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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
March is the month for State MATHCOUNTS competitions! Kudos to everyone who participated in their local chapter competitions and best of luck to all going to State! Join us on March 11th for a Math Jam devoted to our favorite Chapter competition problems! Are you interested in training for MATHCOUNTS? Be sure to check out our AMC 8/MATHCOUNTS Basics and Advanced courses.

Are you ready to level up with Olympiad training? Registration is open with early bird pricing available for our WOOT programs: MathWOOT (Levels 1 and 2), CodeWOOT, PhysicsWOOT, and ChemWOOT. What is WOOT? WOOT stands for Worldwide Online Olympiad Training and is a 7-month high school math Olympiad preparation and testing program that brings together many of the best students from around the world to learn Olympiad problem solving skills. Classes begin in September!

Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

Be sure to mark your calendars for the following events:
[list][*]March 5th (Wednesday), 4:30pm PT/7:30pm ET, HCSSiM Math Jam 2025. Amber Verser, Assistant Director of the Hampshire College Summer Studies in Mathematics, will host an information session about HCSSiM, a summer program for high school students.
[*]March 6th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar on Math Competitions from elementary through high school. Join us for an enlightening session that demystifies the world of math competitions and helps you make informed decisions about your contest journey.
[*]March 11th (Tuesday), 4:30pm PT/7:30pm ET, 2025 MATHCOUNTS Chapter Discussion MATH JAM. AoPS instructors will discuss some of their favorite problems from the MATHCOUNTS Chapter Competition. All are welcome!
[*]March 13th (Thursday), 4:00pm PT/7:00pm ET, Free Webinar about Summer Camps at the Virtual Campus. Transform your summer into an unforgettable learning adventure! From elementary through high school, we offer dynamic summer camps featuring topics in mathematics, language arts, and competition preparation - all designed to fit your schedule and ignite your passion for learning.[/list]
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0 replies
jlacosta
Mar 2, 2025
0 replies
k i A Letter to MSM
Arr0w   23
N Sep 19, 2022 by scannose
Greetings.

I have seen many posts talking about commonly asked questions, such as finding the value of $0^0$, $\frac{1}{0}$,$\frac{0}{0}$, $\frac{\infty}{\infty}$, why $0.999...=1$ or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
[list]
[*]Firstly, the case of $0^0$. It is usually regarded that $0^0=1$, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.

[*]What about $\frac{\infty}{\infty}$? The issue here is that $\infty$ isn't even rigorously defined in this expression. What exactly do we mean by $\infty$? Unless the example in question is put in context in a formal manner, then we say that $\frac{\infty}{\infty}$ is meaningless.

[*]What about $\frac{1}{0}$? Suppose that $x=\frac{1}{0}$. Then we would have $x\cdot 0=0=1$, absurd. A more rigorous treatment of the idea is that $\lim_{x\to0}\frac{1}{x}$ does not exist in the first place, although you will see why in a calculus course. So the point is that $\frac{1}{0}$ is undefined.

[*]What about if $0.99999...=1$? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
\begin{align*}
\sum_{n=1}^{\infty} \frac{9}{10^n}&=9\sum_{n=1}^{\infty}\frac{1}{10^n}=9\sum_{n=1}^{\infty}\biggr(\frac{1}{10}\biggr)^n=9\biggr(\frac{\frac{1}{10}}{1-\frac{1}{10}}\biggr)=9\biggr(\frac{\frac{1}{10}}{\frac{9}{10}}\biggr)=9\biggr(\frac{1}{9}\biggr)=\boxed{1}
\end{align*}
[*]What about $\frac{0}{0}$? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
[/list]
Hopefully all of these issues and their corollaries are finally put to rest. Cheers.

2nd EDIT (6/14/22): Since I originally posted this, it has since blown up so I will try to add additional information per the request of users in the thread below.

INDETERMINATE VS UNDEFINED

What makes something indeterminate? As you can see above, there are many things that are indeterminate. While definitions might vary slightly, it is the consensus that the following definition holds: A mathematical expression is be said to be indeterminate if it is not definitively or precisely determined. So how does this make, say, something like $0/0$ indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that $0/0$ is an indeterminate form.

But we need to more specific, this is still ambiguous. An indeterminate form is a mathematical expression involving at most two of $0$, $1$ or $\infty$, obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are
\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function $f:\mathbb{R}^{+}\cup\{0\}\rightarrow\mathbb{R}$ given by the mapping $x\mapsto \sqrt{x}$ is undefined for $x<0$. On the other hand, $1/0$ is undefined because dividing by $0$ is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context.

