Stay ahead of learning milestones! Enroll in a class over the summer!

Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
Contests & Programs AMC and other contests, summer programs, etc.
AMC and other contests, summer programs, etc.
3 M G
BBookmark  VNew Topic kLocked
G
Topic
First Poster
Last Poster
k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
ranttttt
alcumusftwgrind   40
N a few seconds ago by ZMB038
rant
40 replies
alcumusftwgrind
Apr 30, 2025
ZMB038
a few seconds ago
Goals for 2025-2026
Airbus320-214   78
N 12 minutes ago by Yiyj
Please write down your goal/goals for competitions here for 2025-2026.
78 replies
Airbus320-214
Yesterday at 8:00 AM
Yiyj
12 minutes ago
D1030 : An inequalitie
Dattier   0
an hour ago
Source: les dattes à Dattier
Let $0<a<b<c<d$ reals, and $n \in \mathbb N^*$.

Is it true that $a^n(b-a)+b^n(c-b)+c^n(d-c) \leq \dfrac {d^{n+1}}{n+1}$ ?
0 replies
Dattier
an hour ago
0 replies
[MAIN ROUND STARTS MAY 17] OMMC Year 5
DottedCaculator   43
N an hour ago by vincentwant
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? $\phantom{You lost the game.}$
Do you want to have a chance to win thousands in cash and raffle prizes (no matter your skill level)?

Check out the fifth 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 (6000+ 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.

How hard is it?

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!

How are the problems?

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

How will the test be held?/How do I sign up?

Solo teams?

Test Policy

Timeline:
Main Round: May 17th - May 24th
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 26th - May 28th
The top placing teams will qualify for this invitational round (5-10 questions). The final round consists of 5-10 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.

Scoring:

Prizes:

I have more questions. Whom do I ask?

We hope for your participation, and good luck!

OMMC staff

OMMC’S 2025 EVENTS ARE SPONSORED BY:

[list]
[*]Nontrivial Fellowship
[*]Citadel
[*]SPARC
[*]Jane Street
[*]And counting!
[/list]


