It's February and we'd love to help you find the right course plan!

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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!

Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

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Special AIME Problem Seminar B
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F=ma Problem Series
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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 14 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
14 minutes ago
Sequences Handout
M11100111001Y1R   4
N 18 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
18 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!
Two circles, a tangent line and a parallel
Valentin Vornicu   97
N Feb 10, 2025 by ehuseyinyigit
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP = EQ$.
97 replies
Valentin Vornicu
Oct 24, 2005
ehuseyinyigit
Feb 10, 2025
Two circles, a tangent line and a parallel
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G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
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Valentin Vornicu
7301 posts
#1 • 17 Y
Y by Davi-8191, OlympusHero, FaThEr-SqUiRrEl, lc426, centslordm, Adventure10, Designerd, jhu08, megarnie, Numbertheorydog, HWenslawski, Kanimet0, ImSh95, Zhaom, Mango247, Rounak_iitr, Math_.only.
Two circles $ G_1$ and $ G_2$ intersect at two points $ M$ and $ N$. Let $ AB$ be the line tangent to these circles at $ A$ and $ B$, respectively, so that $ M$ lies closer to $ AB$ than $ N$. Let $ CD$ be the line parallel to $ AB$ and passing through the point $ M$, with $ C$ on $ G_1$ and $ D$ on $ G_2$. Lines $ AC$ and $ BD$ meet at $ E$; lines $ AN$ and $ CD$ meet at $ P$; lines $ BN$ and $ CD$ meet at $ Q$. Show that $ EP = EQ$.
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arqady
30131 posts
#2 • 28 Y
Y by jam10307, shiningsunnyday, claserken, shawnee03, futurestar, Richangles, myh2910, OlympusHero, User582032, starchan, sotpidot, FaThEr-SqUiRrEl, centslordm, Adventure10, megarnie, asdf334, Numbertheorydog, HWenslawski, rayfish, Kanimet0, ImSh95, Mango247, CoC_Ali, Rounak_iitr, Math_.only., ehuseyinyigit, and 2 other users
let MN and AB intesect at K. Then $KA^2=KM*KN=KB^2.$ Hence KA=KB.
CD||AB. Hence PM=QM.
$\angle AEB=\angle ECD=\angle MAB,$ $\angle EBA=\angle EDC=\angle MBA.$
hence$\triangle AMB$ and$\triangle AEB$ congruity.
Hence $EM\perp AB.$ Hence $EM\perp PQ.$
Hence $EP=EQ.$
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yetti
2643 posts
#3 • 9 Y
Y by JasperL, FaThEr-SqUiRrEl, centslordm, Adventure10, Designerd, jhu08, ImSh95, Rounak_iitr, and 1 other user
The radical axis MN of the circles $(G_1), (G_2)$ cuts the tangent length segment AB at its midpoint C. Since $PQ \parallel AB$, the triangles $\triangle ABN \sim \triangle PQN$ are centrally similar with similarity center N. Hence, M is the midpoint of the segment PQ. The radii $G_1A, G_2B$ of the circles $(G_1), (G_2)$ are perpendicular to the tangent AB, i.e, also to the chords AM, DM. Hence they cut these chords at their midpoints, which means that the triangles $\triangle ACM, \triangle BDM$ are isosceles and the angles $\angle ACM = \angle AMC, \angle BDM = \angle BMD$ are equal. Since the lines $AB \parallel CD$ are parallel and $M \in CD$, the angles $\angle MAB = \angle AMC = \angle ACM = \angle EAB$ are equal and similarly, the angles $\angle MBA = \angle BMD = \angle BDM = \angle EBA$ are also equal. Thus the diagonal AB of the quadrilateral AMBE bisects its opposite angles at the vertices A, B, which implies that this quadrilateral is a kite with AE = AM, BE = BM and and its diagonals $AB \perp EM$ are perpendicular to each other. Consequently, the lines $EM \perp CD \equiv PQ$ are also perpendicular to each other. Since M is the midpoint of the segment PQ, the line EM is the perpendicular bisector of this segment. It follows that the triangle $\triangle EPQ$ is isosceles with EP = EQ.
Attachments:
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Arne
3660 posts
#4 • 7 Y
Y by mihajlon, OlympusHero, FaThEr-SqUiRrEl, Adventure10, ImSh95, and 2 other users
An "extra" question:

Prove that $EN$ bisects $\angle{CND}$.
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Zhero
2043 posts
#5 • 4 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247
Since $ AB$ is the common tangent of $ G_1$ and $ G_2$ and $ MN$ is the radical axis of $ G_1$ and $ G_2$, $ MN$ bisects $ AB$. A homothety centered at $ N$ that maps $ A$ to $ P$ and $ B$ to $ Q$ therefore maps the midpoint of $ AB$ to $ PQ$, so we have that $ M$ is the midpoint of $ PQ$.

It is sufficient to show that $ EM \perp CD$. Let $ R$ and $ S$ be the projections of $ A$ and $ B$ onto $ CD$, respectively; since $ AB || CD$, $ AR = BS$. We first note that $ AC = AM$, since $ m \angle AMC = m \angle EAB = m \angle ACM$ (as $ AB$ is tangent to $ G_1$). We can similarly deduce that $ BM = BD$.