WHEN THE WATERS GET MUDDIED

So with this notion of indeterminate and undefined, things get convoluted. First of all, just because something is indeterminate does not mean it is not undefined. For example $0/0$ is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.

One example of making something undefined into something defined is the extended real number line, which we define as
\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let $-\infty\le a\le \infty$ for each $a\in\overline{\mathbb{R}}$ which means that via this order topology each subset has an infimum and supremum and $\overline{\mathbb{R}}$ is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In $\overline{\mathbb{R}}$ it is perfectly OK to say that,
\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,
\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define $\infty \times 0=0$ as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by $0$ is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as
\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, $\tilde{\infty}$ means complex infinity, since we are in the complex plane now. Here's the catch: division by $0$ is allowed here! In fact, we have
\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]where $\tilde{\infty}/\tilde{\infty}$ and $0/0$ are left undefined. We also have
\begin{align*}
z+\tilde{\infty}=\tilde{\infty}, \forall z\ne -\infty\\
z\times \tilde{\infty}=\tilde{\infty}, \forall z\ne 0
\end{align*}Furthermore, we actually have some nice properties with multiplication that we didn't have before. In $\mathbb{C}^*$ it holds that
\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]but $\tilde{\infty}-\tilde{\infty}$ and $0\times \tilde{\infty}$ are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with $\mathbb{C}^*$, by defining them as
\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]They behave in a similar way to the Riemann Sphere, with division by $0$ also being allowed with the same indeterminate forms (in addition to some other ones).
23 replies
Arr0w
Feb 11, 2022
scannose
Sep 19, 2022
k i Marathon Threads
LauraZed   0
Jul 2, 2019
Due to excessive spam and inappropriate posts, we have locked the Prealgebra and Beginning Algebra threads.

We will either unlock these threads once we've cleaned them up or start new ones, but for now, do not start new marathon threads for these subjects. Any new marathon threads started while this announcement is up will be immediately deleted.
0 replies
LauraZed
Jul 2, 2019
0 replies
k i Basic Forum Rules and Info (Read before posting)
jellymoop   368
N May 16, 2018 by harry1234
f (Reminder: Do not post Alcumus or class homework questions on this forum. Instructions below.) f
Welcome to the Middle School Math Forum! Please take a moment to familiarize yourself with the rules.

Overview:
[list]
[*] When you're posting a new topic with a math problem, give the topic a detailed title that includes the subject of the problem (not just "easy problem" or "nice problem")
[*] Stay on topic and be courteous.
[*] Hide solutions!
[*] If you see an inappropriate post in this forum, simply report the post and a moderator will deal with it. Don't make your own post telling people they're not following the rules - that usually just makes the issue worse.
[*] When you post a question that you need help solving, post what you've attempted so far and not just the question. We are here to learn from each other, not to do your homework. :P
[*] Avoid making posts just to thank someone - you can use the upvote function instead
[*] Don't make a new reply just to repeat yourself or comment on the quality of others' posts; instead, post when you have a new insight or question. You can also edit your post if it's the most recent and you want to add more information.
[*] Avoid bumping old posts.
[*] Use GameBot to post alcumus questions.
[*] If you need general MATHCOUNTS/math competition advice, check out the threads below.
[*] Don't post other users' real names.
[*] Advertisements are not allowed. You can advertise your forum on your profile with a link, on your blog, and on user-created forums that permit forum advertisements.
[/list]

Here are links to more detailed versions of the rules. These are from the older forums, so you can overlook "Classroom math/Competition math only" instructions.
Posting Guidelines
Update on Basic Forum Rules
What belongs on this forum?
How do I write a thorough solution?
How do I get a problem on the contest page?
How do I study for mathcounts?
Mathcounts FAQ and resources
Mathcounts and how to learn

As always, if you have any questions, you can PM me or any of the other Middle School Moderators. Once again, if you see spam, it would help a lot if you filed a report instead of responding :)

Marathons!
Relays might be a better way to describe it, but these threads definitely go the distance! One person starts off by posting a problem, and the next person comes up with a solution and a new problem for another user to solve. Here's some of the frequently active marathons running in this forum:
[list][*]Algebra
[*]Prealgebra
[*]Proofs
[*]Factoring
[*]Geometry
[*]Counting & Probability
[*]Number Theory[/list]
Some of these haven't received attention in a while, but these are the main ones for their respective subjects. Rather than starting a new marathon, please give the existing ones a shot first.

