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Inspired by Jackson0423
sqing   1
N 22 minutes ago by sqing
Source: Own
Let $ a, b, c>0 $ and $ a^2 + b^2 =c(a + b). $ Prove that
$$   \frac{b^2 +bc+ c^2}{ a(a +b+  c)} \geq 2\sqrt 3-3$$
1 reply
1 viewing
sqing
37 minutes ago
sqing
22 minutes ago
Combo problem
soryn   1
N 22 minutes ago by soryn
The school A has m1 boys and m2 girls, and ,the school B has n1 boys and n2 girls. Each school is represented by one team formed by p students,boys and girls. If f(k) is the number of cases for which,the twice schools has,togheter k girls, fund f(k) and the valute of k, for which f(k) is maximum.
1 reply
1 viewing
soryn
Today at 6:33 AM
soryn
22 minutes ago
Incredible combinatorics problem
A_E_R   2
N 26 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
3 hours ago
quacksaysduck
26 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 2 hours 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
2 hours 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 3 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
3 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
Projection of vertex onto bisectors
randomusername   8
N Apr 14, 2025 by AshAuktober
Source: ITAMO 2016, Problem 1
Let $ABC$ be a triangle, and let $D$ and $E$ be the orthogonal projections of $A$ onto the internal bisectors from $B$ and $C$. Prove that $DE$ is parallel to $BC$.
8 replies
randomusername
May 11, 2016
AshAuktober
Apr 14, 2025
Projection of vertex onto bisectors
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G H BBookmark kLocked kLocked NReply
Source: ITAMO 2016, Problem 1
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randomusername
1059 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be a triangle, and let $D$ and $E$ be the orthogonal projections of $A$ onto the internal bisectors from $B$ and $C$. Prove that $DE$ is parallel to $BC$.
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bobthesmartypants
4337 posts
#2 • 2 Y
Y by Adventure10, Mango247
Let $I$ be the incenter. Then $\angle IDE = \angle IAE$ from $ADIE$ concyclic. Then $\angle IAE = \angle CAE-\angle CAI = 90-\dfrac{1}{2}\angle C - \dfrac{1}{2}\angle A = \dfrac{1}{2}\angle B$ so $\angle IDE = \angle IBC$, so $DE\parallel BC$.
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randomusername
1059 posts
#4 • 2 Y
Y by MSTang, Adventure10
If $AD,AE$ hits $BC$ at $K,L$, then $K,L$ is the reflection of $A$ on the $B$-bisector and $C$-bisector, resp., and consequently $DE$ is the image of $KL$ after a homothety from $A$ with ratio $1/2$, implying $DE||KL\equiv BC$.
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jayme
9781 posts
#5 • 1 Y
Y by Adventure10
Dear Mathlinkers,
have a look at

http://jl.ayme.pagesperso-orange.fr/Docs/An%20unlikely%20concurrence.pdf p. 3

Sincerely
Jean-Louis
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EulerMacaroni
851 posts
#6 • 3 Y
Y by AlastorMoody, Adventure10, Mango247
We're done by the right angles on intouch chord (well, midline in this case) lemma.
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TheDarkPrince
3042 posts
#7 • 2 Y
Y by AlastorMoody, Adventure10
Let $X,Y$ be feet of $A$ on angle bisectors of $B$ and $C$ respectively. We have \[\angle IXY = \angle IAY = \angle CAY - \angle CAI = 90^{\circ} - \frac{C}{2} - \frac{A}{2} = \frac{B}{2}.\]
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Taha-R
45 posts
#8
Y by
It's enough to prove $ \angle EDB $ = $ \angle DBC $ .
Claim 1 : AEID is cyclic .
Proof : $ \angle AEC $ + $ \angle ADB $ = $ 180^\circ $ .
Hence, $ \angle EDI $ = $ \angle IAE $ = $ \angle BAI $ - $ \angle BAE $ .
$ \angle BAE $ = $ 90^\circ $ - $ \angle AFC $ and $ \angle AFC $ = $ 180^\circ $ - $ \angle BAC $ - $ \angle FCA $ . (from the sum of the angles of the triangle FAC)
We know that $ \angle BAC $ + $ \angle ABC $ +$ \angle ACB $ = $ 180^\circ $ (*)
And $ \angle ABC $ = 2$ \angle ABG $ = 2$ \angle GBC $ ,...(**)
Due to the (*) and (**) , the proof is complete !
Attachments:
This post has been edited 1 time. Last edited by Taha-R, Sep 21, 2023, 6:00 PM
Reason: small problem !
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Rohit-2006
224 posts
#9
Y by
Here is my solution
Note that $\alpha$=$\angle OAE$
Attachments:
This post has been edited 1 time. Last edited by Rohit-2006, Feb 27, 2025, 8:13 PM
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AshAuktober
993 posts
#10
Y by
Note that $\angle EDI =\angle EAI= \frac{\pi-A-C}2=\frac B2 = \angle DBC$, so we're done.
This post has been edited 1 time. Last edited by AshAuktober, Apr 14, 2025, 7:50 AM
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