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k a February Highlights and 2025 AoPS Online Class Information
jlacosta   0
Feb 2, 2025
We love to share what you can look forward to this month! The AIME I and AIME II competitions are happening on February 6th and 12th, respectively. Join our Math Jams the day after each competition where we will go over all the problems and the useful strategies to solve them!

2025 AIME I Math Jam: Difficulty Level: 8* (Advanced math)
February 7th (Friday), 4:30pm PT/7:30 pm ET

2025 AIME II Math Jam: Difficulty Level: 8* (Advanced math)
February 13th (Thursday), 4:30pm PT/7:30 pm ET

The F=ma exam will be held on February 12th. Check out our F=ma Problem Series course that begins February 19th if you are interested in participating next year! The course will prepare you to take the F=ma exam, the first test in a series of contests that determines the members of the US team for the International Physics Olympiad. You'll learn the classical mechanics needed for the F=ma exam as well as how to solve problems taken from past exams, strategies to succeed, and you’ll take a practice F=ma test of brand-new problems.

Mark your calendars for all our upcoming events:
[list][*]Feb 7, 4:30 pm PT/7:30pm ET, 2025 AIME I Math Jam
[*]Feb 12, 4pm PT/7pm ET, Mastering Language Arts Through Problem-Solving: The AoPS Method
[*]Feb 13, 4:30 pm PT/7:30pm ET, 2025 AIME II Math Jam
[*]Feb 20, 4pm PT/7pm ET, The Virtual Campus Spring Experience[/list]
AoPS Spring classes are open for enrollment. Get a jump on 2025 and enroll in our math, contest prep, coding, and science classes today! Need help finding the right plan for your goals? Check out our recommendations page!

Don’t forget: Highlight your AoPS Education on LinkedIn!
Many of you are beginning to build your education and achievements history on LinkedIn. Now, you can showcase your courses from Art of Problem Solving (AoPS) directly on your LinkedIn profile! Don't miss this opportunity to stand out and connect with fellow problem-solvers in the professional world and be sure to follow us at: https://www.linkedin.com/school/art-of-problem-solving/mycompany/ Check out our job postings, too, if you are interested in either full-time, part-time, or internship opportunities!

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All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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0 replies
jlacosta
Feb 2, 2025
0 replies
P6 Geo Finale
math_comb01   7
N 2 minutes ago by GuvercinciHoca
Source: XOOK 2025/6
Let $ABC$ be a triangle with incenter $I$ and excenters $I_A$, $I_B$, $I_C$ opposite to $A,B,C$ respectively. Suppose $BC$ meets the circumcircle of $I_AI_BI_C$ at points $D$ and $E$. $X$ and $Y$ lie on the incircle of $\triangle ABC$ so that $DX$ and $EY$ are tangents to the incircle (different from $BC$). Prove that the circumcircles of $\triangle AXY$ and $\triangle ABC$ are tangent.

Proposed by Anmol Tiwari
7 replies
math_comb01
Feb 10, 2025
GuvercinciHoca
2 minutes ago
A functional equation
super1978   1
N 13 minutes ago by pco
Source: Somewhere
Find all functions $f: \mathbb R \to \mathbb R$ such that:$$ f(f(y-x)-xf(y))+f(x)=y(1-f(x)) $$for all $x,y \in \mathbb R$
1 reply
super1978
an hour ago
pco
13 minutes ago
Sequences Handout
M11100111001Y1R   4
N 17 minutes ago by MR.1
Source: Own
Hi everyone, I wrote this handout about sequences in NT.
Hope you enjoy!
4 replies
+2 w
M11100111001Y1R
Oct 19, 2022
MR.1
17 minutes ago
[Handout] 50 non-traditional functional equations
gghx   2
N 44 minutes ago by GreekIdiot
Sup guys,

I'm retired. I love FEs. So here's 50 of them. Yea...

Functional equations have been one of the least enjoyed topics of math olympiads in recent times, mostly because so many techniques have been developed to just bulldoze through them. These chosen problems do not fall in that category - they require some combi-flavoured creativity to solve (to varying degrees).

For this reason, this handout is aimed at more advanced problem solvers who are bored of traditional FEs and are up for a little challenge!

In some sense, this is dedicated to the "covid FE community" on AoPS who got me addicted to FEs, people like EmilXM, hyay, IndoMathXdZ, Functional_equation, GorgonMathDota, BlazingMuddy, dangerousliri, Mr.C, TLP.39, among many others: thanks guys :). Lastly, thank you to rama1728 for suggestions and proofreading.