You can also view marathons via the Marathon tag.

Think this list is incomplete or needs changes? Let the mods know and we'll take a look.
368 replies
jellymoop
May 8, 2015
harry1234
May 16, 2018
IMO 2023 P2
799786   89
N 17 minutes ago by kaede_Arcadia
Source: IMO 2023 P2
Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$.
89 replies
799786
Jul 8, 2023
kaede_Arcadia
17 minutes ago
combinatorıc
o.k.oo   0
38 minutes ago
A total of 3300 handshakes were made at a party attended by 600 people. It was observed
that the total number of handshakes among any 300 people at the party is at least N. Find
the largest possible value for N.
0 replies
o.k.oo
38 minutes ago
0 replies
Problem 2, Olympic Revenge 2013
hvaz   66
N an hour ago by MonkeyLuffy
Source: XII Olympic Revenge - 2013
Let $ABC$ to be an acute triangle. Also, let $K$ and $L$ to be the two intersections of the perpendicular from $B$ with respect to side $AC$ with the circle of diameter $AC$, with $K$ closer to $B$ than $L$. Analogously, $X$ and $Y$ are the two intersections of the perpendicular from $C$ with respect to side $AB$ with the circle of diamter $AB$, with $X$ closer to $C$ than $Y$. Prove that the intersection of $XL$ and $KY$ lies on $BC$.
66 replies
hvaz
Jan 26, 2013
MonkeyLuffy
an hour ago
Functional equation wrapped in f's
62861   35
N an hour ago by ihatemath123
Source: RMM 2019 Problem 5
Determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
\[f(x + yf(x)) + f(xy) = f(x) + f(2019y),\]for all real numbers $x$ and $y$.
35 replies
62861
Feb 24, 2019
ihatemath123
an hour ago
Amc10 prep question
Shadow6885   13
N 2 hours ago by RandomMathGuy500
My question is how much of the geo and IA textbooks is relevant to AMC 10?
13 replies
Shadow6885
Yesterday at 6:20 AM
RandomMathGuy500
2 hours ago
k Is your state listed?
Chatelet1   408
N 3 hours ago by Eddie_tiger
Multiple states have announced their top students who will advance to the 2025 MATHCOUNTS National Competition in May:

• From Alabama: Henry Gladden of Mobile, Austin Lu of Birmingham, Jessie Shi of Vestavia, and Minlu Wang-He of Auburn.

• From Arkansas: Ryan Fan of Fayetteville, Vivek Kalyankar of Fayetteville, Evan Ning of Fayetteville and Charles Yao of Conway.

• From Connecticut: Hayden Hughes of Newtown, Ethan Shi of Riverside, Alex Svoronos of Greenwich and Elaine Zhou of Hamden.

• From the Department of Defense: Narmin Guliyeva of Ankara, Turkey; Taeyul Kim of Manana, Bahrain; Nathan Liang of Wiesbaden, Germany; and Lucas Sze of Okinawa, Japan.

• From Hawaii: Taehwan Jeon, Hilohak Kwak, Isaac Qian and Thien Tran, all from Honolulu.

• From Kansas: Haidan Anderson & Jayden Xue of Overland Park, Christopher Spencer of Manhattan, and Ruby Jiang of Lawrence.

• From Maine: Ana Kanitkar & Connor Kirkham of Falmouth, Anna McClary of Hermon and Poppy Sandin of Bar Harbor.

• From Massachusetts: Eric Huang of Acton, Shlok Mukund & Brandon Ni of Lexington, and Soham Samanta of Medford.

• From Missouri: Lucas Lai of Columbia, Kevin Shi of St. Louis, Charles Yong & Jay Zhou of Chesterfield.

• From Montana: Titus Gilder of Missoula, Otis Heggem of Billings, Kaleb Houtz of Great Falls and Evan Newcomer of Missoula.

• From Nevada: Solomon Dumont of Las Vegas, Aaron Lei of Reno, Leeoz Nebat of Henderson and Maxwell Tsai of Las Vegas.

• From New Mexico: Mark Goldman, Daniel He, Iris Huang and Patrick McArdle, all from Albuquerque.

• From New York: Derrick Chen of Great Neck, Victor Yang of Great Neck, Hanru Zhang of Jericho and Ryan Zhang of Jericho.

• From Rhode Island: Kahlan Anderson of the Wheeler School, Julian Bernhoft & Colin Hegstrom of Providence, and Theodora Watson of Barrington.

