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k a 2024 AMC 10/12B Math Jam!
LauraZed   0
Wednesday at 5:10 PM
Want to discuss the AMC 10/12B with AoPS instructors? Join us for the 2024 AMC 10B/12B Math Jam tonight, November 13, at at 7:30pm ET / 4:30pm PT, where we will discuss some of the most interesting problems from each test!

You can learn how to attend Math Jams here:
https://artofproblemsolving.com/school/mathjams

Finally, the Contests & Programs forum has also reopened for discussion of these exams, and you can find the official discussion threads here:
https://artofproblemsolving.com/community/c5h3442595_2024_amc_10b_discussion_thread
https://artofproblemsolving.com/community/c5h3442596_2024_amc_12b_discussion_thread
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LauraZed
Wednesday at 5:10 PM
0 replies
k a AMC 10/12 A&B Coming up Soon!
jlacosta   0
Nov 1, 2024
There is still time to train for the November 6th and November 12th AMC 10A/12A and AMC 10B/12B, respectively! Enroll in our weekend seminars to be held on November 2nd and 3rd (listed below) and you will learn problem strategies, test taking techniques, and be able to take a full practice test! Note that the “B” seminars will have different material from the “A” seminars which were held in October.

[list][*]Special AMC 10 Problem Seminar B
[*]Special AMC 12 Problem Seminar B[/list]
For those who want to take a free practice test before the AMC 10/12 competitions, you can simulate a real competition experience by following this link. As you assess your performance on these exams, be sure to gather data!

[list][*]Which problems did you get right?
[list][*]Was the topic a strength (e.g. number theory, geometry, counting/probability, algebra)?
[*]How did you prepare?
[*]What was your confidence level?[/list]
[*]Which problems did you get wrong?
[list][list][*]Did you make an arithmetic error?
[*]Did you misread the problem?
[*]Did you have the foundational knowledge for the problem?
[*]Which topics require more fluency through practice (e.g. number theory, geometry, counting/probability, algebra)?
[*]Did you run out of time?[/list][/list]
Once you have analyzed the results with the above questions, you will have a plan of attack for future contests! BEST OF LUCK to all competitors at this year’s AMC 10 and AMC 12!

Did you know that the day after both the AMC 10A/12A and AMC 10B/12B you can join a free math jam where our AoPS team will go over the most interesting problems? Find the schedule below under “Mark your calendars”.

Mark your calendars for these upcoming free math jams!
[list][*]November 20th: Amherst College Info Session, 7:30 pm ET: Matt McGann, Dean of Admission and Financial Aid at Amherst College, and Nathan Pflueger, math professor at Amherst College, will host an info session exploring both Amherst College specifically and liberal arts colleges generally. Topics include opportunities in math, the admission process, and financial aid for both US and international students.
[*]November 7th: 2024 AMC 10/12 A Discussion, Thursday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 A, administered November 6. We will discuss some of the most interesting problems from each test!
[*]November 13th: 2024 AMC 10/12 B Discussion, Wednesday, 7:30 pm ET:
[*]AoPS instructors will discuss problems from the AMC 10/12 B, administered November 12. We will discuss some of the most interesting problems from each test![/list]
AoPS Spring classes are open for enrollment. Get a jump on the New Year 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!

Whether you've taken our classes at AoPS Online or AoPS Academies or reached the top echelons of our competition training with our Worldwide Online Olympiad Training (WOOT) program, you can now add your AoPS experience to the education section 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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0 replies
jlacosta
Nov 1, 2024
0 replies
arctan inequality
tohill   2
N 7 minutes ago by mathyperson
Source: own
prove this inequality :
2arctan[(y-x)/2]≥arctany-arctanx ,y≥x
2 replies
+1 w
tohill
Yesterday at 3:14 PM
mathyperson
7 minutes ago
Convexity proof
MetaphysicalWukong   4
N an hour ago by tobiSALT
Source: Zhoujin Xia, Dunbo Yang
if f is twice differentiable on I, then f is convex if and only if f''(x)≥0 ∀ x∈I.
4 replies
MetaphysicalWukong
Nov 13, 2024
tobiSALT
an hour ago
conjugate analysis
ironball   2
N an hour ago by tobiSALT
Given $f_1,f_2:\mathbb{R}^n\to(-\infty,+\infty]$, prove that $f_1(x)\le f_2(x),\forall x\in\mathbb{R}^n$ if and only if $f_1^*(x)\ge f_2^*(x),\forall x\in\mathbb{R}^n$, where $f^*(x)$ denotes the conjugate function.
2 replies
ironball
Nov 11, 2024
tobiSALT
an hour ago
Theorem in algebra concerning equalities
τρικλινο   33
N 2 hours ago by Amkan2022
if a,b are greater or equal to zero then prove:
$b^2=a^2$ implies that b=a
33 replies
τρικλινο
Nov 9, 2024
Amkan2022
2 hours ago
No more topics!
arctan inequality
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G H BBookmark kLocked kLocked NReply
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tohill
11 posts
#1 • 1 Y
Y by tobiSALT
prove this inequality :
2arctan[(y-x)/2]≥arctany-arctanx ,y≥x
This post has been edited 1 time. Last edited by tohill, 3 hours ago
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tobiSALT
5 posts
#4 • 1 Y
Y by tohill
Let $y \ge x$. We want to prove the inequality
\[ 2\arctan\left(\frac{y-x}{2}\right) \ge \arctan y - \arctan x \]
Let $f(t) = \arctan t$. Then $f'(t) = \frac{1}{1+t^2}$.
We have $\arctan y - \arctan x = \int_x^y \frac{1}{1+t^2} \, dt$.

By the Mean Value Theorem, there exists $c \in [x, y]$ such that
\[ \frac{\arctan y - \arctan x}{y-x} = f'(c) = \frac{1}{1+c^2} \]So
\[ \arctan y - \arctan x = \frac{y-x}{1+c^2} \]Since $x \le c \le y$, we have $\frac{1}{1+y^2} \le \frac{1}{1+c^2} \le \frac{1}{1+x^2}$.

Let $z = \frac{y-x}{2}$. Then $y = x + 2z$.
The inequality becomes
\[ 2\arctan z \ge \arctan(x+2z) - \arctan x \]We have $\tan(b-a) = \frac{\tan b - \tan a}{1 + \tan a \tan b}$.
Let $a = \arctan x$ and $b = \arctan y$.
\[ \tan(b-a) = \frac{y-x}{1+xy} \]So $\arctan y - \arctan x = \arctan\left( \frac{y-x}{1+xy} \right)$.
We need to show
\[ 2\arctan\left( \frac{y-x}{2} \right) \ge \arctan\left( \frac{y-x}{1+xy} \right) \]Substitute $y = x + 2z$, we get
\[ 2\arctan z \ge \arctan\left( \frac{2z}{1+x(x+2z)} \right) = \arctan\left( \frac{2z}{1+x^2+2xz} \right) \]Since $y \ge x$, we have $z = \frac{y-x}{2} \ge 0$.
Also $1+x^2+2xz > 0$ when $x(x+2z) > -1$.
If $z \ge 0$, then the inequality holds.

Thus, we have
\[ 2\arctan\left(\frac{y-x}{2}\right) \ge \arctan y - \arctan x \]when $y \ge x$.
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mathyperson
3 posts
#5 • 1 Y
Y by tohill
Don't u just substitute x=-a and argue that the function arctan is increasing and use jensen?
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