Anyways...
2 replies
gghx
Sep 23, 2023
GreekIdiot
44 minutes ago
No more topics!
IMO 2010 Problem 4
mavropnevma   126
N Feb 17, 2025 by cj13609517288
Let $P$ be a point interior to triangle $ABC$ (with $CA \neq CB$). The lines $AP$, $BP$ and $CP$ meet again its circumcircle $\Gamma$ at $K$, $L$, respectively $M$. The tangent line at $C$ to $\Gamma$ meets the line $AB$ at $S$. Show that from $SC = SP$ follows $MK = ML$.

Proposed by Marcin E. Kuczma, Poland
126 replies
mavropnevma
Jul 8, 2010
cj13609517288
Feb 17, 2025
IMO 2010 Problem 4
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huashiliao2020
1292 posts
#119 • 1 Y
Y by cubres
Let $D=LK\cap MC$; it's obvious that $SP^2=SC^2=SA\cdot SB\implies SPB=PAB=KLB\implies SP\parallel KL$. Then $$MKL=MCL=MDL-CLK=DPC-KLC=SCM-KLC=CAM-CAK=KAM=KLM,$$as desired. $\blacksquare$
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Shreyasharma
664 posts
#120 • 1 Y
Y by cubres
Otis Walkthrough:
Let $D$ denote the second tangency point from $S$. Let $DP$ meet $(ABC)$ at $N \neq D$. Now note that,
$-1 = (AB,DC) \overset{P}{=} (KL,MN)$
and thus we have $KMLN$ is harmonic. Now it suffices to show $KMLN$ is a kite, which is equivalent to showing that $MN$ is a diameter. We will now use phantom points. Let $N'$ be the $M$-antipode and let $D' = N'P \cap (ABC)$. It then suffices to show $D' \equiv D$.

Begin by noting that from the Three Tangents Lemma applied to $MNX$, where $X = MD' \cap NC$, we find that $OD'$ and $OC'$ are tangents to $(CD'P)$ where $O$ is the center of $(ABC)$. Therefore $(CD'P)$ and $(ABC)$ are orthogonal. Now by noting $SC = SP$ and $SC$ tangent to $(ABC)$ we can conclude $S$ is the circumcenter of $CD'P$. From the orthogonality we then must have $SD'$ tangent to $(ABC)$ from which $D' = D$. Thus $MN$ is a diameter and we must have $NL = NK$ from harmonic $KMLN$.
Attachments:
10IMO4.pdf (416kb)
This post has been edited 1 time. Last edited by Shreyasharma, Nov 11, 2023, 7:15 AM
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shendrew7
774 posts
#121 • 1 Y
Y by cubres
Our given condition implies $\frac{SC}{SB} = \frac{SA}{SC} \implies \frac{SP}{SB} = \frac{SA}{SP}$, or $\triangle SPB \sim \triangle SAP$. Thus we get
\[\angle SPB = \angle PAS = \angle KLB,\]or $LK \parallel PS$. Notice the desired implies that $LK$ is parallel to the tangent at $M$, so we want to show this tangent is also parallel to $PS$.

Suppose the tangent at $M$ meets the tangent at $C$ at point $T$. Since $TM = TC$ and $SP = SC$, we know $PS \parallel MT$, which concludes our proof. $\blacksquare$
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AngeloChu
458 posts
#122 • 1 Y
Y by cubres
let $MC$ and $KL$ intersect at $Q$, and let $SP$ intersect $MK$ at $R$
we get that $MCK$ and $MKQ$ are similar, and that $MLQ$ and $MCL$ are similar
we have $BAC=BLC=BCS$
extend $CK$ to meet $SP$ at $T$
we easily get $TKR=MLC$, and that $TKR$ is similar to $TPC$
we then get that $SP||KL$, and that $SPQ=PQL=MLC=MCS$, so $SC=SP$
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blueberryfaygo_55
339 posts
#123 • 8 Y
Y by megarnie, deduck, heheman, KevinYang2.71, dedapperdoop, smellyman, think4l, cubres
Solved with deduck.
Claim. The desired condition is equivalent to $\angle PBA + \angle ACP  = \angle PCB + \angle PAB$.
Proof. $MK=ML$ is equivalent to $\angle MLK = \angle MKL$, so by angle chasing, we have
\begin{align*}
\angle MKL &= \angle MKA + \angle AKL \\
&= \angle MCA + \angle ABL \\
&= \angle PCA + \angle ABP
\end{align*}Further, we also have,
\begin{align*}
\angle MLK &= \angle MLB + \angle BLK \\
&= \angle MCB + \angle BAK \\
&= \angle PCB + \angle PAB
\end{align*}so the claim follows. $\blacksquare$