• From South Carolina: Yukai Hu of Elgin, Justin Peng of Clemson, Geonhoo Shim of Columbia, and Aaron Wang of Mount Pleasant.

• From South Dakota: Seth Chaplin & Maxwell Wang of Sioux Falls, Laukia Gundewar of Aberdeen, and Cohwen Heimann of Aberdeen.

• From Texas: Shaheem Samsudeen & Ayush Narayan of Plano, Nathan Liu of Richardson, and James Stewart of Southlake.

• From Vermont: Mohid Ali of South Burlington, Vivek Chadive of South Burlington, Joshua Kratze of St. Johnsbury and Albert Zhang of South Burlington.

• From Wisconsin: August Reeder & Lucy Chen of Fitchburg, Junhao Feng of Milwaukee, and Jiyan Singh of River Hills.

===
Updated on 3/15/2025:

• From Colorado: Noah Liu, Christopher Zhu, Neo Luo, and Andrew Zhao.

• From Florida: Arnav Bhatia, Gnaneswar Peddesugari, Edwin Gao, and Rananjay Parmar.

• From Indiana: Roland Li, Hrishabh Bhowmik, Sophia Chen, and Arjun Raman.

• From Kentucky: Sri Shubhaan Vulava, Joyce Liu, Victor Gong, and Brandon Tedja.

• From Maryland: Eric Xie, Angie Zhu, Roger Huang, and Leo Su.

• From Michigan: Arnav Vunnam, Eric Jin, Akshaj Malraj, and Chaithanya Budida.

• From Minnesota: Ahmed Ilyasov, Will Masanz, Anshdeep Singh, and Branden Qiao.

• From New Jersey: Ethan Imanuel, Advait Joshi, Jay Wang, and Easton Wei.

• From North Carolina: Shivank Chintalpati, Steven Wang, Lucas Li, and Leo Hong.

• From Ohio: Henry Lu, Andy Mo, Archishmen Dey, and Caleb Tan.

• From Oregon: Sophia Han, Kevin Cheng, Garud Shah, and Ryan Zhang.
408 replies
1 viewing
Chatelet1
Mar 8, 2025
Eddie_tiger
3 hours ago
probability pentagon contains center
JohnStuckey   3
N 4 hours ago by sadas123
Here's a cute problem:

Consider a regular pentagon. Choose 3 points along the perimeter of the pentagon, and form a triangle with those 3 points. What is the probability that this triangle contains the center of the pentagon.
3 replies
JohnStuckey
Yesterday at 6:27 PM
sadas123
4 hours ago
ohio mathcounts state
Owinner   38
N 4 hours ago by Andyluo
what is the cutoff for cdr for ohio? Is ohio a competitve state?
38 replies
Owinner
Mar 11, 2025
Andyluo
4 hours ago
What was your Mathcounts Chant?
ilovebender   59
N 4 hours ago by wikjay
Hi, I wanted to ask everyone, before the written competition, each state had to do their chant. What chant did you and your team make xd!

Im in WV and I think we did the least cringe chant, and it goes like this:

(After one person says something, the other person following waits ~2 seconds, and then says their line)
P1: Hi
P2: Hi
P3: Hi
P4: Bye

Lol if you were at the national competition you prob remember.
59 replies
ilovebender
May 12, 2022
wikjay
4 hours ago
Problem of the week
evt917   26
N 4 hours ago by PikaPika999
Whenever possible, I will be posting problems twice a week! They will be roughly of AMC 8 difficulty. Have fun solving! Also, these problems are all written by myself!

First problem:

$20^{16}$ has how many digits?
26 replies
evt917
Mar 5, 2025
PikaPika999
4 hours ago
Imposible
maxamc   1
N 6 hours ago by maromex
if 1+1 is 2 then what is the square root of 4 with 100 significant figures?
1 reply
maxamc
6 hours ago
maromex
6 hours ago
9 What should I do..?
Leeoz   6
N Yesterday at 2:13 PM by sadas123
So, there is a very important decision.. and I have decided to asked all of the people on AoPS :P

I am going to MathCounts nats, but there is the MathCon a day before the competition. It there anything I will miss by just going to MathCounts right on the day of the contest, or is it worth it to go to MathCon, even if I will have to leave before the awards.

Just want your ideas and opinions for this :)
also pls say if there is something that I will miss in MathCounts before the actual contest
6 replies
Leeoz
Yesterday at 6:17 AM
sadas123
Yesterday at 2:13 PM
MATHCOUNTS State Preparation
mithu542   27
N Yesterday at 1:16 PM by krish6_9
Hello!