43 replies
DottedCaculator
Apr 26, 2025
vincentwant
an hour ago
Long and wacky inequality
Royal_mhyasd   0
an hour ago
Source: Me
Let $x, y, z$ be positive real numbers such that $x^2 + y^2 + z^2 = 12$. Find the minimum value of the following sum :
$$\sum_{cyc}\frac{(x^3+2y)^3}{3x^2yz - 16z - 8yz + 6x^2z}$$knowing that the denominators are positive real numbers.
0 replies
Royal_mhyasd
an hour ago
0 replies
Vietnamese national Olympiad 2007, problem 4
hien   16
N an hour ago by de-Kirschbaum
Given a regular 2007-gon. Find the minimal number $k$ such that: Among every $k$ vertexes of the polygon, there always exists 4 vertexes forming a convex quadrilateral such that 3 sides of the quadrilateral are also sides of the polygon.
16 replies
hien
Feb 8, 2007
de-Kirschbaum
an hour ago
2n^2+4n-1 and 3n+4 have common powers
bin_sherlo   4
N an hour ago by CM1910
Source: Türkiye 2025 JBMO TST P5
Find all positive integers $n$ such that a positive integer power of $2n^2+4n-1$ equals to a positive integer power of $3n+4$.
4 replies
bin_sherlo
Yesterday at 7:13 PM
CM1910
an hour ago
An interesting geometry
k.vasilev   19
N an hour ago by Ilikeminecraft
Source: All-Russian Olympiad 2019 grade 10 problem 4
Let $ABC$ be an acute-angled triangle with $AC<BC.$ A circle passes through $A$ and $B$ and crosses the segments $AC$ and $BC$ again at $A_1$ and $B_1$ respectively. The circumcircles of $A_1B_1C$ and $ABC$ meet each other at points $P$ and $C.$ The segments $AB_1$ and $A_1B$ intersect at $S.$ Let $Q$ and $R$ be the reflections of $S$ in the lines $CA$ and $CB$ respectively. Prove that the points $P,$ $Q,$ $R,$ and $C$ are concyclic.
19 replies
k.vasilev
Apr 23, 2019
Ilikeminecraft
an hour ago
sum (a+b)/(a^2+ab+b^2) <=2 if 1/a+1/b+1/c =3 for a,b,c>0
parmenides51   15
N 2 hours ago by AylyGayypow009
Source: 2020 Greek JBMO TST p2
Let $a,b,c$ be positive real numbers such that $\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}=3$. Prove that
$$\frac{a+b}{a^2+ab+b^2}+ \frac{b+c}{b^2+bc+c^2}+ \frac{c+a}{c^2+ca+a^2}\le 2$$When is the equality valid?
15 replies
parmenides51
Nov 14, 2020
AylyGayypow009
2 hours ago
Bosnia and Herzegovina JBMO TST 2016 Problem 3
gobathegreat   3
N 2 hours ago by Sh309had
Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2016
Let $O$ be a center of circle which passes through vertices of quadrilateral $ABCD$, which has perpendicular diagonals. Prove that sum of distances of point $O$ to sides of quadrilateral $ABCD$ is equal to half of perimeter of $ABCD$.
3 replies
gobathegreat
Sep 16, 2018
Sh309had
2 hours ago
Power Of Factorials
Kassuno   180
N 2 hours ago by maromex
Source: IMO 2019 Problem 4
Find all pairs $(k,n)$ of positive integers such that \[ k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}). \]Proposed by Gabriel Chicas Reyes, El Salvador
180 replies
1 viewing
Kassuno
Jul 17, 2019
maromex
2 hours ago
100 card with 43 having odd integers on them
falantrng   7
N 2 hours ago by Just1
Source: Azerbaijan JBMO TST 2018, D2 P4
In the beginning, there are $100$ cards on the table, and each card has a positive integer written on it. An odd number is written on exactly $43$ cards. Every minute, the following operation is performed: for all possible sets of $3$ cards on the table, the product of the numbers on these three cards is calculated, all the obtained results are summed, and this sum is written on a new card and placed on the table. A day later, it turns out that there is a card on the table, the number written on this card is divisible by $2^{2018}.$ Prove that one hour after the start of the process, there was a card on the table that the number written on that card is divisible by $2^{2018}.$
7 replies
falantrng
Aug 1, 2023
Just1
2 hours ago
GCD and LCM operations
BR1F1SZ   1
N 3 hours ago by WallyWalrus
Source: 2025 Francophone MO Juniors P4
Charlotte writes the integers $1,2,3,\ldots,2025$ on the board. Charlotte has two operations available: the GCD operation and the LCM operation.
[list]
[*]The GCD operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{gcd}(a, b)$.
[*]The LCM operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{lcm}(a, b)$.
[/list]
An integer $N$ is called a winning number if there exists a sequence of operations such that, at the end, the only integer left on the board is $N$. Find all winning integers among $\{1,2,3,\ldots,2025\}$ and, for each of them, determine the minimum number of GCD operations Charlotte must use.

Note: The number $\operatorname{gcd}(a, b)$ denotes the greatest common divisor of $a$ and $b$, while the number $\operatorname{lcm}(a, b)$ denotes the least common multiple of $a$ and $b$.
1 reply
BR1F1SZ
Saturday at 11:24 PM
WallyWalrus
3 hours ago
9 JMO<200?
DreamineYT   2
N 4 hours ago by xHypotenuse
Just wanted to ask
2 replies
1 viewing
DreamineYT
May 10, 2025
xHypotenuse
4 hours ago
Prove a polynomial has a nonreal root
KevinYang2.71   46
N Apr 23, 2025 by megarnie
Source: USAMO 2025/2
Let $n$ and $k$ be positive integers with $k<n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$ is zero. Prove that $P(x)$ has a nonreal root.
46 replies
KevinYang2.71
Mar 20, 2025
megarnie
Apr 23, 2025
Prove a polynomial has a nonreal root
G H J
G H BBookmark kLocked kLocked NReply
Source: USAMO 2025/2
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
golue3120
58 posts
#36 • 2 Y
Y by Ilikeminecraft, Math4Life2020
Here, we have a solution without WLOG $k=n-1$, without Descartes rule of signs, without Rolle, but is somehow not terrible?