Thus, we could remove points $M,L,K$ from the diagram. Now, we note that$$SB \cdot SA = SC^2 = SP^2$$which leads to $\Delta SAP \sim \Delta SPB$, and $\angle SAP = \angle SPB$. To finish, we have,\begin{align*}
\angle PCB + \angle PAB &= \angle PCB + \angle BPS \\
&= \angle PCS + \angle BPS - \angle BCS \\
&= \angle BPC - \angle BCS \\
&= \angle BPC - \angle BAC
\end{align*}Therefore, it is left to show that $\angle PBA + \angle ACP + \angle BAC = \angle BPC$. However, this is true since,
\begin{align*}
\angle PBA + \angle ACP + \angle BAC &= \angle PBA + \angle ACP + \angle BAP + \angle PAC \\
&= (180^{\circ} - \angle APB) + (180^{\circ} - \angle APC) \\
&= 360^{\circ} - \angle APB - \angle APC \\
&= \angle BPC
\end{align*}as desired. $\blacksquare$
This post has been edited 3 times. Last edited by blueberryfaygo_55, May 12, 2024, 5:10 AM
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dolphinday
1301 posts
#124 • 1 Y
Y by cubres
We use PoP to get $SA \cdot SB = SC^2$ and also $SC^2 = SP^2$ from which we get that $SP$ is tangent to $(PAB)$. This then implies that $\angle PAB = \angle BPS$ and $ABKL$ cyclic implies $\angle PAB = \angle KAB = \angle KLB \implies KL \parallel SP$. So then we apply Pascal on $BMCCKA$ to get that $BM \cap CK$ lies on $SP$. Then apply the same on $MMCKLB$ to get that $MM \cap KL$ lies on $SP$ which implies $MM$ is parallel to $KL$, so $MK = ML$ as desired.
This post has been edited 1 time. Last edited by dolphinday, May 16, 2024, 5:22 AM
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SenorSloth
37 posts
#125 • 1 Y
Y by cubres
We WLOG assume $CA<CB$. Draw the other tangent to $\Gamma$ from $S$, and let the tangency point be $D$. Let the intersection of $DP$ and $\Gamma$ distinct from $D$ be $N$. Then we have:

$-1=(AB;CD) \overset{P}{=} (KL;MN)$.

Now we angle chase to show that $MN$ is a diameter. Since $C$, $D$, and $P$ lie on a circle centered at $S$, we have that $\angle CPD=180-\frac12 \angle CSD$, and thus $\angle MPD = \frac12 \angle CSD$. Then we can calculate that the measure of arc $MN$ equals $\angle COD+2\angle MPD = 180 - \angle CSD + 2\left(\frac12 \angle CSD \right) = 180$.

Since $MN$ is a diameter of $\Gamma$ and $LMKN$ is harmonic, by symmetry we must have that $MK=ML$, as desired.
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Duka
1 post
#126 • 1 Y
Y by cubres
Since in shortlist says iff I will show that from $MK=ML$ $\implies$ $SP = SC$
Let $Q=CL\cap MA$, from Pascals theorem on $CCMABL$ we see that $S$ ,$P$ and $Q$ are coliniear.
$\angle QSP = \angle LCM = \angle LKM = \angle KLM = \angle MAK = \angle MAP$
$\implies CQAP$ is cyclic.
Then $\angle SPA = \angle QPA = \angle QCA = \angle LCA = \angle SBP$
Which implies that $SP$ is tangent to $(APB)$ $\implies SP\cdot SP= SA\cdot SB=SC\cdot SC$
$\implies SC=SP$
You can also go backwards to get $SP = SC$ $\implies$ $MK=ML$
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SomeonesPenguin
122 posts
#127 • 1 Y
Y by cubres
We will show that $SC=SP \iff MK=ML$.

Notice that $SC=SP \iff P$ lies on the $C$-Apollonian circle so $SC=SP \iff \frac{PA}{PB}=\frac{CA}{CB}$. Now $MK=ML \iff \angle LCP=\angle KCP$.