I'm going to prepare for Mathcounts state soon. I want some advice on what to do. I am in 7th grade, and I want to make it to nationals. I know I should obviously take practice tests, but should I do something else other than that, or just grind all (or most) practice tests from previous years? Also, how much should I focus on Countdown round relative to the other tests?

(For reference, I got 43 on school, and 41 on chapter. Last year, I got 16/116 rank in state. Since then, I have done the following courses from aops:
Intro: algebra b, number theory, c&p, geometry
Intermediate: algebra, number theory, c&p)
27 replies
mithu542
Feb 14, 2025
krish6_9
Yesterday at 1:16 PM
9 Pi or Tau
jkim0656   100
N Yesterday at 5:19 AM by jkim0656
Hey Aops!
Pi = Circumfrence/Diameter
Tau = Circumfrence/Radius
I have noticed a lot of sites, including Khan Academy, in support of tau over pi...
so what do you think?
https://www.scientificamerican.com/article/let-s-use-tau-it-s-easier-than-pi/
However i am still in support of the good ol pi :)
(btw this is my first aops poll) :-D

EDIT: 50 votes!!! :play_ball:
EDIT: 100 votes!!! :jump:
EDIT: 150 votes! :trampoline:
EDIT: 200 votes! ;)
Edit: 250 votes !!!! yaya :gathering:

If u support pi pls upvote :)
100 replies
jkim0656
Mar 14, 2025
jkim0656
Yesterday at 5:19 AM
Yet another classical inequality
Valentin Vornicu   43
N Dec 7, 2019 by sqing
Source: Balkan MO 2006, Problem 1
Let $ a,b,c>0$ be real numbers. Prove that
\[ \frac 1{a(1+b)}+\frac 1{b(1+c)}+\frac 1{c(1+a)}\geq \frac 3{ 1+abc}. \]
Greece

EDIT BY DARIJ:

Please use this topic to discuss the sources and history of the above inequality,
and use http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 to discuss the inequality itself (i. e. different proofs and generalizations).
43 replies
Valentin Vornicu
Apr 29, 2006
sqing
Dec 7, 2019
Yet another classical inequality
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G H BBookmark kLocked kLocked NReply
Source: Balkan MO 2006, Problem 1
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Valentin Vornicu
7301 posts
#1 • 2 Y
Y by Adventure10 and 1 other user
Let $ a,b,c>0$ be real numbers. Prove that
\[ \frac 1{a(1+b)}+\frac 1{b(1+c)}+\frac 1{c(1+a)}\geq \frac 3{ 1+abc}. \]
Greece

EDIT BY DARIJ:

Please use this topic to discuss the sources and history of the above inequality,
and use http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 to discuss the inequality itself (i. e. different proofs and generalizations).
This post has been edited 1 time. Last edited by Valentin Vornicu, May 3, 2006, 3:18 PM
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silouan
3952 posts
#2 • 3 Y
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See this http://www.mathlinks.ro/Forum/viewtopic.php?t=161059

and I have no comments . How some people give a problem very-very-well-known in a big contest as BMO
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dule_00
246 posts
#3 • 3 Y
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yeah, like functional eqaution which was twice on BMO?
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mzdanowicz
2 posts
#4 • 3 Y
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Now I know, where I saw this nice method...
What a pity, it's just the same problem ;/

[Moderator edit: The "nice method" referred to here is Mzdanovicz's solution in http://www.mathlinks.ro/Forum/viewtopic.php?t=161059 .]
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Arne
3660 posts
#5 • 2 Y
Y by Adventure10 and 1 other user
I can't believe they give this problem at a BMO. It's just so incredibly well known :huh:
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Valentin Vornicu
7301 posts
#6 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Arne wrote:
I can't believe they give this problem at a BMO. It's just so incredibly well known :huh:
It was problem 1. I assume they knew it was kind of known but they put it there to be the easy problem in the competition ... :?
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Arne
3660 posts
#7 • 3 Y
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What makes it even worse is the fact that this is VERY hard for people who've never seen it before.

It's just unfair!
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Peter
3615 posts
#8 • 2 Y
Y by Adventure10 and 1 other user
That can be said of 30% of the IMC problems...
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Valentin Vornicu
7301 posts
#9 • 2 Y
Y by Adventure10 and 1 other user
Arne wrote:
What makes it even worse is the fact that this is VERY hard for people who've never seen it before.

It's just unfair!
Yeah, that can also be said about usage of ISL problems in TSTs around the world, given that some of the students might know some of the problems ahead ... :?

