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a+b+c=3 ine
jokehim   0
42 minutes ago
Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that $$\color{black}{\frac{a\left(b+c\right)}{bc+3}+\frac{b\left(c+a\right)}{ca+3}+\frac{c\left(a+b\right)}{ab+3}\le \frac{3}{2}.}$$Proposed by Phan Ngoc Chau
0 replies
1 viewing
jokehim
42 minutes ago
0 replies
Inequalities
sqing   3
N an hour ago by sqing
Let $ a,b\geq 2  . $ Prove that
$$ (1-a^2)(1-b^2) (ab-1)^2-108  ab \geq-351 $$$$(1-a^2)(1-b^2) (ab-1)(a^2b^2-1)-  621 ab \geq-2079 $$
3 replies
sqing
an hour ago
sqing
an hour ago
[Registration Open] Mustang Math Tournament 2025
MustangMathTournament   31
N Today at 4:00 AM by ethan2011
Mustang Math is excited to announce that registration for our annual tournament, MMT 2025, is open! This year, we are bringing our tournament to 9 in-person locations, as well as online!

Locations include: Colorado, Norcal, Socal, Georgia, Illinois, Massachusetts, New Jersey, Nevada, Washington, and online. For registration and more information, check out https://mustangmath.com/competitions/mmt-2025.

MMT 2025 is a math tournament run by a group of 150+ mathematically experienced high school and college students who are dedicated to providing a high-quality and enjoyable contest for middle school students. Our tournament centers around teamwork and collaboration, incentivizing students to work with their teams not only to navigate the challenging and interesting problems of the tournament but also to develop strategies to master the unique rounds. This includes a logic puzzle round, a strategy-filled hexes round, a race-like gallop round, and our trademark ‘Mystery Mare’ round!

Awards:
[list]
[*] Medals for the top teams
[*] Shirts, pins, stickers and certificates for all participants
[*] Additional awards provided by our wonderful sponsors!
[/list]

We are also holding a free MMT prep seminar from 3/15-3/16 to help students prepare for the upcoming tournament. Join the Google Classroom! https://classroom.google.com/c/NzQ5NDUyNDY2NjM1?cjc=7sogth4
31 replies
MustangMathTournament
Mar 8, 2025
ethan2011
Today at 4:00 AM
JSMCR Results
FuturePanda   14
N Today at 2:40 AM by Pengu14
Hi everyone,

Did anyone get their JSMCR decisions back yet? They were supposed to release on 2/28

Thanks!
14 replies
FuturePanda
Mar 1, 2025
Pengu14
Today at 2:40 AM
AMC 10.........
BAM10   8
N Today at 2:19 AM by ChickensEatGrass
I'm in 8th grade and have never taken the AMC 10. I am currently in alg2. I have scored 20 on AMC 8 this year and 34 on the chapter math counts last year. Can I qualify for AIME. Also what should I practice AMC 10 next year?
8 replies
BAM10
Mar 2, 2025
ChickensEatGrass
Today at 2:19 AM
Tennessee Math Tournament (TMT) Online 2025
TennesseeMathTournament   31
N Today at 2:16 AM by NashvilleSC
Hello everyone! We are excited to announce a new competition, the Tennessee Math Tournament, created by the Tennessee Math Coalition! Anyone can participate in the virtual competition for free.

The testing window is from March 22nd to April 5th, 2025. Virtual competitors may participate in the competition at any time during that window.

The virtual competition consists of three rounds: Individual, Bullet, and Team. The Individual Round is 60 minutes long and consists of 30 questions (AMC 10 level). The Bullet Round is 20 minutes long and consists of 80 questions (Mathcounts Chapter level). The Team Round is 30 minutes long and consists of 16 questions (AMC 12 level). Virtual competitors may compete in teams of four, or choose to not participate in the team round.

To register and see more information, click here!

If you have any questions, please email connect@tnmathcoalition.org or reply to this thread!
31 replies
TennesseeMathTournament
Mar 9, 2025
NashvilleSC
Today at 2:16 AM
hcssim application question
enya_yurself   2
N Today at 1:35 AM by Magnetoninja
do they send the Interesting Test to everyone who applied or do they read the friendly letter first and only send to the kids they like?
2 replies
enya_yurself
Yesterday at 11:13 PM
Magnetoninja
Today at 1:35 AM
The return of American geo
brianzjk   77
N Today at 1:17 AM by Ilikeminecraft
Source: USAJMO 2023/6
Isosceles triangle $ABC$, with $AB=AC$, is inscribed in circle $\omega$. Let $D$ be an arbitrary point inside $BC$ such that $BD\neq DC$. Ray $AD$ intersects $\omega$ again at $E$ (other than $A$). Point $F$ (other than $E$) is chosen on $\omega$ such that $\angle DFE = 90^\circ$. Line $FE$ intersects rays $AB$ and $AC$ at points $X$ and $Y$, respectively. Prove that $\angle XDE = \angle EDY$.

Proposed by Anton Trygub
77 replies
brianzjk
Mar 23, 2023
Ilikeminecraft
Today at 1:17 AM
Apply for Team USA at the International Math Competition (IMC)!
peace09   53
N Today at 1:17 AM by stjwyl
The International Math Competition (IMC) is essentially the elementary and middle school equivalent of the IMO, with individual and team rounds featuring both short-answer and proof-based problems. See past problems here.

Team USA is looking for 6th graders and below with AIME qualification or AMC 8 DHR (or equivalent), and for 9th graders and below with JMO or Mathcounts Nationals qualification. If you think you meet said criteria, fill out the initial form here.

