Scary Binomial Coefficient Sum
by EpicBird08, Mar 21, 2025, 11:59 AM
Determine, with proof, all positive integers 
such that 
is an integer for every positive integer 
such that 
is an integer for every positive integer 
This post has been edited 2 times. Last edited by EpicBird08, Mar 21, 2025, 12:06 PM
what the yap
by KevinYang2.71, Mar 20, 2025, 12:00 PM
Alice the architect and Bob the builder play a game. First, Alice chooses two points 
and 
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
if and only if the following condition holds:
Note:
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
, 
to 
, and 
to 
.
and 
in the plane and a subset 
of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
if and only if the following condition holds:
For every city 
distinct from 
and 
, there exists 
such
distinct from 
and 
, there exists 
such
that 
is directly similar to either 
or 
.
Alice wins the game if (i) the resulting roads allow for travel between any pair of cities via a finite sequence of roads and (ii) no two roads cross. Otherwise, Bob wins. Determine, with proof, which player has a winning strategy.
is directly similar to either 
or 
.
Note:
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
, 
to 
, and 
to 
.[TEST RELEASED] Mock Geometry Test for College Competitions
by Bluesoul, Feb 24, 2025, 9:42 AM
This post has been edited 12 times. Last edited by Bluesoul, Today at 3:37 AM
2024 AMC 10B Discussion Thread
by LauraZed, Nov 13, 2024, 5:09 PM
This post has been edited 7 times. Last edited by LauraZed, Nov 13, 2024, 6:20 PM
[TEST RELEASED] OMMC Year 4
by DottedCaculator, Apr 23, 2024, 2:31 PM
FINAL LEADERBOARD: https://docs.google.com/spreadsheets/u/0/d/12RamVH-gQIPN4wibYZVqkx1F2JQuy5Li_8IJ8TqVEyg/htmlview#gid=409219165
Hello to all creative problem solvers,
Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists?
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?
Check out the fourth annual iteration of the
Online Monmouth Math Competition!
Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.
Our website: https://www.ommcofficial.org/
Our Discord (5000+ members): https://tinyurl.com/joinommc
Test portal: https://ommc-test-portal.vercel.app/
This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.
We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!
You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc
Timeline:
Main Round: May 19th - May 26th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.
Final Round: May 28th - May 30th
The top placing teams will qualify for this invitational round (7 questions). The final round consists of 7 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.
Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.
We hope for your participation, and good luck!
OMMC staff
OMMC’S 2024 EVENTS ARE SPONSORED BY:
Hello to all creative problem solvers,
Do you want to work on a fun, untimed team math competition with amazing questions by MOPpers and IMO & EGMO medalists?
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?
Check out the fourth annual iteration of the
Online Monmouth Math Competition!
Online Monmouth Math Competition, or OMMC, is a 501c3 accredited nonprofit organization managed by adults, college students, and high schoolers which aims to give talented high school and middle school students an exciting way to develop their skills in mathematics.
Our website: https://www.ommcofficial.org/
Our Discord (5000+ members): https://tinyurl.com/joinommc
Test portal: https://ommc-test-portal.vercel.app/
This is not a local competition; any student 18 or younger anywhere in the world can attend. We have changed some elements of our contest format, so read carefully and thoroughly. Join our Discord or monitor this thread for updates and test releases.
We plan to raffle out a TON of prizes over all competitors regardless of performance. So just submit: a few minutes of your time will give you a great chance to win amazing prizes!
You can check out our past problems and sample problems here:
https://www.ommcofficial.org/sample
https://www.ommcofficial.org/2022-documents
https://www.ommcofficial.org/2023-documents
https://www.ommcofficial.org/ommc-amc
Timeline:
Main Round: May 19th - May 26th
Test Portal Released. The Main Round of the contest is held. The Main Round consists of 25 questions that each have a numerical answer. Teams will have the entire time interval to work on the questions. They can submit any time during the interval. Teams are free to edit their submissions before the period ends, even after they submit.
Final Round: May 28th - May 30th
The top placing teams will qualify for this invitational round (7 questions). The final round consists of 7 proof questions. Teams again will have the entire time interval to work on these questions and can submit their proofs any time during this interval. Teams are free to edit their submissions before the period ends, even after they submit.
Conclusion of Competition: Early June
Solutions will be released, winners announced, and prizes sent out to winners.
We hope for your participation, and good luck!
