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
I need this. Pls help me
Giahuytls2326   12
N 6 minutes ago by Giahuytls2326
Source: Own
Given $\triangle ABC$ , orthocenter $H$, $E$ is the center of Euler's circle of $\triangle ABC$. $X,Y,Z$ is the midpoint of $AH,BH,CH$. From $X,Y,Z$ draw the tangent of $(E)$, cut a line through $E$ perpendicular to sides $BC, CA,AB$ at $M,N,P$ .Prove that $AM,BN,CP$ concurrent
12 replies
Giahuytls2326
Thursday at 4:59 PM
Giahuytls2326
6 minutes ago
Thanks u!
Ruji2018252   4
N 7 minutes ago by sqing
Let $a,b,c>0$ and $a+b+c=abc$
Find minimum (and prove)
\[P=(ab-1)(bc+1)^6(ca-1)\]
4 replies
Ruji2018252
Yesterday at 2:38 PM
sqing
7 minutes ago
Functional Equation
AnhQuang_67   4
N 7 minutes ago by gordian.knot
Find all functions $f:\mathbb{R} \to \mathbb{R}$ satisfying: $$x^2f(x)+f(1-x)=2x-x^4, \forall x \in \mathbb{R} $$
4 replies
AnhQuang_67
Yesterday at 4:02 PM
gordian.knot
7 minutes ago
thanks u!
Ruji2018252   1
N 27 minutes ago by sqing
Let $a,b,c\in \mathbb{R},a,b,c\ne 0$ and $a+b+c=0.$ Find minimum (and prove)
\[C=\dfrac{b^2+c^2-a^2}{b^2+c^2}+\dfrac{c^2+a^2-b^2}{c^2+a^2}+\dfrac{a^2+b^2-c^2}{a^2+b^2}\]
1 reply
1 viewing
Ruji2018252
3 hours ago
sqing
27 minutes ago
No more topics!
best bounds of f(a,b,c)=4(1/a+1/b+1/c)-1/abc, when a+b+c=1, 0<a,b,c<1/2
parmenides51   7
N Dec 20, 2024 by iStud
Source: Balkan BMO Shortlist 2017 A4
Let $M = \{(a,b,c)\in R^3 :0 <a,b,c<\frac12$ with $a+b+c=1 \}$ and $f: M\to  R$ given as $$f(a,b,c)=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{abc}$$Find the best (real) bounds $\alpha$ and $\beta$ such that $f(M) = \{f(a,b,c): (a,b,c)\in M\}\subseteq [\alpha,\beta]$ and determine whether any of them is achievable.
7 replies
parmenides51
Aug 1, 2019
iStud
Dec 20, 2024
best bounds of f(a,b,c)=4(1/a+1/b+1/c)-1/abc, when a+b+c=1, 0<a,b,c<1/2
G H J
G H BBookmark kLocked kLocked NReply
Source: Balkan BMO Shortlist 2017 A4
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
parmenides51
30618 posts
#1 • 4 Y
Y by jhu08, Adventure10, Mango247, ehuseyinyigit
Let $M = \{(a,b,c)\in R^3 :0 <a,b,c<\frac12$ with $a+b+c=1 \}$ and $f: M\to  R$ given as $$f(a,b,c)=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{1}{abc}$$Find the best (real) bounds $\alpha$ and $\beta$ such that $f(M) = \{f(a,b,c): (a,b,c)\in M\}\subseteq [\alpha,\beta]$ and determine whether any of them is achievable.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mihaig
7338 posts
#2 • 2 Y
Y by jhu08, Adventure10
Beautiful one.
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
hakN
428 posts
#3 • 4 Y
Y by jhu08, teomihai, Assassino9931, Mathlover_1
Note that $(1-2a)(1-2b)(1-2c) > 0 \implies 4\sum ab > 8abc + 1$.
So, $f(a,b,c) = \frac{4\sum ab - 1}{abc} > 8$, and equality cannot be achieved.

We will prove $f(a,b,c) \leq 9$.
It suffices to show $$\frac{4\sum ab - (\sum a)^3}{abc} \leq 9 \iff \sum a^3 + 3\sum_{\text{sym}} a^2b + 15abc \geq 4(\sum ab)(\sum a)$$$$\iff \sum a^3 + 3abc \geq \sum_{\text{sym}} a^2b$$But the last inequality is Schur's inequality. Thus equality is achieved if and only if $a=b=c=\frac{1}{3}$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Iora
194 posts
#4
Y by
Define $\overline{M}= \{ (a,b,c \in \mathbb{R} : 0 \le a,b,c \le \frac{1}{2} : a+b+c=1 \}$. Note that $\overline{M}$ is bounded and closed, meaning it is a compact set. Hence $f: \overline{M} \rightarrow \mathbb{R}$ has minimum/maximum value in constraint $a+b+c=1$. Let's say $f$ gets its minimum/maximum value at the boundary. Therefore we have either $a=0$ or $a=\frac{1}{2}$. Since $\frac{1}{0}$ is undefined, we can still calculate the value by limits.It is not hard to see that we have to check $a=b=0,c=1$,$a=0,b=c=\frac{1}{2}$ case. $c=1$ case is clearly impossible. therefore for second case which is $a \rightarrow 0, b,c \rightarrow \frac{1}{2}$ we can see $f \rightarrow 8$

If $a \rightarrow \frac{1}{2}$ we have to check $b \rightarrow \frac{1}{2},c \rightarrow 0$ and $b,c\rightarrow \frac{1}{4}$. The first case is same as above.The second case gives $f \rightarrow 8$. Therefore we have finished for this case.

