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

G
Topic
First Poster
Last Poster
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.

Introductory: Grades 5-10

Prealgebra 1
Monday, Feb 3 - May 19
Sunday, Mar 2 - Jun 22
Friday, Mar 28 - Jul 18
Sunday, Apr 13 - Aug 10

Prealgebra 1 Self-Paced

Prealgebra 2
Sunday, Feb 16 - Jun 8
Tuesday, Mar 25 - Jul 8
Sunday, Apr 13 - Aug 10

Prealgebra 2 Self-Paced

Introduction to Algebra A
Sunday, Feb 16 - Jun 8 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)
Sunday, Mar 23 - Jul 20
Monday, Apr 7 - Jul 28

Introduction to Algebra A Self-Paced

Introduction to Counting & Probability
Sunday, Feb 9 - Apr 27 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)
Sunday, Mar 16 - Jun 8
Wednesday, Apr 16 - Jul 2

Introduction to Counting & Probability Self-Paced

Introduction to Number Theory
Sunday, Feb 16 - May 4
Monday, Mar 17 - Jun 9
Thursday, Apr 17 - Jul 3

Introduction to Algebra B
Thursday, Feb 13 - May 29
Sunday, Mar 2 - Jun 22
Monday, Mar 17 - Jul 7
Wednesday, Apr 16 - Jul 30

Introduction to Algebra B Self-Paced

Introduction to Geometry
Friday, Feb 14 - Aug 1
Tuesday, Mar 4 - Aug 12
Sunday, Mar 23 - Sep 21
Wednesday, Apr 23 - Oct 1

Intermediate: Grades 8-12

Intermediate Algebra
Wednesday, Feb 12 - Jul 23
Sunday, Mar 16 - Sep 14
Tuesday, Mar 25 - Sep 2
Monday, Apr 21 - Oct 13

Intermediate Counting & Probability
Monday, Feb 10 - Jun 16
Sunday, Mar 23 - Aug 3

Intermediate Number Theory
Thursday, Feb 20 - May 8
Friday, Apr 11 - Jun 27

Precalculus
Tuesday, Feb 25 - Jul 22
Sunday, Mar 16 - Aug 24
Wednesday, Apr 9 - Sep 3

Advanced: Grades 9-12

Olympiad Geometry
Wednesday, Mar 5 - May 21

Calculus
Friday, Feb 28 - Aug 22
Sunday, Mar 30 - Oct 5

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Tuesday, Feb 4 - Apr 22
Sunday, Mar 23 - Jun 15
Wednesday, Apr 16 - Jul 2

MATHCOUNTS/AMC 8 Advanced
Sunday, Feb 16 - May 4
Friday, Apr 11 - Jun 27

AMC 10 Problem Series
Sunday, Feb 9 - Apr 27
Tuesday, Mar 4 - May 20
Monday, Mar 31 - Jun 23

AMC 10 Final Fives
Sunday, Feb 9 - Mar 2 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)

AMC 12 Problem Series
Sunday, Feb 23 - May 11

AMC 12 Final Fives
Sunday, Feb 9 - Mar 2 (3:30 - 5:00 pm ET/12:30 - 2:00 pm PT)

Special AIME Problem Seminar B
Sat & Sun, Feb 1 - Feb 2 (4:00 - 7:00 pm ET/1:00 - 4:00 pm PT)

F=ma Problem Series
Wednesday, Feb 19 - May 7

Programming

Introduction to Programming with Python
Sunday, Feb 16 - May 4
Monday, Mar 24 - Jun 16

Intermediate Programming with Python
Tuesday, Feb 25 - May 13

USACO Bronze Problem Series
Thursday, Feb 6 - Apr 24

Physics

Introduction to Physics
Friday, Feb 7 - Apr 25
Sunday, Mar 30 - Jun 22

Physics 1: Mechanics
Sunday, Feb 9 - Aug 3
Tuesday, Mar 25 - Sep 2

Relativity
Sat & Sun, Apr 26 - Apr 27 (4:00 - 7:00 pm ET/1:00 - 4:00pm PT)
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
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 1, IMO Shortlist 2000, G2
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
774 posts
#89
Y by
Note $MN$ bisects $AB$ by radical axis. Because $AB \parallel PQ$, there exists a homothety mapping $AB$ to $PQ$ centered at $N$ which also maps the midpoint of $AB$ to $M$. Hence $M$ is the midpoint of $PQ$.