OMMC staff
OMMC’S 2024 EVENTS ARE SPONSORED BY:
- Nontrivial Fellowship
- Citadel
- SPARC
- Jane Street
- And counting!
This post has been edited 5 times. Last edited by DottedCaculator, Jul 31, 2024, 1:21 AM
Cyclic to symmetric conversion
by MathPassionForever, Sep 4, 2019, 7:54 AM
A comparatively much smaller post (as of now), as I don't have much information on this as of now.
We can clearly see that the SD theorems are strong. But then, some inequalities are cyclic, sigh. So the basic objective of this post is to serve as a collection of inequalities which can make cyclic inequalities symmetric (and these results need to be strong themselves,for our inequality shouldn't become false after their use
)
Here is the first one:
Proof: This inequality is obviously cyclic. So we may work with just 2 cases:
and 
In either case, we have:
Luckily, I have a question to post
Question: For nonnegative reals
, prove that
And in case someone doesn't find it that sharp, we have the famous and super strong Vasc inequality here!
![\[ (a^2 + b^2 + c^2)^2 \geq 3(a^3b + b^3c + c^3a) \ \forall a,b,c \in \mathbb{R} \]](//latex.artofproblemsolving.com/7/4/6/74654a43f2ccd3f8db453256b7b230baa274c910.png)
I shall take leave here. Bye, see you soon!
We can clearly see that the SD theorems are strong. But then, some inequalities are cyclic, sigh. So the basic objective of this post is to serve as a collection of inequalities which can make cyclic inequalities symmetric (and these results need to be strong themselves,for our inequality shouldn't become false after their use
)Here is the first one:
Proof: This inequality is obviously cyclic. So we may work with just 2 cases:
and 
In either case, we have:
Luckily, I have a question to post
Question: For nonnegative reals
, prove that
![\[ (a^2 + b^2 + c^2)^2 \geq (\sqrt{2}-1)(a+b+c)(\sqrt{2}\left(a^3+b^3+c^3)+\dfrac{4}{27}(a+b+c)^3\right)\]](http://latex.artofproblemsolving.com/e/d/1/ed1caa4cef3ce1702be9fe4c4f33e485135ec567.png)
Well well, looks ugly. But what is the SD for? The first check is 
and the second one is 
.
,And in case someone doesn't find it that sharp, we have the famous and super strong Vasc inequality here!
![\[ (a^2 + b^2 + c^2)^2 \geq 3(a^3b + b^3c + c^3a) \ \forall a,b,c \in \mathbb{R} \]](http://latex.artofproblemsolving.com/7/4/6/74654a43f2ccd3f8db453256b7b230baa274c910.png)
I shall take leave here. Bye, see you soon!This post has been edited 5 times. Last edited by MathPassionForever, Sep 5, 2019, 8:32 PM
Symmetric Polynomial Inequalities in Three Variables
by MathPassionForever, Sep 3, 2019, 1:30 PM
Hey guys! I shall present here, contrary to what you would normally expect me to, some strong INEQUALITY STUFF! Here we go:
We'll start off with with something which has REALLY ugly looks, but get along with it, it gets MUCH nicer in the end.
Define
Theorem 1: Given
, there exist corresponding 
iff:
![\begin{align*}
&(i)u^2 \geq v^2\\
&(ii)w \in\left[3uv^2 - 2u^3 - 2\sqrt{(u^2 - v^2)^3}, 3uv^2 - 2u^3 + 2\sqrt{(u^2-v^2)^3}\right]
\end{align*}](//latex.artofproblemsolving.com/a/e/8/ae839bfe7a2c3092664568d61663cd641da814e3.png)
Proof:
Define
. Clearly the roots of this equation are .
Lemma 1:
One way is clear. The other way is the one we need to focus on.
Let
and let it denote the complex number 
. Since the coefficients of 
are real, there must be another nonreal root, WLOG 
, which is 
. Plug into our expression to get 
.
After our lemma, the problem looks like this:
Given
, there exist corresponding iff 
I'll skip some computational pain for you and present right away that thing thing becomes
Condition (i) clearly is necessary here (It is of course necessary, but here we learn why it is part of criteria that make it sufficient). And Condition (ii) is mere rewriting of what I just presented!
Theorem 2:
Proof: One direction is obvious, the other direction is what I shall prove i.e. if at least one of is negative, then so is the case for 
.
The case when all three are negative gives us negative
.
The case where one is negative gives negative
.
Now we assume that exactly two are negative, say
. Also assume that 
is nonnegative i.e. 
