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Solve All 6 IMO 2024 Problems (42/42), New Framework Looking for Feedback
Blackhole.LightKing   2
N a few seconds ago by DottedCaculator
Hi everyone,

I’ve been experimenting with a different way of approaching mathematical problem solving — a framework that emphasizes recursive structures and symbolic alignment rather than conventional step-by-step strategies.

Using this method, I recently attempted all six problems from IMO 2024 and was able to arrive at what I believe are valid full-mark solutions across the board (42/42 total score, by standard grading).

However, I don’t come from a formal competition background, so I’m sure there are gaps in clarity, communication, or even logic that I’m not fully aware of.

If anyone here is willing to take a look and provide feedback, I’d appreciate it — especially regarding:

The correctness and completeness of the proofs

Suggestions on how to make the ideas clearer or more elegant

Whether this approach has any broader potential or known parallels

I'm here to learn more and improve the presentation and thinking behind the work.

You can download the Solution here.

https://agi-origin.com/assets/pdf/AGI-Origin_IMO_2024_Solution.pdf


Thanks in advance,
— BlackholeLight0


2 replies
Blackhole.LightKing
3 hours ago
DottedCaculator
a few seconds ago
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circle geometry showing perpendicularity
Kyj9981   4
N 7 minutes ago by cj13609517288
Two circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$. A line through $B$ intersects $\omega_1$ and $\omega_2$ at points $C$ and $D$, respectively. Line $AD$ intersects $\omega_1$ at point $E \neq A$, and line $AC$ intersects $\omega_2$ at point $F \neq A$. If $O$ is the circumcenter of $\triangle AEF$, prove that $OB \perp CD$.
4 replies
1 viewing
Kyj9981
Mar 18, 2025
cj13609517288
7 minutes ago
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Prove excircle is tangent to circumcircle
sarjinius   8
N 17 minutes ago by Lyzstudent
Source: Philippine Mathematical Olympiad 2025 P4
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on side $BC$. Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$, the extension of $AE$ past $E$, and the extension of $AF$ past $F$. Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$.
8 replies
sarjinius
Mar 9, 2025
Lyzstudent
17 minutes ago
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Number Theory
AnhQuang_67   1
N 22 minutes ago by GreekIdiot
Source: HSGSO 2024
Let $p$ be an even prime number and a sequence $\{a_n\}_{n=1}^{+\infty}$ satisfy $$a_1=1, a_2=2$$and $$a_{n+2}=2\cdot a_{n+1}+3\cdot a_n, \forall n \geqslant 1$$Prove that always exists positive integer $k$ satisfying for all positive integers $n$, then $a_n \ne k \mod{p}$.

P/s: $\ne$ is "not congruence"
1 reply
AnhQuang_67
an hour ago
GreekIdiot
22 minutes ago
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IMO Shortlist 2014 N6
hajimbrak   28
N 43 minutes ago by MajesticCheese
Let $a_1 < a_2 < \cdots <a_n$ be pairwise coprime positive integers with $a_1$ being prime and $a_1 \ge n + 2$. On the segment $I = [0, a_1 a_2 \cdots a_n ]$ of the real line, mark all integers that are divisible by at least one of the numbers $a_1 , \ldots , a_n$ . These points split $I$ into a number of smaller segments. Prove that the sum of the squares of the lengths of these segments is divisible by $a_1$.

Proposed by Serbia
28 replies
hajimbrak
Jul 11, 2015
MajesticCheese
43 minutes ago
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3 knightlike moves is enough
sarjinius   3
N an hour ago by JollyEggsBanana
Source: Philippine Mathematical Olympiad 2025 P6
An ant is on the Cartesian plane. In a single move, the ant selects a positive integer $k$, then either travels [list]
[*] $k$ units vertically (up or down) and $2k$ units horizontally (left or right); or
[*] $k$ units horizontally (left or right) and $2k$ units vertically (up or down).
[/list]
Thus, for any $k$, the ant can choose to go to one of eight possible points.
Prove that, for any integers $a$ and $b$, the ant can travel from $(0, 0)$ to $(a, b)$ using at most $3$ moves.
3 replies
sarjinius
Mar 9, 2025
JollyEggsBanana
an hour ago
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Why is the old one deleted?
EeEeRUT   15
N an hour ago by Tuvshuu
Source: EGMO 2025 P1
For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
15 replies
EeEeRUT
Apr 16, 2025
Tuvshuu
an hour ago
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My problem that I could not find(NT)
Nuran2010   0
an hour ago
Source: Own
While I was thinking on some other geometry problem, a NT problem came to my mind. Despite some tries(which were mostly order), I could not find a way to solve the problem. As I searched, this problem has never been posted before. Here is the problem.

