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Solve All 6 IMO 2024 Problems (42/42), New Framework Looking for Feedback
Blackhole.LightKing   0
an hour ago
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


0 replies
1 viewing
Blackhole.LightKing
an hour ago
0 replies
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Excircle Tangency Points Concyclic with A
tastymath75025   35
N an hour ago by bin_sherlo
Source: USA Winter TST for IMO 2019, Problem 6, by Ankan Bhattacharya
Let $ABC$ be a triangle with incenter $I$, and let $D$ be a point on line $BC$ satisfying $\angle AID=90^{\circ}$. Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to $\overline{BC}$ at $A_1$. Define points $B_1$ on $\overline{CA}$ and $C_1$ on $\overline{AB}$ analogously, using the excircles opposite $B$ and $C$, respectively.

Prove that if quadrilateral $AB_1A_1C_1$ is cyclic, then $\overline{AD}$ is tangent to the circumcircle of $\triangle DB_1C_1$.

Ankan Bhattacharya
35 replies
tastymath75025
Jan 21, 2019
bin_sherlo
an hour ago
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Domain swept by a parabola
Kunihiko_Chikaya   1
N 2 hours ago by Mathzeus1024
Source: 2015 The University of Tokyo entrance exam for Medicine, BS
For a positive real number $a$, consider the following parabola on the coordinate plane.
$C:\ y=ax^2+\frac{1-4a^2}{4a}$
When $a$ ranges over all positive real numbers, draw the domain of the set swept out by $C$.
1 reply
Kunihiko_Chikaya
Feb 25, 2015
Mathzeus1024
2 hours ago
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Show that XD and AM meet on Gamma
MathStudent2002   91
N 2 hours ago by IndexLibrorumProhibitorum
Source: IMO Shortlist 2016, Geometry 2
Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I$ and let $M$ be the midpoint of $\overline{BC}$. The points $D$, $E$, $F$ are selected on sides $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ such that $\overline{ID} \perp \overline{BC}$, $\overline{IE}\perp \overline{AI}$, and $\overline{IF}\perp \overline{AI}$. Suppose that the circumcircle of $\triangle AEF$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\Gamma$.

Proposed by Evan Chen, Taiwan
91 replies
MathStudent2002
Jul 19, 2017
IndexLibrorumProhibitorum
2 hours ago
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2013 China girls' Mathematical Olympiad problem 7
s372102   5
N 3 hours ago by Nari_Tom
As shown in the figure, $\odot O_1$ and $\odot O_2$ touches each other externally at a point $T$, quadrilateral $ABCD$ is inscribed in $\odot O_1$, and the lines $DA$, $CB$ are tangent to $\odot O_2$ at points $E$ and $F$ respectively. Line $BN$ bisects $\angle ABF$ and meets segment $EF$ at $N$. Line $FT$ meets the arc $\widehat{AT}$ (not passing through the point $B$) at another point $M$ different from $A$. Prove that $M$ is the circumcenter of $\triangle BCN$.
5 replies
s372102
Aug 13, 2013
Nari_Tom
3 hours ago
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Equal angles with midpoint of $AH$
Stuttgarden   2
N 4 hours ago by HormigaCebolla
Source: Spain MO 2025 P4
Let $ABC$ be an acute triangle with circumcenter $O$ and orthocenter $H$, satisfying $AB<AC$. The tangent line at $A$ to the circumcicle of $ABC$ intersects $BC$ in $T$. Let $X$ be the midpoint of $AH$. Prove that $\angle ATX=\angle OTB$.
2 replies
Stuttgarden
Mar 31, 2025
HormigaCebolla
4 hours ago
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<DPA+ <AQD =< QIP wanted, incircle circumcircle related
parmenides51   41
N Today at 6:02 AM by Ilikeminecraft
Source: IMo 2019 SL G6
Let $I$ be the incentre of acute-angled triangle $ABC$. Let the incircle meet $BC, CA$, and $AB$ at $D, E$, and $F,$ respectively. Let line $EF$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle DPA + \angle AQD =\angle QIP$.

(Slovakia)
41 replies
parmenides51
Sep 22, 2020
Ilikeminecraft
Today at 6:02 AM
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Parallelity and equal angles given, wanted an angle equality
BarisKoyuncu   5
N Today at 5:53 AM by SleepyGirraffe
Source: 2022 Turkey JBMO TST P4
Given a convex quadrilateral $ABCD$ such that $m(\widehat{ABC})=m(\widehat{BCD})$. The lines $AD$ and $BC$ intersect at a point $P$ and the line passing through $P$ which is parallel to $AB$, intersects $BD$ at $T$. Prove that
$$m(\widehat{ACB})=m(\widehat{PCT})$$
5 replies
1 viewing
BarisKoyuncu
Mar 15, 2022
SleepyGirraffe
Today at 5:53 AM
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Cyclic points and concurrency [1st Lemoine circle]
shobber   10
N Today at 4:47 AM by Ilikeminecraft
Source: China TST 2005
Let $\omega$ be the circumcircle of acute triangle $ABC$. Two tangents of $\omega$ from $B$ and $C$ intersect at $P$, $AP$ and $BC$ intersect at $D$. Point $E$, $F$ are on $AC$ and $AB$ such that $DE \parallel BA$ and $DF \parallel CA$.
(1) Prove that $F,B,C,E$ are concyclic.