Let $e_k$ denote the degree $k$ elementary symmetric polynomial.

Suppose for the sake of contradiction that all roots of $P(x)$ were real. Let the roots be $r_1,r_2,\dotsc,r_n$. Let $R$ be the set of roots. Then for every subset $S$ of $k$ roots, $e_i(S)=0$ for some $1\le i\le k-1$.

Define two subsets of $k$ elements to be adjacent if they share $k-1$ elements.
Lemma. If $\mathcal F$ is a family of $k$-element subsets of $R$ such that no two are adjacent, then $|\mathcal F|\le\frac{1}{k+1}\binom nk$.
Proof. Every set $T$ in $\mathcal F$ is adjacent to exactly $k(n-k)$ sets outside of $\mathcal F$ because we may pick any element of $T$ to remove and any root not in $T$ to add. Meanwhile, any set $U$ not in $\mathcal F$ is adjacent to at most $n-k$ sets in $\mathcal F$ because when we remove an element of $U$ and add a root not in $U$, the additional roots for the adjacent sets in $\mathcal F$ must be distinct. Counting then gives the desired result.

Now define $\mathcal F_i$ to be the family of $k$-element subsets of roots such that $e_i$ vanishes. Then there are $k-1$ such families, and every size $k$ subset of the roots is in at least one family, so one family must have size at least $\frac{1}{k-1}\binom nk$. Therefore, at least one family must contain two adjacent subsets. This implies that there are roots $r_1,r_2,\dotsc,r_k,r_k'$ such that
\[e_i(r_1,r_2,\dotsc,r_k)=e_i(r_1,r_2,\dotsc,r_{k-1},r_k')=0.\]Now we expand these by writing
\[e_i(r_1,r_2,\dotsc,r_{k-1})+r_ke_{i-1}(r_1,r_2,\dotsc,r_{k-1})=e_i(r_1,r_2,\dotsc,r_{k-1})+r_k'e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0,\]and since $r_k\neq r_k'$, we have
\[e_i(r_1,r_2,\dotsc,r_{k-1})=e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0.\]
Suppose that $\{r_1,r_2,\dotsc,r_{k-1}\}$ is a set of real numbers such that $e_i(r_1,r_2,\dotsc,r_{k-1})=e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0$ for some $i$, but no strict subset has this property. It suffices to show that such a set cannot exist.