.Now, 
, and we're done.
Okay, this was quite boring so far. But, just as I said, stay with me, please.
Hey, have you noticed that in most positive real inequalities, equality cases are where three, or more generally speaking, two variables are equal, or one of them is zero? Well, the following theorem explains why!
Tej's theroem: There are three parts of this theorem. I'll write all the three statements, but present the proof to just one, for each of them has nearly the same proof. Well, let me get straight to the point. That's an exercise for the readers.
(i) Fix
and let there exist a value of 
such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former is attained only when two of are equal to each other and the latter is attained only when two of are equal or one of is zero.
(ii) Fix
and let there exist a value of 
such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former and latter are both attained only when two of are equal.
(iii) Fix
and let there exist a value of such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former and latter are both attained only when two of are equal.
I shall prove only the former.
We know that. Now substitute 
and write it in the form of a polynomial of 
, i.e. 
. While and can be hideous, what is important is that 
.So as tends to infinity, 
is definitely negative. And we assumed that there exists a value of for which is positive. This means that has a positive root. Let 
be the largest root and 
be the smallest positive root. Equality is clearly achieved iff the polynomial is zero i.e. 
or 
, i.e.
. Take your time to realise this.
Aha! this powerful theorem gives us the nuke we were waiting for! * Insert devious laughter *
SD-3,4,5 theorem(For all nonnegative
tho).
To verify if a symmetric homogeneous 3 variable polynomial
of degree is nonnegative for nonnegative , we just need to verify the cases:
(i)
(ii)
Told you we'd get to nice stuff
Proof:
Every polynomial function in 3 variables of degree
can be written as 
. Fixing we maximise or minimise to find extrema of function. And extrema are achieved when two variables are equal or one is zero, and we're done!
Based on similar logic, just that you need to focus on the Tej's theorem reasoning,to find this:
Let be sides of a possibly degenerate triangle instead. Then we need to verify this just for:
(i)![$f(x,1,1) \geq 0 \forall x \in [0,2]$](//latex.artofproblemsolving.com/3/9/e/39eaeeae6b9500155245b4c1cf1ba844777a2ea4.png)
(ii)
We also have a non-homogenous variant!
Non-homogenous variant: Let satisfy all criteria that it did earlier, except homogeneity. We can still check for just the cases:
(i)
(ii)
Some crucial facts that help you do manipulations as per your need:
(i)If the degree is
, you can make it a polynomial in instead and check that 
only from the second part of Tej's theorem!
(ii) We used the fact that has 
as functions. But they can be ANY functions, not just polynomials!
Edit: Hey guys, back with part of what I promised!
SD-6 Theorem: Let be degree 6 instead (and follows all other criteria as mentioned earlier). Then, it can be written as
![\[ f(a,b,c) = Aw^6 + B(u,v^2)w^3 + C(u,v^2) \]](//latex.artofproblemsolving.com/c/9/9/c992e31cbd796e035d9da84c0f2f439bc9688876.png)
After noting that is a constant, see that there are two cases:
(i)
: Just check if and . The same logic as used in Tej's theorem.
(ii) Um..... this is the bad part. For
, we need to prove
![\[ f(a,b,c) \geq A\left(w^3 + Ku^3 + Luv^2 + M\dfrac{v^4}{u}\right)^2 \]](//latex.artofproblemsolving.com/8/c/d/8cd30bcc953376100064a7fa43022afad049259b.png)
So yeah, we need to find 
ourselves. The point of doing this is to make 
to apply (i) now.
Adding problems soon!
Dec 27:
Edit: Ok maybe not so soon
April 1:
Edit: Can we act like I never said this?
We'll start off with with something which has REALLY ugly looks, but get along with it, it gets MUCH nicer in the end.
Define
Theorem 1: Given
, there exist corresponding 
iff:![\begin{align*}
&(i)u^2 \geq v^2\\
&(ii)w \in\left[3uv^2 - 2u^3 - 2\sqrt{(u^2 - v^2)^3}, 3uv^2 - 2u^3 + 2\sqrt{(u^2-v^2)^3}\right]
\end{align*}](http://latex.artofproblemsolving.com/a/e/8/ae839bfe7a2c3092664568d61663cd641da814e3.png)
Proof:Define
. Clearly the roots of this equation are .Lemma 1:
One way is clear. The other way is the one we need to focus on.
