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Advanced topics in Inequalities
va2010   23
N 6 minutes ago by Novmath
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
23 replies
va2010
Mar 7, 2015
Novmath
6 minutes ago
JBMO TST Bosnia and Herzegovina 2022 P3
Motion   7
N 11 minutes ago by cafer2861
Source: JBMO TST Bosnia and Herzegovina 2022
Let $ABC$ be an acute triangle. Tangents on the circumscribed circle of triangle $ABC$ at points $B$ and $C$ intersect at point $T$. Let $D$ and $E$ be a foot of the altitudes from $T$ onto $AB$ and $AC$ and let $M$ be the midpoint of $BC$. Prove:
A) Prove that $M$ is the orthocenter of the triangle $ADE$.
B) Prove that $TM$ cuts $DE$ in half.
7 replies
1 viewing
Motion
May 21, 2022
cafer2861
11 minutes ago
hard problem
Cobedangiu   5
N 23 minutes ago by arqady
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$
5 replies
Cobedangiu
Yesterday at 1:51 PM
arqady
23 minutes ago
AGI-Origin Solves Full IMO 2020–2024 Benchmark Without Solver (30/30) beat Alpha
AGI-Origin   1
N 31 minutes ago by mannshah1211
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
1 reply
1 viewing
AGI-Origin
an hour ago
mannshah1211
31 minutes ago
Radical Axes and circles
mathprodigy2011   4
N an hour ago by spiderman0
Can someone explain how to do this purely geometrically?
4 replies
mathprodigy2011
Today at 1:58 AM
spiderman0
an hour ago
density over modulo M
SomeGuy3335   3
N an hour 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.
an hour ago
Diophantine equation !
ComplexPhi   5
N 2 hours 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.
2 hours ago
Combo problem
soryn   0
2 hours 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
2 hours ago
0 replies
Parity and sets
betongblander   7
N 2 hours ago by ihategeo_1969
Source: Brazil National Olympiad 2020 5 Level 3
Let $n$ and $k$ be positive integers with $k$ $\le$ $n$. In a group of $n$ people, each one or always
speak the truth or always lie. Arnaldo can ask questions for any of these people
provided these questions are of the type: “In set $A$, what is the parity of people who speak to
true? ”, where $A$ is a subset of size $ k$ of the set of $n$ people. The answer can only
be $even$ or $odd$.
a) For which values of $n$ and $k$ is it possible to determine which people speak the truth and
which people always lie?
b) What is the minimum number of questions required to determine which people
speak the truth and which people always lie, when that number is finite?
7 replies
betongblander
Mar 18, 2021
ihategeo_1969
2 hours ago
Mount Inequality erupts on a sequence :o
GrantStar   88
N 2 hours ago by Nari_Tom
Source: 2023 IMO P4
Let $x_1,x_2,\dots,x_{2023}$ be pairwise different positive real numbers such that
\[a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}\]is an integer for every $n=1,2,\dots,2023.$ Prove that $a_{2023} \geqslant 3034.$
88 replies
GrantStar
Jul 9, 2023
Nari_Tom
2 hours ago
JBMO Shortlist 2022 N1
Lukaluce   8
N 2 hours ago by godchunguus
Source: JBMO Shortlist 2022
Determine all pairs $(k, n)$ of positive integers that satisfy
$$1! + 2! + ... + k! = 1 + 2 + ... + n.$$
8 replies
Lukaluce
Jun 26, 2023
godchunguus
2 hours ago
geometry parabola problem
smalkaram_3549   10
N Apr 13, 2025 by ReticulatedPython
How would you solve this without using calculus?
10 replies
smalkaram_3549
Apr 11, 2025
ReticulatedPython
Apr 13, 2025
geometry parabola problem
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smalkaram_3549
168 posts
#1 • 1 Y
Y by Kizaruno
How would you solve this without using calculus?
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ShadowDragonRules
366 posts
#2 • 1 Y
Y by Kizaruno
hmm... I would suggest to find a relationship of graphing this through first finding the parabola's graph to find the eventual diameter. BTW
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mathmax001
14 posts
#3 • 1 Y
Y by Kizaruno
smalkaram_3549 wrote:
How would you solve this without using calculus?