Find all positive integers $a,b$ such that:
$a+b|2^{ab}+1$

Moreover, I wonder if there is a way to solve the question in this variant:

Find all positive integers $a,b,n$ such that:
$a+b|n^{ab}+1$
0 replies
Nuran2010
an hour ago
0 replies
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Classic graph theory lemma?
eulerleonhardfan   1
N an hour ago by eulerleonhardfan
$n \in \mathbb{N}$ is given, $A$, $B$ are graphs on the same set of $n$ nodes, having $a, b$ connected components respectively. Prove that $A \cup B$ has at least $a+b-n$ connected components.
1 reply
eulerleonhardfan
an hour ago
eulerleonhardfan
an hour ago
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Min Number of Subsets of Strictly Increasing
taptya17   5
N an hour ago by kotmhn
Source: India EGMO TST 2025 Day 1 P1
Let $n$ be a positive integer. Initially the sequence $0,0,\cdots,0$ ($n$ times) is written on the board. In each round, Ananya choses an integer $t$ and a subset of the numbers written on the board and adds $t$ to all of them. What is the minimum number of rounds in which Ananya can make the sequence on the board strictly increasing?

Proposed by Shantanu Nene
5 replies
taptya17
Dec 13, 2024
kotmhn
an hour ago
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Nice inequality
sqing   3
N 2 hours ago by Oksutok
Source: WYX
Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be real numbers . Prove that : There exist positive integer $k\in \{1,2,\cdots,n\}$ such that $$\sum_{i=1}^{n}\{kx_i\}(1-\{kx_i\})<\frac{n-1}{6}.$$Where $\{x\}=x-\left \lfloor x \right \rfloor.$
3 replies
sqing
Apr 24, 2019
Oksutok
2 hours ago
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Inspired by 2024 Fall LMT Guts
sqing   2
N 2 hours ago by Jackson0423
Source: Own
Let $x$, $y$, $z$ are pairwise distinct real numbers satisfying $x^2+y =y^2 +z = z^2+x. $ Prove that
$$(x+y)(y+z)(z+x)=-1$$Let $x$, $y$, $z$ are pairwise distinct real numbers satisfying $x^2+2y =y^2 +2z = z^2+2x. $ Prove that
$$(x+y)(y+z)(z+x)=-8$$
2 replies
sqing
2 hours ago
Jackson0423
2 hours ago
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Dividing Pairs
Jackson0423   2
N 2 hours ago by Jackson0423
Source: Own
Let \( a \) and \( b \) be positive integers.
Suppose that \( a \) is a divisor of \( b^2 + 1 \) and \( b \) is a divisor of \( a^2 + 1 \).
Find all such pairs \( (a, b) \).
2 replies
Jackson0423
Apr 13, 2025
Jackson0423
2 hours ago
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configurational geometry as usual
GorgonMathDota   11
N Apr 1, 2025 by ratavir
Source: Indonesia National Math Olympiad 2021 Problem 7 (INAMO 2021/7)
Given $\triangle ABC$ with circumcircle $\ell$. Point $M$ in $\triangle ABC$ such that $AM$ is the angle bisector of $\angle BAC$. Circle with center $M$ and radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively, $(B \not= D, B \not= E)$. Let $P$ be the midpoint of arc $BC$ in $\ell$ that didn't have $A$. Prove that $AP$ angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.
11 replies
GorgonMathDota
Nov 9, 2021
ratavir
Apr 1, 2025
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configurational geometry as usual
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Reveal topic tags
geometry
if and only if
circumcircle
angle bisector
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G H BBookmark kLocked kLocked NReply
Source: Indonesia National Math Olympiad 2021 Problem 7 (INAMO 2021/7)
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GorgonMathDota
1063 posts
GorgonMathDota
#1 h  Nov 9, 2021, 5:10 AM • 1 Y
Y by Rounak_iitr
Given $\triangle ABC$ with circumcircle $\ell$. Point $M$ in $\triangle ABC$ such that $AM$ is the angle bisector of $\angle BAC$. Circle with center $M$ and radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively, $(B \not= D, B \not= E)$. Let $P$ be the midpoint of arc $BC$ in $\ell$ that didn't have $A$. Prove that $AP$ angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.
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BolaOne
20 posts
BolaOne
#2 h  Nov 9, 2021, 5:35 AM
Y by
$DMEP$ is cyclic. Then it's just simple angle chasing.
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somebodyyouusedtoknow
255 posts
somebodyyouusedtoknow
#3 h  Nov 9, 2021, 6:58 AM
Y by
I heard that this is the most difficult problem on the test, by far.
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Timmy456
92 posts
Timmy456
#5 h  Nov 10, 2021, 3:09 AM • 3 Y
Y by PRMOisTheHardestExam, Siddharth03, Rounak_iitr
Since $AM$ is the angle bisector of $\angle BAC$ and $P$ is the midpoint of arc $BC$, it is obvious that $P$ is located at the second intersection of $AM$ and $\ell$.