Let
and let it denote the complex number 
. Since the coefficients of 
are real, there must be another nonreal root, WLOG 
, which is 
. Plug into our expression to get 
.After our lemma, the problem looks like this:
Given
, there exist corresponding iff 
I'll skip some computational pain for you and present right away that thing thing becomes
Condition (i) clearly is necessary here (It is of course necessary, but here we learn why it is part of criteria that make it sufficient). And Condition (ii) is mere rewriting of what I just presented!
Theorem 2:
Proof: One direction is obvious, the other direction is what I shall prove i.e. if at least one of
is negative, then so is the case for 
.The case when all three are negative gives us negative
.The case where one is negative gives negative
.Now we assume that exactly two are negative, say
. Also assume that 
is nonnegative i.e. 
.Now, 
, and we're done.Okay, this was quite boring so far. But, just as I said, stay with me, please.
Hey, have you noticed that in most positive real inequalities, equality cases are where three, or more generally speaking, two variables are equal, or one of them is zero? Well, the following theorem explains why!
Tej's theroem: There are three parts of this theorem. I'll write all the three statements, but present the proof to just one, for each of them has nearly the same proof. Well, let me get straight to the point. That's an exercise for the readers.
(i) Fix
and let there exist a value of 
such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former is attained only when two of are equal to each other and the latter is attained only when two of are equal or one of is zero.(ii) Fix
and let there exist a value of 
such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former and latter are both attained only when two of are equal.(iii) Fix
and let there exist a value of such that there exist corresponding positive reals . Then has both a global maximum and a global minimum. The former and latter are both attained only when two of are equal.I shall prove only the former.
We know that
. Now substitute 
and write it in the form of a polynomial of 
, i.e. 
. While and can be hideous, what is important is that 
.So as tends to infinity, 
is definitely negative. And we assumed that there exists a value of for which is positive. This means that has a positive root. Let 
be the largest root and 
be the smallest positive root. Equality is clearly achieved iff the polynomial is zero i.e. 
or 
, i.e.
. Take your time to realise this.Aha! this powerful theorem gives us the nuke we were waiting for! * Insert devious laughter *
SD-3,4,5 theorem(For all nonnegative
tho).To verify if a symmetric homogeneous 3 variable polynomial
of degree is nonnegative for nonnegative , we just need to verify the cases:(i)
(ii)
Told you we'd get to nice stuff
Proof:
Every polynomial function in 3 variables of degree
can be written as 
. Fixing we maximise or minimise to find extrema of function. And extrema are achieved when two variables are equal or one is zero, and we're done!Based on similar logic, just that you need to focus on the Tej's theorem reasoning,to find this:
Let
be sides of a possibly degenerate triangle instead. Then we need to verify this just for:(i)
![$f(x,1,1) \geq 0 \forall x \in [0,2]$](http://latex.artofproblemsolving.com/3/9/e/39eaeeae6b9500155245b4c1cf1ba844777a2ea4.png)
(ii)
We also have a non-homogenous variant!
Non-homogenous variant: Let
satisfy all criteria that it did earlier, except homogeneity. We can still check for just the cases:(i)
(ii)
Some crucial facts that help you do manipulations as per your need:
(i)If the degree is
, you can make it a polynomial in instead and check that 
only from the second part of Tej's theorem!(ii) We used the fact that
has 
as functions. But they can be ANY functions, not just polynomials!Edit: Hey guys, back with part of what I promised!
SD-6 Theorem: Let
be degree 6 instead (and follows all other criteria as mentioned earlier). Then, it can be written as![\[ f(a,b,c) = Aw^6 + B(u,v^2)w^3 + C(u,v^2) \]](http://latex.artofproblemsolving.com/c/9/9/c992e31cbd796e035d9da84c0f2f439bc9688876.png)
After noting that is a constant, see that there are two cases:(i)
: Just check if and . The same logic as used in Tej's theorem.(ii) Um..... this is the bad part. For
, we need to prove![\[ f(a,b,c) \geq A\left(w^3 + Ku^3 + Luv^2 + M\dfrac{v^4}{u}\right)^2 \]](http://latex.artofproblemsolving.com/8/c/d/8cd30bcc953376100064a7fa43022afad049259b.png)
So yeah, we need to find 
ourselves. The point of doing this is to make 
to apply (i) now.Adding problems soon!
Dec 27:
Edit: Ok maybe not so soon
April 1:
Edit: Can we act like I never said this?
This post has been edited 14 times. Last edited by MathPassionForever, Apr 1, 2020, 12:54 PM
September 2019