The second case is when $a,b,c$ lies on open subset $M$ only, we can bash this with lagrange multipliers since all the functions are differentiable in set $M$. The equation $f_a=f_b=f_c$ gives obvious answer $a=b=c=\frac{1}{3}$. By Graphing the function estimately, we can see there is no such $a\neq b \neq c$ satisfying $f_a=f_b=f_c$. For case $a=b$ we can let $c=1-2b$ and analyze new 1 variable function, which again gives $a=b=c$.
and $f(\frac{1}{3},\frac{1}{3},\frac{1}{3})=9$

Therefore $8 < f \le 9$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mihaig
7338 posts
#5
Y by
See also https://artofproblemsolving.com/community/c6h1886311p12854498
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
amogususususus
369 posts
#6
Y by
Very fun! , we'll prove that $f(M) = (8,9]$. To show this we'll need to do three things, proving the upper and lower bound while also proving that every value is attainable.
Upper Bound. $f(a,b,c) \le 9$ for all $(a,b,c) \in M$
Proof. First, let's use $(a+b+c)^2=1$ to transform
$$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}=\frac{1-2(a^2+b^2+c^2)}{abc}$$. What we want to prove is
$$2(a^2+b^2+c^2)+9abc\ge 1$$, using $a+b+c=1$ to homogenize the inequality yields
$$2(a^2+b^2+c^2)(a+b+c)+9abc\ge (a+b+c)^3$$$$\Rightarrow 2\sum_{cyc}a^3+2\sum_{sym}a^2b+9abc \ge \sum_{cyc}a^3 + 3\sum_{sym}a^2b +6ab$$$$\Rightarrow \sum_{cyc}a^3+3abc\ge \sum_{sym}a^2b$$. Which is true by Schur's Inequality. As desired.
Lower Bound. $f(a,b,c) > 8$ for all $(a,b,c) \in M$
Proof. Note that
$$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}$$What we want to prove is
$$1+8abc<4(ab+bc+ca)$$. Homogenizing the inequality and cancelling yields (same as before)
$$(a+b+c)^3+8abc<4(ab+bc+ca)(a+b+c)$$$$\Rightarrow \sum_{cyc}a^3+ 3\sum_{sym}a^2b+14abc<4\sum_{cyc}a^2b+12abc$$$$\Rightarrow \sum_{cyc}a^3+2abc < \sum_{sym}a^2b$$. Note that $a,b,c$ are the sides of a triangle ($b+c>a$, and similarly for others), let $a=x+y,b=y+z,c=z+x$ where $x,y,z>0$. We have,
$$ \sum_{cyc}a^3+2abc=\sum_{cyc} x^3+3x^2y+3xy^2+y^3 + 2\sum_{sym}x^2y+4xyz $$$$\Rightarrow \sum_{cyc}a^3+2abc=2\sum_{cyc}x^3 +5\sum_{sym}x^2y+4xyz$$. While
$$\sum_{sym}a^2b = \sum_{sym}y^3+3x^2y+y^2z+zx^2+2xyz$$$$\Rightarrow \sum_{sym}a^2b = 2\sum_{cyc}x^3+5\sum_{sym}x^2y+12xyz$$. So,
$$\Rightarrow \sum_{cyc}a^3+2abc < \sum_{sym}a^2b$$. As desired.
Attainability. Every value in $(8,9]$ is attainable
Proof. Pick $a=b=c=\frac{1}{3}$ to get $f(a,b,c)=9$. Next, pick $a=b=x$ and $c=1-2x$ where $x \in \left( \frac{1}{3},\frac{1}{2} \right)$, it obeys the restriction $0<a,b,c <\frac{1}{2}$. Then,
$$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}=\frac{-12x^2+8x-1}{x^2(1-2x)}=\frac{6x-1}{x^2}$$. Define $g(x)=\frac{6x-1}{x^2}$ where $x \in \left( \frac{1}{3},\frac{1}{2} \right)$ , then
$$g'(x)=-6x^{-2}+2x^{-3}=2x^{-2}(x^{-1}-3)<0$$. So $g$ is strictly decreasing and clearly continuous at $\left( \frac{1}{3},\frac{1}{2} \right)$. We have $g\left( \frac{1}{3}\right)=9$ and $g\left( \frac{1}{2} \right)=8$. Hence every value in $(8,9)$ is attainable by $g$, and our job is finished :showoff: .
Hence $\alpha=8$ and $\beta=9$ where $\alpha$ is an infimum while $\beta$ is a maximum.