The problem reduces to proving $EM \perp CD$, or $EM \perp AB$. At this point, points $M$, $N$, $P$, and $Q$ are irrelevant. We can angle chase to get
\[\angle BAM = \angle AMC = \angle ACM = \angle EAB,\]\[\angle ABM = \angle BMD = \angle BDM = \angle EBA.\]
Consequently, $EAMB$ is a kite, giving our desired conclusion.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cursed_tangent1434
523 posts
#90
Y by
First, notice that we have
$$\angle CMA = \angle BAM = \angle ACM = \angle EAB$$and similarly
$$\angle BMD = \angle ABM = \angle BDM = \angle EBA$$Let $X$ be the intersection of $EM$ and $AB$. Now, notice that $$CA = AM = AE \text{ and } BM = BD = EB$$And by congruency we have that
$$\angle AXE = \angle AXM = 90^\circ$$Thus, $\angle CME = 90^\circ$ as well using Parallel Lines. Now, notice that $MN$ bisects $AB$ and thus using the parallel lines, $MN$ bisects $PQ$ as well. Thus, $PM=MQ$. Therefore, $\triangle EPM \cong \triangle EQM$ which gives us that $EP=EQ$ as required.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
kamatadu
461 posts
#91 • 1 Y
Y by GeoKing
Wow I never posted soln for this problem.

$\angle EAB=\angle ECD=\angle ACM=\angle BAM$ and similarly $\angle EBA = \angle ABM \implies AEBM$ is kite $\implies EM\perp AB\implies EM \perp CD$. Now $M$ is midpoint of $PQ$ due to homothety at $N$ mapping $AB\mapsto PQ$. Thus $\triangle EPQ$ is isosceles $\implies EP=EQ$. :yoda:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dolphinday
1301 posts
#92
Y by
Let $\overline{EN} \cap \overline{AB} = K$.
$\newline$
Notice that
\[\angle{ACM} = \angle{AMC} = \angle{MAB}\]and similarly
\[\angle{MBA} = \angle{EBA}\]so $\triangle EBA \cong \triangle MBA$ are directly similar $\implies EBMA$ is a kite.
This implies that $EN \perp AB$. Since $K$ lies on the radical axis of $G_1$ and $G_2$, $KA = KB$ so the homothety sending $AB$ to $PQ$ also sends $K$ to $M$.
$\newline$
Since the $E$-altitude and the $E$-median of $\triangle EPQ$ coincide(they intersect $PQ$ at $M$), $\triangle EPQ$ is isosceles, specifically $EP = EQ$, which finishes. $\blacksquare$
This post has been edited 1 time. Last edited by dolphinday, Feb 2, 2024, 9:15 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lelouchvigeo
171 posts
#93
Y by
My first problem after 1 month of INMO
It is well known that $M$ is altitude. Therefore $ EM\perp CD$. Also well known that $NM$ bisects $AB$. Therfore we get $MP=MQ.$ Now we get our desired result. :D :D
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
blueberryfaygo_55
339 posts
#94
Y by
Let $O$ be the center of $G_2$.
Claim 1: $EM \parallel BO$
Proof. Let $EM$ intersect $G_2$ at a second point $X \neq M$. First, we have $AB \parallel CD$, and angle chasing leads to $$\angle BMD = \angle ABM = \angle BDM$$It follows that $$\angle MXD = m\overset{\Large\frown}{BD} = \angle BQD$$and the claim follows. $\blacksquare$

Since we have $OB \perp AB$ and $AB \parallel CD$, we obtain $EM \perp CD$.