Now we perform a bit of further expansion
\[e_i(r_1,r_2,\dotsc,r_{k-2})+r_{k-1}e_{i-1}(r_1,r_2,\dotsc,r_{k-2})=e_{i-1}(r_1,r_2,\dotsc,r_{k-2})+r_{k-1}e_{i-2}(r_1,r_2,\dotsc,r_{k-2})=0,\]from which we may deduce that
\[e_i(r_1,r_2,\dotsc,r_{k-2})e_{i-2}(r_1,r_2,\dotsc,r_{k-2})=e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2.\]
By minimality, none of $e_i$, $e_{i-2}$, or $e_{i-1}$ can vanish, for then two consecutive elementary symmetric polynomials must vanish. By "On the nonnegativity of generalized discriminants of quadratics",
\[e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2\ge \frac{i(k-i)}{(i-1)(k-i-1)}e_i(r_1,r_2,\dotsc,r_{k-2})e_{i-2}(r_1,r_2,\dotsc,r_{k-2}).\]Since $e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2$ is strictly positive, we have a contradiction.
This post has been edited 1 time. Last edited by golue3120, Mar 21, 2025, 1:32 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MathLuis
1525 posts
#37 • 2 Y
Y by KevinYang2.71, Pengu14
This problem is kinda funny, I guess giving a rating is hard.
So FTSOC $P$ has $n$ real roots and WLOG it is monic and WLOG $k=n-1$ by simply shifting with the other monomials then define $P(x)$ as $(x-r_1) \cdots (x-r_n)$ then it happens that all of $\frac{P(x)}{x-r_i}$ for $1 \le i \le n$ happen to have one coefficient equal to $0$, since there is at most $n-2$ choices for these so by pigeonhole there exists $i \ne j$ such that $\frac{P(x)}{x-r_i}$ and $\frac{P(x)}{x-r_j}$ are missing the same coefficient $x^{\ell}$ for $\ell \ge 1$ (this is because none of the roots are equal and none can be zero), let these be called $A(x), B(x)$ respectively and let $i,j$ be $n,n-1$ (it doesn't matter anyway lmao), then we have that $A(x)=(x-r_1) \cdots (x-r_{n-1})$ and $B(x)=(x-r_1) \cdots (x-r_{n-2})(x-r_n)$ which the means that $A(x)-B(x)=(r_n-r_{n-1})(x-r_1) \cdots (x-r_{n-2})$ is missing the $x^{\ell}$ coefficient but also notice that $r_nA(x)-r_{n-1}B(x)=(r_n-r_{n-1})x(x-r_1) \cdots (x-r_{n-2})$ is missing the $x^{\ell}$ coefficient but since $r_n \ne r_{n-1}$ then we do in fact have that $A(x)-B(x)$ is missing both the $x^{\ell}$ and $x^{\ell-1}$ coefficient. (Now we can wlog say $\ell$ is minimal).
Call it $C(x)$ then it has $n-2$ real roots, is $\deg C=n-2$ but also it has two consecutive missing coefficients so write it as $x^2D(x)+E(x)$ (essy to check that if $n=2$ this is trivial and if $n=3$ then you have cuadratics of the form $x^2+b$ which shows pairwise sum of roots is zero and thus all are zero, a contradiction!), so for $n \ge 4$ notice that $n-4=\deg D$ and that you can set it so that least degree term on $D$ is greater than $\deg E$.
Now from derivative analytical definition and propeties we can check graphically that counting multiplicity a polynomial has all its roots real if and only if the derivative does as well, so now take the $\deg E+1$-th derivative of $Q$, then $Q^{(\deg E+1)}(0)=0$ but also $Q^{(\deg E+2)}(0)=0$ and these polynomials must have all of their roots real by chaining the fact that $Q$ does follow this so because of minimality of $\ell$, it happens that $E$ has a non-zero leading coefficient and thus if it were the zero polynomial we are done by the same argument as if it weren't which we will see now.
So let $R(x)=Q^{(\deg E)}(x)=c(x-q_1) \cdots (x-q_m)$, now focus on the product of monomials and call it $R_1(x)$, then it happens that $\frac{R_1'(x)}{R_1(x)}=\sum_{i=1}^{m} \frac{1}{x-q_i}$ and thus by taking derivative w.r.t. $x$ it happens that $\frac{R_1'(x)^2-R_1''(x)R_1(x)}{R_1(x)^2}=-\sum_{i=1}^{m} \frac{1}{(x-q_i)^2}$ for all $x \ne q_j$ for $1 \le j \le m$, however clearly none of the roots are zero once again but replacing $x=0$ we have LHS is zero while RHS isn't, contradiction!.
Therefore $P$ must have a non-real root thus we are done :cool:.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Pomansq
12 posts
#38
Y by
How many points for proving it suffices to prove k=n-1? I made a mistake for the part where I prove k=n-1 works, so I don’t think I’m getting any credit there.
This post has been edited 2 times. Last edited by Pomansq, Mar 21, 2025, 6:56 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
atdaotlohbh
186 posts
#39
Y by
This problem is certainly based on the following lemma:
Lemma: A polynomial with only real distinct roots can't have two consecutive zero coefficients
Proof: Let the polynomial be $P$ and the coefficients be for $x^k,x^{k+1}$. Then $P^{(k)}$ is a polynomial which must have only real distinct roots, but it is divisible by $x^2$, a contradiction.