I belive we got very close to that point where almost all problems are known, and those which aren't are sent to the IMO :D
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pvthuan
844 posts
#10 • 3 Y
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Arne wrote:
What makes it even worse is the fact that this is VERY hard for people who've never seen it before.

It's just unfair!

I can NOT believe that. Why don't they hire some inequality experts from Mathlinks? :lol: ;)
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Arne
3660 posts
#11 • 3 Y
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I guess all team leaders knew this problem. Why didn't they stop it from being selected? :(
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pvthuan
844 posts
#12 • 3 Y
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That can tell us something about Balkan Olympiad. In the IMO, this way is forbidden.
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Lovasz
868 posts
#13 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Is not this problem from Old and new inequalities?
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Arne
3660 posts
#14 • 3 Y
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Might be, although the problem is quite old.

I think the BMO is making a joke of itself this way...
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esti
83 posts
#15 • 3 Y
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Arne wrote:
Might be, although the problem is quite old.

I think the BMO is making a joke of itself this way...
I got feeling long time ago, that competition among only a few countries, like BMO ,has been a joke ever since.
However,reasons like making a friends and visiting beutiful parts of Mediteranian make it sufficient to keep it going every year. :)
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Centy
260 posts
#16 • 2 Y
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I think when the same question in the BMO appeared twice in different years, eyebrows must have been raised.

This question appeared last year on the UK Correspondence Course and I assume it therefore a well publicised inequality.

Still I guess these regional Olympiads should also be about making friends and meeting new people. If it were just about the questions, then why isn't the whole thing done in two days?
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Arne
3660 posts
#17 • 3 Y
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Of course, but I just can't believe the problem committee cannot find any better or new questions.
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Peter
3615 posts
#18 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Maybe... *starts writing the plot of a new detective novell*

this question just got proposed

all team leaders knew that their students knew the question

but none knew it was well-known to the other countries too

so they let is pass :diablo: :D
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Michael
99 posts
#19 • 3 Y
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Valentin Vornicu wrote:
Let $a,b,c>0$ be real numbers. Prove that \[ \frac 1{a(1+b)} + \frac 1{b(1+c)} + \frac 1{c(1+a)} \geq \frac 3 { 1+abc}.  \]

This problem appeared some years ago in Crux Mathematicorum. It was proposed by Mohammed Aassila.
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Arne
3660 posts
#20 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Might be, it appeared like everywhere. I'm sure you can find at least 10 threads on this forum about this problem.
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stergiu
1648 posts
#21 • 3 Y
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I have this inequality in my book on Inequalities(written by Skombris and me, in Greek). As source we give there << Komal 1993 >> , so it was known before Aassila sent it to CRUX , vol 1998. This nice solution, sent by Silouan , is to be found there, too(by a chinese).Of course the solution was discused in ML very wide.

My opinion is that the problem should not be proposed in the BMO 2006 , although very good. problem. Competition problems must be original( this is very difficult) or clever and nice variations of nice problems.

Babis
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Tiks
1144 posts
#22 • 3 Y
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Guys if you want to know,this problem even was proposed on Armenian National Olimpiad 1996 :lol: ,I think they have found most pupular ineq. in the world(my be after AM-GM :D ) and proposed it for BMO :oops: .
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Michael
99 posts
#23 • 2 Y
Y by Adventure10, Mango247
stergiu wrote:
I have this inequality in my book on Inequalities(written by Skombris and me, in Greek). As source we give there << Komal 1993 >> , so it was known before Aassila sent it to CRUX , vol 1998. This nice solution, sent by Silouan , is to be found there, too(by a chinese).Of course the solution was discused in ML very wide.

My opinion is that the problem should not be proposed in the BMO 2006 , although very good. problem. Competition problems must be original( this is very difficult) or clever and nice variations of nice problems.

Babis

Dear Babis Stergiou,

I have your book, and it unfortunately contains several imprecisions. The inequality appeared later on in Komal (but not in 1993). This inequality was first proposed by the Russian mathematician D. P. Mavlo in the Bulgarian journal Mathematica (number 4, 1987). A little later, it was published in Kvant (problem M1136, number 11-12, 1988). It was also a problem given at the training of the USA IMO team in 1993, and also was used in the training of the South African IMO team in 1994,....etc..etc
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ambros
301 posts
#24 • 3 Y
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Guys i wish to see if this solution is correct and tell me as soon as possible please
thank you
Let $\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq x$ where x is the best real .