Case $1$, If we know that $\angle B = 90^{\circ}$
By angle chasing,
$\angle DEM = \frac{180^{\circ}-\angle DME}{2} = 90^{\circ}-\frac{\angle DME}{2} = 90^{\circ}-\angle DBE$ $= 90^{\circ}-(180^{\circ}-\angle DPC) = 90^{\circ}-(180^{\circ}-(\angle DPM + 90^{\circ}))$
$= 90^{\circ}-(90^{\circ}-\angle DPM) = \angle DPM$
Hence, $DPEM$ is a cyclic quadrilateral. Since $DM = ME$ and $DPEM$ is a cyclic quadrilateral, it is obvious that $\angle DPM = \angle MPE \implies AP$ is the angle bisector of $\angle DPE$.

Case $2$, If we know that $AP$ is the angle bisector of $\angle DPE$
That means, $DPEM$ is a cyclic quadrilateral. By angle chasing,
$\angle ABC = \angle APC = 180^{\circ}-(\angle DBE + \angle DPM) = 180^{\circ}-(\frac{\angle DME}{2} + \angle DEM)$
$= 180^{\circ}-(\frac{180^{\circ}-\angle MDE-\angle DEM}{2} + \angle DEM) = 180^{\circ}-(\frac{180^{\circ}-2\angle DEM}{2} + \angle DEM)$
$= 180^{\circ}-(90^{\circ}-\angle DEM + \angle DEM) = 180^{\circ}-90^{\circ} = 90^{\circ}$, as desired.

Hence, we have proven that $AP$ is the angle bisector of $\angle DPE$ if and only if $\angle B = 90^{\circ}$.
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Wildabandon
507 posts
Wildabandon
#6 h  Nov 13, 2021, 3:05 PM
Y by
For case 2, can you explain why $DPEM$ cyclic? I think thats doesn't work if $DP=PE$.
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Timmy456
92 posts
Timmy456
#7 h  Nov 14, 2021, 4:26 PM
Y by
@above I think I fake solved case 2. Do you have a solution?
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Timmy456
92 posts
Timmy456
#8 h  Nov 14, 2021, 4:30 PM
Y by
The following picture matches the criteria for case 2, but why isn't $\angle ABC = 90^{\circ}$
https://i.imgur.com/YGY21Tk.png
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Timmy456
92 posts
Timmy456
#9 h  Nov 14, 2021, 4:31 PM
Y by
Did I miss any given information?
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Wildabandon
507 posts
Wildabandon
#10 h  Nov 15, 2021, 5:20 AM
Y by
I also didn't find a solution if $DP=PE$. I also tried using geogebra with $15$ digit accuracy, but couldn't find the correct construction .-.
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Timmy456
92 posts
Timmy456
#11 h  Nov 15, 2021, 5:30 AM
Y by
This problem only works for case 1
I wonder what the official solution says for case 2
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somebodyyouusedtoknow
255 posts
somebodyyouusedtoknow
#12 h  Jan 20, 2022, 4:23 PM
Y by
Here's a fixed version of this problem.

Let the circumcircle of $\triangle{ABC}$ be $\ell$. There exists a point $M$ in $\triangle{ABC}$ such that $AM$ is the angle bisector of $\angle{BAC}$. The circle centred at $M$ with radius $MB$ intersects $\ell$ and $BC$ at $D$ and $E$ respectively ($B \neq D, B\neq E$). Let $P$ be the midpoint of arc $BC$ in $\ell$ not containing $A$. Prove that if $\triangle{DPE}$ is not isosceles, line $AP$ acts as the angle bisector of $\angle{DPE}$ if and only if $\angle{B} = 90^{\circ}$.
This post has been edited 1 time. Last edited by somebodyyouusedtoknow, Jan 20, 2022, 4:23 PM
Reason: I forgot that AoPS doesn't use \textbf
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ratavir
1 post
ratavir
#13 h  Apr 1, 2025, 10:22 PM
Y by
@above
was this the one written in the official problem?
This post has been edited 1 time. Last edited by ratavir, Apr 1, 2025, 10:22 PM
Reason: tagging people
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