Now to the problem. We consider the contrary. Take any $k+1$ of this $n$ roots, say they are $r_1,\ldots,r_{k+1}$. Consider the polynomials of the form $\frac{(x-r_1)\ldots (x-r_{k+1})}{x-r_i}$. $P$ is divisible by it, hence it has a zero coefficient. But its coefficients are elementary symmetric polynomials in variables $r_1,r_2,\ldots,r_{k+1}$. Also, it is not the product since none of the roots is zero. So there are only $k-1$ options for the size of the elementary polynomial, and $k+1$ polynomials, by Pigeonhole there is a repetition. So $\sigma_i(r_1,\ldots,r_{k-1},r_k)=\sigma_i(r_1,\ldots,r_{k-1},r_{k+1})=0$. But $\sigma_i(r_1,\ldots,r_{k-1},x)=x\sigma_{i-1}(r_1,\ldots,r_{k-1})+\sigma_i(r_1,\ldots,r_{k-1})$, and as it is a linear function with two distinct roots, we must have $\sigma_{i-1}(r_1,\ldots,r_{k-1})=\sigma_i(r_1,\ldots,r_{k-1})=0$. This contradicts the Lemma. Thus, $P$ must have a nonreal root.
This post has been edited 1 time. Last edited by atdaotlohbh, Mar 29, 2025, 9:02 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
laniakea0
33 posts
#40
Y by
How much do I get if I proved everything but assumed n=k+1 without justifying? I was a bit confused by the wording
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
awesomeming327.
1719 posts
#41
Y by
Assume on the contrary. We prove the result for $k=n-1$. This is equivalent to the original problem because we can simply remove some of the roots of $P$ to reduce to this new problem anyway.

Let $P(x)=(x-r_1)(x-r_2)\dots(x-r_n)$ for some distinct nonzero roots $r_1$, $r_2$, $\dots$, $r_n$. Let $P_i(x)$ be $(x-r_1)\dots(x-r_{i-1})(x-r_{i+1})\dots(x-r_n)$, which must have a zero coefficient. Note that there are $n$ polynomials $P_1$, $P_2$, $\dots$, $P_n$ and $n-2$ coefficients at which this zero coefficient can appear. Therefore, there exists two polynomials (WLOG let them be $P_n$ and $P_{n-1}$) such that $P_n$ and $P_{n-1}$ both have a zero as an $x^i$ coefficient.

Let $(x-r_1)(x-r_2)\dots(x-r_{n-2})$ be $Q(x)$. We have
\begin{align*}P_n-P_{n-1} &= (x-r_{n-1})Q(x)-(x-r_n)Q(x)=(r_n-r_{n-1})Q(x) \\
r_nP_n-r_{n-1}P_{n-1} &= (r_nx-r_{n-1}r_n)Q(x)-(r_{n-1}x-r_{n-1}r_n)Q(x)=(r_n-r_{n-1})xQ(x)\end{align*}both of which have zero as an $x^i$ coefficient, which means that $Q(x)$ has a zero in both its $x^i$ and $x^{i-1}$ coefficients. We now prove the following claim, which finishes the problem:

Claim 1: Let $Q(x)=(x-r_1)(x-r_2)\dots(x-r_n)$ be a polynomial with distinct nonzero real roots. Then $Q(x)$ may not have two consecutive zero coefficients.
Let $R(x)$ be a polynomial with $r$ distinct roots. Then, $R'(x)$ has $r-1$ distinct roots because between each two roots of $R(x)$, there will be a relative extrema. Therefore, if we keep taking derivatives of $Q$, there will always be $\operatorname{deg} Q$ distinct roots. But since there are two consecutive zero coefficients, eventually some multiple derivative of $Q$ will have both its constant and linear coefficients zero, which means that it has a double root at $0$, impossible.
This post has been edited 2 times. Last edited by awesomeming327., Mar 22, 2025, 2:58 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mysteriouxxx
8 posts
#42
Y by
...........
This post has been edited 1 time. Last edited by Mysteriouxxx, Apr 8, 2025, 8:38 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
yanling
1 post
#43
Y by
KevinYang2.71 wrote:
scannose wrote:
tfw you wrote up a fakesolve that turned out to be correct
KevinYang2.71 wrote:
Claim. $A(x)$ cannot have $n-2$ distinct real roots.
how many points for claiming this but not proving it

