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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
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density over modulo M
SomeGuy3335   3
N 12 minutes ago by ja.
Let $M$ be a positive integer and let $\alpha$ be an irrational number. Show that for every integer $0\leq a < M$, there exists a positive integer $n$ such that $M \mid \lfloor{n \alpha}\rfloor-a$.
3 replies
SomeGuy3335
Apr 20, 2025
ja.
12 minutes ago
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AGI-Origin Solves Full IMO 2020–2024 Benchmark Without Solver (30/30) beat Alpha
AGI-Origin   0
33 minutes ago
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
0 replies
+4 w
AGI-Origin
33 minutes ago
0 replies
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Diophantine equation !
ComplexPhi   5
N an hour ago by aops.c.c.
Source: Romania JBMO TST 2015 Day 1 Problem 4
Solve in nonnegative integers the following equation :
$$21^x+4^y=z^2$$
5 replies
ComplexPhi
May 14, 2015
aops.c.c.
an hour ago
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Combo problem
soryn   0
an hour ago
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.
0 replies
soryn
an hour ago
0 replies
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construct a triangle given circumcenter, incenter and one excenter
parmenides51   1
N Aug 5, 2018 by NikoIsLife
Source: Quantum - English M59
Given three points $O,I$, and $E$ in the plane construct a triangle such that its circumcenter is at $O$, its incenter is at $I$, and one of its excenters is at $E$.

(B. Martynov)
1 reply
parmenides51
Aug 5, 2018
NikoIsLife
Aug 5, 2018
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construct a triangle given circumcenter, incenter and one excenter
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Reveal topic tags
geometry
construction
excenter
Circumcenter
incenter
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Source: Quantum - English M59
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parmenides51
30630 posts
parmenides51
#1 h  Aug 5, 2018, 10:38 AM • 2 Y
Y by Adventure10, Mango247
Given three points $O,I$, and $E$ in the plane construct a triangle such that its circumcenter is at $O$, its incenter is at $I$, and one of its excenters is at $E$.

(B. Martynov)
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NikoIsLife
9657 posts
NikoIsLife
#3 h  Aug 5, 2018, 10:49 AM • 2 Y
Y by Adventure10, Mango247
1) Let the midpoint of $IE$ by $M$.
2) Draw a circle centered at $O$ which passes through $M$. Call this circle $\Omega$.
3) Extend line $MI$ to meet $\Omega$ again at $A$.
4) Draw a circle centered at $M$ which passes through $I$. Call this circle $\omega$.
5) Circles $\omega$ and $\Omega$ should intersect at two points. Call these intersections $B$ and $C$.

The triangle that you wanted to construct was $\Delta ABC$.

You can easily verify this by the Incenter-Excenter Theorem.
This post has been edited 1 time. Last edited by NikoIsLife, Aug 5, 2018, 10:50 AM
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