Here are a couple quick links for further information:
[list=disc]
[*] Dr. Tao Hong's website, which contains a detailed recap of the 2024 competition (and previous years'), as well as Team USA's historical results. (You may recognize a couple names... @channing421 @vrondoS et al.: back me up here :P)
[*] My journal, which gives an insider's perspective on the camp :ninja:
[/list]
53 replies
peace09
Aug 13, 2024
stjwyl
Today at 1:17 AM
OTIS Mock AIME 2025 airs Dec 19th
v_Enhance   39
N Today at 1:06 AM by MonkeyLuffy
Source: https://web.evanchen.cc/mockaime.html
Satisfactory. Keep cooking.
IMAGE

Problems are posted at https://web.evanchen.cc/mockaime.html#current now!

Like last year, we're running the OTIS Mock AIME 2025 again, except this time there will actually be both a I and a II because we had enough problems to pull it off. However, the two versions will feel quite different from each other:

[list]
[*] The OTIS Mock AIME I is going to be tough. It will definitely be harder than the actual AIME, by perhaps 2 to 4 problems. But more tangibly, it will also have significant artistic license. Problems will freely assume IMO-style background throughout the test, and intentionally stretch the boundary of what constitutes an “AIME problem”.
[*] The OTIS Mock AIME II is meant to be more practically useful. It will adhere more closely to the difficulty and style of the real AIME. There will inevitably still be some more IMO-flavored problems, but they’ll appear later in the ordering.
[/list]
Like last time, all 30 problems are set by current and past OTIS students.

Details are written out at https://web.evanchen.cc/mockaime.html, but to highlight important info:
[list]
[*] Free, obviously. Anyone can participate.
[*]Both tests will release sometime Dec 19th. You can do either/both.
[*]If you'd like to submit for scoring, you should do so by January 20th at 23:59 Pacific time (same deadline for both). Please hold off on public spoilers before then.
[*]Solutions, statistics, and maybe some high scores will be published shortly after that.
[/list]
Feel free to post questions, hype comments, etc. in this thread.
39 replies
1 viewing
v_Enhance
Dec 6, 2024
MonkeyLuffy
Today at 1:06 AM
They mixed up USAJMO and AIME I guess
Math4Life7   54
N Today at 12:36 AM by littlefox_amc
Source: USAJMO 2024/1
Let $ABCD$ be a cyclic quadrilateral with $AB = 7$ and $CD = 8$. Point $P$ and $Q$ are selected on segment $AB$ such that $AP = BQ = 3$. Points $R$ and $S$ are selected on segment $CD$ such that $CR = DS = 2$. Prove that $PQRS$ is a cyclic quadrilateral.

Proposed by Evan O'Dorney
54 replies
Math4Life7
Mar 20, 2024
littlefox_amc
Today at 12:36 AM
USAMO vs USAJMO Prestige
elasticwealth   47
N Today at 12:36 AM by axusus
Just curious, what does everyone think about the prestige of a USAMO qualification vs a USAJMO qual? Obv USAMO > USAJMO but how much?

And while we’re at it, what about amc 10 dhr vs amc 12 dhr? Thoughts?

Edit: also how much is a perfect score on 10/12 worth? Asking for a friend….
47 replies
elasticwealth
Feb 17, 2025
axusus
Today at 12:36 AM
Inequality with integers and indices
Michael Niland   6
N 3 hours ago by allisok100
Without using a calculator, show that $2^{\frac{1}{2}}$ $< 3^{\frac{1}{3}}$, and for a positive integer $n\geq3$, prove that

$n^{\frac{1}{n}}$ $>(n+1)^{\frac{1}{n+1}}$.
6 replies
Michael Niland
Yesterday at 6:52 AM
allisok100
3 hours ago
Inequality with integers and indices
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Michael Niland
681 posts
#1
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Without using a calculator, show that $2^{\frac{1}{2}}$ $< 3^{\frac{1}{3}}$, and for a positive integer $n\geq3$, prove that

$n^{\frac{1}{n}}$ $>(n+1)^{\frac{1}{n+1}}$.
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invisibleman
7 posts
#2
Y by
In order to solve the first problem, I have the following idea: raise both sides to the 6th power, and you're done! Right?
In the general case, consider the function f(x)=lnx/x, whose derivative is negative for n>2, so the function is decreasing. Let x=n and x=n+1, and compare them! I think you understand, if not, write and I'll answer!
This post has been edited 2 times. Last edited by invisibleman, Yesterday at 1:10 PM
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krish6_9
18 posts
#3
Y by
The inequality is equivalent to $n^{n+1} > (n+1)^n$. Taking natural logarithm on both sides, this is equivalent to $(n+1) \ln n > n \ln (n+1).$ Dividing, this is equivalent to $\frac{n+1}{\ln (n+1)} > \frac{n}{\ln n}$. Since $\frac{x}{\ln x}$ is increasing (obviously linear functions increase more quickly than logarithms, or using the derivative like @above said), we are done by definition.

i hope this is right! :D :ddr:
well someone taught me the solution hehe
This post has been edited 2 times. Last edited by krish6_9, Today at 12:31 AM
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Speedysolver1
75 posts
#4
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Help me with a forum
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krish6_9
18 posts
#5
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who, me? wut does that mean
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ohiorizzler1434
704 posts
#6
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Bro! Let's show n^(n+1) > (n+1)^n , or (1+1/n)^n < n. Note that (1+1/n)^n < e because this limits upwards towards e. So for n >= 3, this is obviously true as e<3. Bruh! Now that's rizz!
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allisok100
1 post
#7
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f(x)=\frac{\ln x}{x}\Rightarrow f'(x)=\frac{1-\ln x}{{{x}^{2}}}<0,\forall x>e\Rightarrow f(n)>f(n+1),\forall n\ge 2
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