e7b7ec1e277f4a4ce85a99434fdb54eb02f3.png)
25 replies
Combo with cyclic sums
oVlad 1
N
4 hours ago
by ja.
Source: Romania EGMO TST 2017 Day 1 P4
In
of the vertices of the regular polygon
we write the number
and in the remaining ones we write the number
Let
be the number written on the vertex
A vertex is good if
for any integers
and
such that
Note that the indices are taken modulo
Determine the greatest possible value of
such that, regardless of numbering, there always exists a good vertex.






![\[x_i+x_{i+1}+\cdots+x_j>0\quad\text{and}\quad x_i+x_{i-1}+\cdots+x_k>0,\]](http://latex.artofproblemsolving.com/b/f/4/bf4cf984af7f310c1eb90b6b37126bfca9181305.png)





1 reply
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
be a scalene triangle,
be the median through
, and
be the incircle. Let
touch
at point
and segment
meet
for the second time at point
. Let
be the triangle formed by lines
and
and the tangent to
at
. Prove that the incircle of triangle
is tangent to the circumcircle of triangle
.

















11 replies