Let $a(b+1)+b(c+1)+c(a+1)\geq y$ where y is the best contsnt .

By multipliing these we have and using B-C-S
$[\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}][a(b+1)+b(c+1)+c(a+1)]\geq xy\geq 9$

From Cauchy -AM-GM we have $a(b+1)+b(c+1)+c(a+1)\geq 6\sqrt{abc}$

so $y=6\sqrt{abc}$ and so $6x\sqrt{abc}\geq 9$ and so

$x\geq\frac{3}{2\sqrt{abc}}\geq\frac{3}{abc+1}$
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Albanian Eagle
1693 posts
#25 • 3 Y
Y by Adventure10, Mango247, and 1 other user
actually the problem appears also in the book "mathematical miniatures" by T.Andreescu

The rules were broken since the moment when the leaders allowed two problems from the same country!

(If you want more check the 4th problem)
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DuLac
54 posts
#26 • 2 Y
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To: Albanian Eagle

The Leaders do not know the origin of the problems. This is a rule of IMO too. The proposing countries are revealed after the final selsction. In this BMO there was a real danger of having three problems proposed by the same country since the only Combinatorics problem was proposed by Greece! The Problem finally was eliminated with the intervention of the Selection Committee for exactly this reason. I got all this on the grapevine. I was not there!
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ambros
301 posts
#27 • 2 Y
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Dear albanian eagle
1 congratulations
2 which problem is written in this book that you mention cause i have it ?
3 could you please see and think what i've written above and answer to me as soon as possible?
thank you
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Cezar Lupu
1906 posts
#28 • 2 Y
Y by Adventure10, Mango247
I'm quite sure that this inequality was proposed by Sotiris Louridas. What I do not understand is the fact that why inequalities are proposed by someone at BMO and JBMO and by someone at IMO??? :?
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DuLac
54 posts
#29 • 1 Y
Y by Adventure10
Wrong. The real author told me that he started from another known inequality and then transformed it by a to 1/a, b to 1/b and c to 1/c and arrived to this (known) inequality! The Problem Selection Committee and the Jury detected it after the Competition! (the problem was being used in training at MIT).
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stergiu
1648 posts
#30 • 2 Y
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Michel ,

:) thank you for your nice information.You re better informed. I will use the history of the problem in the future.It is sure now that my source was not the older. I think , I have seen Bulgarian Olympiads 1987 and the name Mavlo is known to me.He must have proposed an ieq. problem in Bulgarian olympiad. Any way , probably - as Dulac said - the greek proposer created the same problem trying to create a new problem. This hapens often. If Louridas is the proposer(Lupu wrote this) , then I'm sure that he gave it as a new problem - in his opinion. He never likes - is a friend - to propose discussed proplems in competitions.
Babis
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DuLac
54 posts
#31 • 1 Y
Y by Adventure10
Louridas is definitely not the author of the problem! The perpetrator is well-known to the Greek Math Competitions scene.
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Peter
3615 posts
#32 • 1 Y
Y by Adventure10
DuLac wrote:
Source: Deep throat
lol @ hidden message in post #33 :D
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exalibur
17 posts
#33 • 2 Y
Y by Adventure10, Mango247
Also by eding 3/1+abc to both parts of the inequality and regrouping by AM-GM you can found the solution
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ambros
301 posts
#34 • 2 Y
Y by Adventure10, Mango247
HELLO GUYS HAS ANYBODY SEEN MY SOLUTION THAT I VE WRITTEN ABOVE? PLEASE SEE IF IT IS CORRECT AND SEN ME A PERSONAL MESSAGE OK? THANK YOU FOR YOUR UNDERSTANDING
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Peter
3615 posts
#35 • 2 Y
Y by Adventure10, Mango247
Please do not shout.

Your proves looks not entirely rigorous to me, although it might be easy to fix (no time to investigate).

You define y as the best constant, and you conclude $y=y=6\sqrt{abc}$, which is not a constant.
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ambros
301 posts
#36 • 2 Y
Y by Adventure10, Mango247
MY FRIEND
FISRTLY I DO NO SHOUT I WRITE THIS WAY BECAUSE I AM USED TO
SECONDLY WHAT DOES RIGOROUS MEAN?
THIRDLY I DID NOT SAY THAT WE TRY TO FIND Y AND AFTERWARDS TO FIND THE LEAST POSSIBLE VALUE FOR X WHICH IS GREATER THAN THE NUMBER ASKED IN THE PROBLEM OK?
THANK YOU PLEASE ANSWER TO ME WHEN YOU SEE THAT AND FIND IF THIS IS CORRECT OR WRONG OK?
THANK YOU
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Arne
3660 posts
#37 • 3 Y
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If you ever make another post using capitals only, it will be deleted. :mad:
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ambros
301 posts
#38 • 3 Y
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Sorry my friend you do not need to bwe so angry :blush: now could you please see my solution and tellme if it is correct ? i have a kind of battle with someone and i have to know from people like you if my solution is correct :D
Thank you
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paladin8
3237 posts
#39 • 2 Y
Y by Adventure10 and 1 other user
ambros wrote:
Guys i wish to see if this solution is correct and tell me as soon as possible please
thank you
Let $\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}\geq x$ where x is the best real .