(2) Denote $A_{1}$ the centre of the circle passing through $F,B,C,E$. $B_{1}$, $C_{1}$ are difined similarly. Prove that $AA_{1}$, $BB_{1}$, $CC_{1}$ are concurrent.
10 replies
shobber
Jun 27, 2006
Ilikeminecraft
Today at 4:47 AM
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Vertices of a convex polygon if and only if m(S) = f(n)
orl   12
N Today at 4:04 AM by Maximilian113
Source: IMO Shortlist 2000, C3
Let $ n \geq 4$ be a fixed positive integer. Given a set $ S = \{P_1, P_2, \ldots, P_n\}$ of $ n$ points in the plane such that no three are collinear and no four concyclic, let $ a_t,$ $ 1 \leq t \leq n,$ be the number of circles $ P_iP_jP_k$ that contain $ P_t$ in their interior, and let \[m(S)=a_1+a_2+\cdots + a_n.\]Prove that there exists a positive integer $ f(n),$ depending only on $ n,$ such that the points of $ S$ are the vertices of a convex polygon if and only if $ m(S) = f(n).$
12 replies
orl
Aug 10, 2008
Maximilian113
Today at 4:04 AM
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Radical Axis and Circumcenter
azagorod101   5
N Jul 2, 2016 by Reynan
Source: BMO 2012/6
Let $ABC$ be a triangle. Let $S$ be the circle through B tangent to $CA$ at $A$ and let $T$ be the circle through C tangent to $AB$ at $A$. Circles $S$ and $T$ intersect at $A$ and $D$. Let $E$ be where $AD$ meets circle $ABC$. Prove that $D$ is the midpoint of $AE$.
5 replies
azagorod101
May 17, 2016
Reynan
Jul 2, 2016
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Radical Axis and Circumcenter
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Source: BMO 2012/6
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azagorod101
194 posts
azagorod101
#1 h  May 17, 2016, 6:39 PM • 2 Y
Y by Adventure10, Mango247
Let $ABC$ be a triangle. Let $S$ be the circle through B tangent to $CA$ at $A$ and let $T$ be the circle through C tangent to $AB$ at $A$. Circles $S$ and $T$ intersect at $A$ and $D$. Let $E$ be where $AD$ meets circle $ABC$. Prove that $D$ is the midpoint of $AE$.
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ThE-dArK-lOrD
4071 posts
ThE-dArK-lOrD
#2 h  May 17, 2016, 7:06 PM • 2 Y
Y by Adventure10, Mango247
Let $\angle{BAD}=x=\angle{ACD}=\angle{BCE},\angle{DAC}=y=\angle{ABD}=\angle{EBC}$
We need to prove $\frac{DE}{BD} \cdot \frac{BD}{AD} =1$ $\Leftrightarrow$ $\frac{sin(\angle{EBD})}{sin(\angle{BEA})} \cdot \frac{sin(x)}{sin(y)}=1$
Note that $\angle{EBD} =\angle{ABC}$ and $\angle{BEA} =\angle{ACB}$
So we need $\frac{AC}{AB} \cdot \frac{BE}{CE} =1$
Note that $\triangle{BEC} \sim \triangle{BDA}$ give $\frac{BE}{CE} =\frac{BD}{AD}$ and $\frac{BD}{AD} =\frac{AB}{AC}$ since $\triangle{BDA} \sim \triangle{ADC}$
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nikolapavlovic
1246 posts
nikolapavlovic
#3 h  May 17, 2016, 7:11 PM • 2 Y
Y by Adventure10, Mango247
$D$ is the center of spiral similarity sending $AC$ to $BA$ so $AD$ is the symmedian.Hence $BC$ is the symmedian in $\triangle EBA$ so we need to prove
$\angle EBC=\angle ACD$ but $\angle EBC=\angle CAE=\angle ACD$ by the above similarity
This post has been edited 2 times. Last edited by nikolapavlovic, May 17, 2016, 7:14 PM
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Ghost_rider
35 posts
Ghost_rider
#4 h  Jul 1, 2016, 9:55 PM • 2 Y
Y by Adventure10, Mango247
Let $AD$ $\cap$ $BC$ = $M$. Since $\bigtriangleup$ $BAD$ $\sim$ $\bigtriangleup$ $ADC$ ( $\frac{AB}{AD}$=$\frac{AC}{DC}$ , $\frac{AB}{BD}$=$\frac{AC}{AD}$) and $\angle$ $BDE$ = $\angle$ $EDC$ so $\frac{AB^2}{AC^2}$=$\frac{BM}{MC}$ $\longrightarrow$ $AM$ is symmedian. Let the tangent to $B$ and $C$ intersect in $N$ so $A$,$D$,$E$ and $N$ are collinear. Is well-know that $AB.EC$ = $BE.AC$ and $\frac{AM}{ME}$=$\frac{AB.AC}{BE.EC}$ so $\frac{AC^2}{CE^2}$=$\frac{AM}{ME}$ $\longrightarrow$ $MC$ is symmedian. Then $AD$ = $DE$.
This post has been edited 1 time. Last edited by Ghost_rider, Jul 1, 2016, 9:55 PM
Reason: type
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jayme
9782 posts
jayme
#5 h  Jul 2, 2016, 6:53 AM • 2 Y
Y by Adventure10, Mango247
Dear Mathlinkers,

see

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=597823

Sincerely
Jean-Louis
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Reynan
634 posts
Reynan
#6 h  Jul 2, 2016, 7:41 AM • 2 Y
Y by Adventure10, Mango247
let $EB$ intersect $S$ again at $F$ and $EC$ intersect $T$ again at $G$. $EB\cdot EF= ED\cdot EA= EC\cdot EG$ makes $FBCG$ cyclic. $\angle ADF=\angle ABF=\angle ACE=180-\angle ACG=180 - \angle ADG$ makes $FDG$ collinear. $\angle AFD=\angle CAD = \angle CGD = \angle EGD$ makes $AF\parallel EG$. Analog $AG\parallel EF$. These make $AEFG$ parallelogram with centre $D$. DONE:)
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