is it $ R=\frac{\sqrt{11}}{2} $ ?
I found it solving a quadratic equation.
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rchokler
2965 posts
#4
Y by
Solution 1:

$y'=2x$, so the normal line through $(a,a^2)$ is $y=a^2-\frac{x-a}{2a}$,

Put $(x,y)=(0,3)$ to get $a^2+\frac{1}{2}=3\implies a^2=\frac{5}{2}$.

So the points of tangency are $\left(\pm\frac{\sqrt{10}}{2},\frac{5}{2}\right)$.

$r=\sqrt{\left(\frac{\sqrt{10}}{2}-0\right)^2+\left(\frac{5}{2}-3\right)^2}=\sqrt{\frac{5}{2}+\frac{1}{4}}=\frac{\sqrt{11}}{2}$.

Solution 2:

$r(t)=\text{dist}[(0,3),(t,t^2)]=\sqrt{t^2+(t^2-3)^2}=\sqrt{t^4-5t^2+9}=\sqrt{\left(t^2-\frac{5}{2}\right)^2+\frac{11}{4}}\geq\frac{\sqrt{11}}{2}$ with equality when $t^2=\frac{5}{2}$.
This post has been edited 1 time. Last edited by rchokler, Apr 11, 2025, 11:43 PM
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smalkaram_3549
168 posts
#5 • 1 Y
Y by Kizaruno
mathmax001 wrote:
smalkaram_3549 wrote:
How would you solve this without using calculus?

is it $ R=\frac{\sqrt{11}}{2} $ ?
I found it solving a quadratic equation.

yes it is. How did you do that? I got it using calculus but can't figure out the algebraic method
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joeym2011
493 posts
#6
Y by
We let the circle equation be $x^2+(y-3)^2=r^2$. We can solve for $y$:
$$y^2-6y+9+y=r^2.$$Due to tangency, there is only one solution of $y$, and we have a double root and
$$5^2=4\left(9-r^2\right)\implies r=\boxed{\frac{\sqrt{11}}2}.$$@rchokler's solution 2 is also algebraic and works well.
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ReticulatedPython
571 posts
#7
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Solution
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mathmax001
14 posts
#8
Y by
ReticulatedPython wrote:
Solution

This is exactly the method I used .

If you don't understand the meaning of this sentence " Since we want the roots to be negations of each other, the discriminant is equal to $0.$ " , I'll explain it more :
If the discriminant $ 5^2-4(9-r^2) $ were $ \neq 0 $ then the equation would have 2 different solutions for $ x^2 $ ( because the discriminant is positive ) , that means four 4 points of tangency which is not true .
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jb2015007
1915 posts
#9 • 2 Y
Y by Kizaruno, ReticulatedPython
sol

also hi reticulated python!

also sorry for a bad sol
i didnt have time to explain everything clearly
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smbellanki
180 posts
#11
Y by
The equation of the circle is \( x^2 + (y - 3)^2 = r^2 \). Subbing in \( y = x^2 \) to find the intersection gets \( x^2 + (x^2 - 3)^2 = r^2 \) which is \( x^4 - 5x^2 + 9 - r^2 = 0 \) now since, the circle only intersects the parabola twice there must be 2 double roots so we consider the determinant which is \( 5^2 - 4(9 - r^2) = 0 \) which becomes \( 4r^2 - 11 = 0 \) so \( r = \sqrt{11}/2 \) only since the radius can't be negative.
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ReticulatedPython
571 posts
#12
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jb2015007 wrote:
sol

also hi reticulated python!

also sorry for a bad sol
i didnt have time to explain everything clearly

Looks good to me!
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