Lemma (Trapezoid Lemma): In trapezoid $ABCD$ ($AB \parallel CD$), let $N$ be the midpoint of $CD$, and $X$ be the intersection of $AD$ and $BC$. The intersection $M$ of $XN$ and $AB$ is the midpoint of $AB$.
Proof. By similar triangles, we obtain $$\dfrac{AM}{DN} = \dfrac{XM}{XN} = \dfrac{MB}{NC}$$However, $N$ is the midpoint of $CD$, so it follows that $AM=MB$, or $M$ is the midpoint of $AB$. $\blacksquare$

Returning to the original problem, let $K = MN \cap AB$. Since $K$ lies on the radical axis of $G_1$ and $G_2$, we have $\text{Pow}_{G_1} (K) = \text{Pow}_{G_2} (K)$, which implies that $KA=KB$, or $K$ is the midpoint of $AB$. Applying the lemma above on trapezoid $ABQP$, we have $M$ is the midpoint $PQ$. Thus, $\Delta EPQ$ is isoceles with $EP=EQ$, as desired. $\blacksquare$
This post has been edited 1 time. Last edited by blueberryfaygo_55, Apr 26, 2024, 5:41 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
crazyConic
3 posts
#95 • 2 Y
Y by ATGY, Math_.only.
Fun problem, and my first ever post on AoPS about an actual oly problem! Diagram attached.
Claim 1: $MP = MQ$

Proof:

Call $K$ the point where radical axis $MN$ intersects line $AB$. We understand that as $K$ lies upon the radical axis of $G_1$ and $G_2$. This means that the power of $K$ to both circles is equal.

$(KA)^2 = (KB)^2$
$\rightarrow KA = KB$

Therefore, $K$ is the midpoint of $AB$.

Now, as $\triangle NAB \sim \triangle NPQ$, we also know that $M$ must be the midpoint of PQ as well, giving us this step of the proof.

Claim 2: $EM \perp CD$

Proof:

First, we prove that $\triangle EAB \sim \triangle ECD$ by a scale factor of 2.
We can easily see that the two triangles are similar. As to their scale factor, we can see that from the Perpendicular Bisector of a Chord Theorem, segment $CD$ is twice as long as segment $AB$.

Now, we use this fact to see that, after constructing the perpendicular from $E$ to $CD$, the foot of the altitude must be $M$ by similar triangles.

Finish:

Now, we have the following facts:

$MP = MQ$ and $EM \perp CD$. This easily gives us that $EP = EQ$ by congruence. :laugh:
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Math_.only.
20 posts
#96
Y by
arqady wrote:
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 $EP=EQ.$

Why angles AEB and ECD are equal ?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MagicalToaster53
159 posts
#97
Y by
Notice that $MN$ bisects $\overline{AB},$ as the point where they intersect has equivalent powers to the respective circles. From this we discover that $M$ is the midpoint of $\overline{PQ}$ due to the homothety at $N$ mapping $\overline{PQ} \mapsto \overline{AB}.$ It then suffices to show $EM \perp PQ$.

Claim: $\triangle EAB \cong \triangle MAB.$
Proof: Observe that $AB$ bisects $\angle EAM:$ \[\angle EAB = \angle ECM = \angle ACM = \angle BAM. \]Similarly by symmetry, we find that $AB$ bisects $\angle EBM.$ Then by SAS, we conclude that $\triangle EAB \cong \triangle MAB,$ as was to be shown. $\square$