This post has been edited 5 times. Last edited by amogususususus, Dec 22, 2024, 6:01 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
XuNuo
26 posts
#7
Y by
First,let's use $a+b+c=1$ to transform
$$f(a,b,c)=6+\sum_{cyc}\frac{a^2+b^2-c^2}{ab}$$Note that $a,b,c$ are the lengths of the sides of an arbitrary triangle,
so $cos C=\frac{a^2+b^2-c^2}{2ab}$,
therefore $$f(a,b,c)=6+\sum_{cyc}cos A$$and in a triangle $1 < \sum_{cyc}cos A \le \frac{3}{2}$
then $8 < f(a,b,c) \le 9$
This post has been edited 3 times. Last edited by XuNuo, Dec 21, 2024, 1:38 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
iStud
243 posts
#8
Y by
amogususususus wrote:
Very fun! , we'll prove that $f(M) = (8,9]$. To show this we'll need to do three things, proving the upper and lower bound while also proving that every value is attainable.
Upper Bound. $f(a,b,c) \le 9$ for all $(a,b,c) \in M$
Proof. First, let's use $(a+b+c)^2=1$ to transform
$$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}=\frac{1-2(a^2+b^2+c^2}{abc}$$. What we want to prove is
$$2(a^2+b^2+c^2)+9abc\ge 1$$, using $a+b+c=1$ to homogenize the inequality yields
$$2(a^2+b^2+c^2)(a+b+c)+9abc\ge (a+b+c)^3$$$$\Rightarrow 2\sum_{cyc}a^3+2\sum_{sym}a^2b+9abc \ge \sum_{cyc}a^3 + 3\sum_{sym}a^2b +6ab$$$$\Rightarrow \sum_{cyc}a^3+3abc\ge \sum_{sym}a^2b$$. Which is true by Schur's Inequality. As desired.
Lower Bound. $f(a,b,c) > 8$ for all $(a,b,c) \in M$
Proof. Note that
$$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}$$What we want to prove is
$$1+8abc<4(ab+bc+ca)$$. Homogenizing the inequality and cancelling yields (same as before)
$$(a+b+c)^3+8abc<4(ab+bc+ca)(a+b+c)$$$$\Rightarrow \sum_{cyc}a^3+ 3\sum_{sym}a^2b+14abc<4\sum_{cyc}a^2b+12abc$$$$\Rightarrow \sum_{cyc}a^3+2abc < \sum_{sym}a^2b$$. Note that $a,b,c$ are the sides of a triangle ($b+c>a$, and similarly for others), let $a=x+y,b=y+z,c=z+x$ where $x,y,z>0$. We have,
$$ \sum_{cyc}a^3+2abc=\sum_{cyc} x^3+3x^2y+3xy^2+y^3 + 2\sum_{sym}x^2y+4xyz $$$$\Rightarrow \sum_{cyc}a^3+2abc=2\sum_{cyc}x^3 +5\sum_{sym}x^2y+4xyz$$. While
$$\sum_{sym}a^2b = \sum_{sym}y^3+3x^2y+y^2z+zx^2+2xyz$$$$\Rightarrow \sum_{sym}a^2b = 2\sum_{cyc}x^3+5\sum_{sym}x^2y+12xyz$$. So,
$$\Rightarrow \sum_{cyc}a^3+2abc < \sum_{sym}a^2b$$. As desired.
Attainability. Every value in $(8,9]$ is attainable
Proof. Pick $a=b=c=\frac{1}{3}$ to get $f(a,b,c)=9$. Next, pick $a=b=x$ and $c=1-2x$ where $x \in \left( \frac{1}{3},\frac{1}{2} \right)$, it obeys the restriction $0<a,b,c <\frac{1}{2}$. Then,
$$f(a,b,c)=\frac{4(ab+bc+ca)-1}{abc}=\frac{-12x^2+8x-1}{x^2(1-2x)}=\frac{6x-1}{x^2}$$. Define $g(x)=\frac{6x-1}{x^2}$ where $x \in \left( \frac{1}{3},\frac{1}{2} \right)$ , then
$$g'(x)=-6x^{-2}+2x^{-3}=2x^{-2}(x^{-1}-3)<0$$. So $g$ is strictly decreasing and clearly continuous at $\left( \frac{1}{3},\frac{1}{2} \right)$. We have $g\left( \frac{1}{3}\right)=9$ and $g\left( \frac{1}{2} \right)=8$. Hence every value in $(8,9)$ is attainable by $g$, and our job is finished :showoff: .
Hence $\alpha=8$ and $\beta=9$ where $\alpha$ is an infimum while $\beta$ is a maximum.

can you give a formal proof of function $g$'s continuity? i think you should prove that the limit exists
Z K Y
N Quick Reply
G
H
=
a