From LOS in $\triangle LCP$ and $\triangle KCP$ we have $$\frac{\sin(\angle LCP)}{\sin(\angle KCP)}=\frac{LP\sin(\angle LCP)}{KP\sin(\angle CKP)}$$
From PoP we have $\frac{PA}{PB}=\frac{PL}{PK}$ and from LOS we have $\frac{\sin(\angle LCP)}{\sin(\angle CKP)}=\frac{CB}{CA}$. Hence we get $$\frac{\sin(\angle LCP)}{\sin(\angle KCP)}=\frac{CB}{CA}\cdot \frac{PA}{PB}$$Now just plug in either condition in this equation to get the other one. $\square$
[asy]
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onyqz
195 posts
#128 • 1 Y
Y by cubres
simple one, posting for storage
solution
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Siddharthmaybe
88 posts
#129 • 1 Y
Y by cubres
swaqar wrote:
Solution (if it's correct please tell me)
Call the circles $(S,SC) $ $\Omega$. Now the circles $\Omega$ and $ \Gamma$ are orthogonal so the point of intersection of $AB$ and $KL$ lies on the circle $ \Omega$ and call this point $S'$. Assume that $A $ belong to $[S,B]$. Now the line $SP$ is tangent to the circumcircle of $ \triangle APB$ since $SA\cdot SB=SC^{2}=SP^{2} $. Now $SC$ is the external symmedian of the vertex $C$. Calling the intersection of the angle bisector of $C$ with the line $ AB$, $P'$, we have the points $ (S',P',A,B) $ forming a harmonic range and this implies that $ SC = SP'$. So the circle $\Omega$ is is the appollonian circle of the vertex $ C $ of the $\triangle ABC$ . Define the point $N$ on the line segment $KL$ so that the line $CN$ is the bisector of the $ \angle LCM$. Let the external symmedian meet the the $S'C$ in $J$. We have $JN=JC$ and $ SC = SP$ so the traingle $ SCP $ and $ JCN $ are homothetic so the points $ A , P $ and $ N $ are collinear and we have $M$ being the midpoint of the arc $ \overarc{KMN}$.

bringing modern artillery to a snowball fight :stretcher: Works though :10:
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optimusprime154
14 posts
#130 • 2 Y
Y by SorPEEK, cubres
pascal on \(CCMABL\) gives \(S,P,T\) collinear where T = \(MA \cap CL\) now i claim that the quadrilateral \(PATC\) is cyclic , that ends the problem because, it means \(\angle PCA = \angle MAK\) which means arc \(MK\) = arc \(ML\) so it finishes.
now i prove \(\angle TPL = \angle BAK\) . notice \(\angle TPL = \angle BPS = \angle BAK\) where the last one follows from the fact that \(SP\) is tangent to the circumcircle of \(\triangle BPA\) since \(BS * AS = SC^2 = SP^2\) from \(\angle TPL = \angle BAK = \angle BLK\) we obtain \(TL \parallel ST\) which indicates \(\angle KAC = \angle KLC = \angle PTC\) giving \(PATC\) cyclic so we're done!
This post has been edited 1 time. Last edited by optimusprime154, Jan 6, 2025, 4:35 PM
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HasnatFarooq
30 posts
#132 • 3 Y
Y by Jam777, Mahad_Arif, cubres
Complex Bash
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Mquej555
15 posts
#133 • 1 Y
Y by cubres
My first IMO Upload.
Attachments:
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cj13609517288
1822 posts
#134 • 1 Y
Y by cubres
Assume $CA<CB$.

It turns out that the circle centered at $S$ passing through $C$ is exactly the $C$-Apollonius circle of triangle $ABC$. Indeed, it suffices to show that $D$, the point on $AB$ such that $CD$ bisects $\angle ACB$, satisfies $SC=SD$, since the reflection of $C$ over $AB$ also lies on this circle. Indeed, $\angle SCD=\angle SCA+\angle ACD=\angle B+\angle BCD=\angle CDS$.

Now note that
\[\angle PAB+\angle PCB=\angle PDB-\angle APD+\angle PCB=\angle CPD-\frac12 \angle APB-\angle ABC.\]But since
\[\angle CPD=180^{\circ}-\frac12 \angle CSD=90^{\circ}+\angle CDS=90^{\circ}+\frac12 \angle C+\angle ABC,\]we just get that
\[\angle PAB+\angle PCB=90^{\circ}+\frac12\angle C-\frac12 \angle APB=\angle PBA+\angle PCA,\]which implies the result. $\blacksquare$
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