Now $EAMB$ is a kite, so that $EM \perp AB \implies EM \perp PQ,$ as $AB \parallel PQ$. Hence $\triangle EPQ$ is isosceles, as desired. $\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
lnzhonglp
120 posts
#98
Y by
Observe that $$\measuredangle EBA = \measuredangle EDC = \measuredangle ABM$$and $$\measuredangle EAB = \measuredangle ECM = \measuredangle MAB,$$so $\triangle EAB \cong \triangle MAB$ and $EM \perp AB$. From radical axis we see that $MN$ bisects $AB$, so by homothety at $N$ we find that $M$ is the midpoint of $PQ$. Then since $EM \perp PQ$ it follows that $EP = EQ$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Likeminded2017
391 posts
#99
Y by
Let $MN$ and $AB$ intersect at $K.$ As $K$ lies on the radical axis we have $KA=KB.$ Taking the homothety centered at $N$ that maps $M$ to $K$ we find that $MP=MQ.$ Then $\angle BAM=\angle MCA=\angle BAE$ and similarly $\angle EBA=\angle ABM$ so by ASA congruence $\triangle EAB \cong \triangle MAB.$ Thus $EM$ is perpendicular to $AB$ and thus $PQ$ so $EP=EQ.$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ezpotd
1241 posts
#100
Y by
Letting $AB$ meet $MN$ at $K$, we see $KA^2 = KM \cdot KN = KB^2$, so $K$ is the midpoint of $AB$, taking the homothety from $AB$ to $PQ$ this gives $MP = MQ$, so it suffices to prove $EM \perp PQ$, or $EM \perp AB$. However, angle chasing gives $\angle EAB = \angle ECM = \angle AMC = \angle BAM, \angle EBA = \angle EDM = \angle BMD = \angle ABM$, so $\triangle EAB \cong \angle MAB$, so $E,M$ are reflections about $AB$, done.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
optimusprime154
14 posts
#101 • 1 Y
Y by SorPEEK
let \(Z \)= \(MN  \cap AB\) from PoP we know \(ZB^2 = ZM * ZN\) and \(ZA^2 = ZM * ZN\) which means \(ZA = ZB\) and since \(PQBA\) is a trapezoid, \(PM = MQ\)
now we prove \(EM \perp CD\) and we're done. label \(\angle ACM = x, \angle BDM = y\) we know from \(AB \parallel CD\) that \(\angle EAB = x\) and \(\angle EDC = y\)
we also know from \(AB\) being tangent that \(\angle MBA\ = y\) and \(\angle MAB = x\) the above facts indicate the similarity of \(\triangle MAB , \triangle EAB \) so \(EM \perp AB\) => \(EM \perp CD\) and we're done
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
430 posts
#102
Y by
Extend $MN$ to intersect $AB$ at $K,$ then clearly $AK=BK.$ Therefore $PM=MQ$ by the parallel lines.

Now, notice that $\angle EAB = \angle ACM = \angle BAM = \angle AMC.$ Therefore $AB$ bisects $\angle EAM$ and $AC=AM.$ Similarly $AB$ bisects $EBM$ and $BD=BM.$ Therefore, $\triangle EAB \cong \triangle MAB$ so $A, B$ are the circumcenters of $\triangle EMC, \triangle EMD$ respectively so by Thales' Theorem it follows that $\angle EMP = \angle EMQ = 90^\circ.$ Hence $EM$ is the perpendicular bisector of $PQ,$ and we are done. QED
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ehuseyinyigit
769 posts
#104
Y by
WELL-KNOWN. Since $AB\parallel CD$, we have $$\angle ACM=\angle AMC=\angle MAB=\angle BAE$$Thus, $AC=AM$ and $AB$ bisects $\angle EAM$. Similarly $BD=BM$ and $AB$ bisects $\angle EBM$ hence $ABME$ is a kite $\Rightarrow AE=AM=AC$ implying $EM\perp CD$. On the other hand, for point $F$ being intersection of lines $AB=MN$, $AF=BF$ implies $PM=MQ$. We obtain $EP=EQ$.

WELL-KNOWN. We will show $EN$ bisects $\angle CND$. This is true because $$\angle ENB=\angle EAB=\angle ENC$$and $\angle ANE=\angle ABE=\angle BND$. Thus, $EN$ bisects $\angle CND$.
Z K Y
N Quick Reply
G
H
=
a