It also follows from this that $ CM = 2CR$ and $ DM = 2DS$. Let $ E'$ be the point such that $ E'M \perp CD$ and $ E'M = 2AR = 2BS$. A homothety centered at $ C$ that maps $ R$ to $ M$ must therefore map $ A$ to $ E'$, and a homothety centered at $ D$ must therefore map $ B$ to $ E'$. But this means that $ E'$ lies on both $ AC$ and $ BD$, implying that $ E' = E$. It follows that $ EM \perp CD$, so we are done.
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Pikachu!!!
58 posts
#6 • 5 Y
Y by S117, FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247
Arne wrote:
An "extra" question:

Prove that $ EN$ bisects $ \angle{CND}$.

This extra question is also beautiful one.

First, from what we know Let $ \angle{ANM}=\angle{BAM}=\angle{BAE}=x$ and $ \angle{BNM}=\angle{ABM}=\angle{ABE}=y$

Therefore, $ \angle{AEB}+\angle{ANB}= (\pi-x-y)+(x+y)=\pi$ $ \longrightarrow$ $ A,N,B,E$ is concyclic.

It follows that $ \angle{BNE}=\angle{BAE}=x$ and it is known that $ \angle{BND}=\angle{BNM}=y$
We get $ \angle{DNE}=x+y$

Similarly, we also have $ \angle{CNE}=x+y$ :lol: :lol: :lol:
This is what we want to prove.
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arshakus
769 posts
#7 • 4 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247
$K$ is the midpint of $AB$. $NM$ is radical axis for $G_1,G_2$ circles. also we know that radical axis passes through the midpoints of comment tangents. Thus $N,M,K$ lies on the same line. $PQ||AB=>M$ is the midpoint of $PQ$.
Also it is easy to note this number of equations.
$\angle{PCA}=\angle{ANM}=\angle{BAE}=\angle{MAB}=\angle{PMA}=\angle{CNA}=\angle{ENB}$[*] the last equation follows from the fact that $AENB$ is cyclic.
Similarly we can get in the opposite side. Thus $\triangle {AMB}=\triangle {AEB}=>EM\perp PQ=>EP=EQ$
@Arne
from [*] it is easy to note it.
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arshakus
769 posts
#8 • 4 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247
One more "extra" question!)
Prove that $2AB=CD$
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siavosh
29 posts
#9 • 4 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247
arshakus wrote:
One more "extra" question!)
Prove that $2AB=CD$
In your solution we prove that $ \triangle{AMB}=\triangle{AEB}=> EB=BM=BD$
and $ CD\parallel AB =>  \frac{EB}{ED}=\frac{AB}{CD}=2 $
:)
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JSGandora
4216 posts
#10 • 5 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, Rounak_iitr
Let $\angle ACM=\alpha$, then since $AB||CD$, we have $\angle EAB=\angle ACM=\alpha$ and since $AB$ is tangent to circle $G_1$, we have $\angle BAM=\angle ACM=\alpha$. Thus $\angle EAB=\angle BAM=\alpha$. Similarly, if we let $\angle BDM=\beta$, we find $\angle EBA=\angle ABM=\beta$.

Now let the intersection of $EM$ and $AB$ be $F$, then $\angle AFE=90^\circ$ and $\angle CME=\angle AFE=90^\circ$ and therefore proving $EP=EQ$ is equivalent to proving $PM=QM$ since $EM\perp PQ$.

Let the intersection of ray $NM$ and $AB$ be point $G$, then $GA=GB$ because $G$ is on the radical axis of the two circles and thus $GA^2=GB^2\implies GA=GB$. Additionally, since $AB||CD$, we have $\triangle AGN~\triangle PMN$ and $\triangle BGN~\triangle QMN$ and therefore $PM=PQ$ as desired. $\blacksquare$
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BBAI
563 posts
#11 • 3 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10
As the tangent is parallel to the chords So it follows that $ \triangle ACM$ and $ \triangle BDM$ are isosceles and $EA=AM=AC$ so $EM$ is perpendicular to $PQ$. Define $Z=MN \cap AB$.$Z$ lies on radical axis ,so $Z$ is the midpoint of $AB$. And hence $M$ is the midpoint of $PQ$. So done.
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subham1729
1479 posts
#12 • 3 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10
OMG!! got a proof without bashing :O - if $MN$ meets $AB$ at $K$ then $AK^2=BK^2\implies AK=KB\implies PM=MQ\implies \angle{CMN}=\frac {\pi}{2}\implies EP=EQ$
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BBAI
563 posts
#13 • 3 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10
I think $E,M,N$ are not at all collinear.
I have drawn a ggb diagram. and checked it.
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sunken rock
4364 posts
#14 • 5 Y
Y by FaThEr-SqUiRrEl, ImSh95, Adventure10, Mango247, and 1 other user
Yet additional requirement:

the quadrilateral $ANBE$ is harmonic!


Best regards,
sunken rock
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exmath89
2572 posts
#15 • 6 Y
Y by Math-Star., FaThEr-SqUiRrEl, ImSh95, Adventure10, Rounak_iitr, and 1 other user
Solution
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