Let $a(b+1)+b(c+1)+c(a+1)\geq y$ where y is the best contsnt .

By multipliing these we have and using B-C-S
$[\frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}][a(b+1)+b(c+1)+c(a+1)]\geq xy\geq 9$

From Cauchy -AM-GM we have $a(b+1)+b(c+1)+c(a+1)\geq 6\sqrt{abc}$

so $y=6\sqrt{abc}$ and so $6x\sqrt{abc}\geq 9$ and so

$x\geq\frac{3}{2\sqrt{abc}}\geq\frac{3}{abc+1}$

I don't think it works. Let $A = \frac{1}{a(b+1)}+\frac{1}{b(c+1)}+\frac{1}{c(a+1)}$ and $B = a(b+1)+b(c+1)+c(a+1)$.

Just because $AB \ge xy$ and $AB \ge 9$ does not imply that $xy \ge 9$. Even if you specify that $x,y$ are the "best" constants, you have $y = 6 \sqrt{abc}$, which is not a constant, so I don't think it holds.
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Peter
3615 posts
#40 • 3 Y
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So now the test for paladin, what if we want to know it for all numbers for which $abc=K$ for a certain constant $K$? ;)

[ because if that's true for all constants K, we have proven the original problem ]
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paladin8
3237 posts
#41 • 3 Y
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Peter VDD wrote:
So now the test for paladin, what if we want to know it for all numbers for which $abc=K$ for a certain constant $K$? ;)

[ because if that's true for all constants K, we have proven the original problem ]

Hmm... well we still have to show that $y = 6\sqrt{K}$ would be the best constant, right? Which is equivalent to showing that there is always an equality case on the AM-GM for any $K$. But I don't think this is true, because the AM-GM has equality only when $a = b = c = 1$ so it only works for $K = 1$. Correct me if I'm wrong, but the proof doesn't seem to work.
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ambros
301 posts
#42 • 3 Y
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Excuse me guys but i want to clarify two things
frist we do not imply xy>=9 we claim it and we search for which x,y is in charge that xy>=9
secondsice y is got form cauchy we know that for each a,b,c $y=6\sqrt{abc}$ so we get that for wach a,b,c $6x\sqrt{abc}\geq 9$ and so on
now that i was clarified could you please resee that and tell me ? thank you for all the time that you spend for this i am obliged
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sqing
41050 posts
#43 • 1 Y
Y by Adventure10
Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive real numbers . Prove that $$\frac{1}{a_1(1+a_2)}+\frac{1}{a_2(1+a_3)}+\cdots+\frac{1}{a_{n-1}(1+ a_n)}+\frac{1}{a_n( 1+a_1)}\geq\frac{n}{1+a_1a_2\cdots  a_n}.$$h
This post has been edited 1 time. Last edited by sqing, Aug 24, 2019, 2:09 AM
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sqing
41050 posts
#45 • 2 Y
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sqing wrote:
Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive real numbers . Prove that $$\frac{1}{a_1(1+a_2)}+\frac{1}{a_2(1+a_3)}+\cdots+\frac{1}{a_{n-1}(1+ a_n)}+\frac{1}{a_n( 1+a_1)}\geq\frac{n}{1+a_1a_2\cdots  a_n}.$$
https://artofproblemsolving.com/community/c6h161059p10340314

Let $a_1,a_2,\cdots,a_n  (n\ge 3)$ be positive real numbers . Prove that $$\frac{1}{a_1(n-2+a_2)}+\frac{1}{a_2(n-2+a_3)}+\cdots+\frac{1}{a_{n-1}(n-2+ a_n)}+\frac{1}{a_n( n-2+a_1)}\geq\frac{n}{n-2+a_1a_2\cdots  a_n}.$$
This post has been edited 1 time. Last edited by sqing, Dec 7, 2019, 12:42 PM
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