2 for everything before using derivative maybe

if only prove the claim,including Rolle,but havent solved others,how many point can i get,thanks
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dno1467
5 posts
#44
Y by
this problem has so much aura

argue by contradiction.
consider $k+1$ real roots $a_1, a_2, .., a_k$ and the $k+1$ factors of degree $k$. Clearly two of these factors have a 0 in the same place. This means that the gcd of these two factors have two consecutive zeroes.

We can prove by induction that no polynomial with all real roots has two consecutive zeroes. If $Q$ has all real roots with single multiplicity, then $Q'$ does too, as each root of $Q'$ lies in between two roots of $Q$. But if $Q$ has two consecutive zeroes, so does $Q'$, contradiction.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
sixoneeight
1138 posts
#45 • 1 Y
Y by Ilikeminecraft
Trivial by Advanced Academic Course Precalculus (PNI Chart)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Math4Life2020
2964 posts
#46
Y by
golue3120 wrote:
Here, we have a solution without WLOG $k=n-1$, without Descartes rule of signs, without Rolle, but is somehow not terrible?

Let $e_k$ denote the degree $k$ elementary symmetric polynomial.

Suppose for the sake of contradiction that all roots of $P(x)$ were real. Let the roots be $r_1,r_2,\dotsc,r_n$. Let $R$ be the set of roots. Then for every subset $S$ of $k$ roots, $e_i(S)=0$ for some $1\le i\le k-1$.

Define two subsets of $k$ elements to be adjacent if they share $k-1$ elements.
Lemma. If $\mathcal F$ is a family of $k$-element subsets of $R$ such that no two are adjacent, then $|\mathcal F|\le\frac{1}{k+1}\binom nk$.
Proof. Every set $T$ in $\mathcal F$ is adjacent to exactly $k(n-k)$ sets outside of $\mathcal F$ because we may pick any element of $T$ to remove and any root not in $T$ to add. Meanwhile, any set $U$ not in $\mathcal F$ is adjacent to at most $n-k$ sets in $\mathcal F$ because when we remove an element of $U$ and add a root not in $U$, the additional roots for the adjacent sets in $\mathcal F$ must be distinct. Counting then gives the desired result.

Now define $\mathcal F_i$ to be the family of $k$-element subsets of roots such that $e_i$ vanishes. Then there are $k-1$ such families, and every size $k$ subset of the roots is in at least one family, so one family must have size at least $\frac{1}{k-1}\binom nk$. Therefore, at least one family must contain two adjacent subsets. This implies that there are roots $r_1,r_2,\dotsc,r_k,r_k'$ such that
\[e_i(r_1,r_2,\dotsc,r_k)=e_i(r_1,r_2,\dotsc,r_{k-1},r_k')=0.\]Now we expand these by writing
\[e_i(r_1,r_2,\dotsc,r_{k-1})+r_ke_{i-1}(r_1,r_2,\dotsc,r_{k-1})=e_i(r_1,r_2,\dotsc,r_{k-1})+r_k'e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0,\]and since $r_k\neq r_k'$, we have
\[e_i(r_1,r_2,\dotsc,r_{k-1})=e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0.\]
Suppose that $\{r_1,r_2,\dotsc,r_{k-1}\}$ is a set of real numbers such that $e_i(r_1,r_2,\dotsc,r_{k-1})=e_{i-1}(r_1,r_2,\dotsc,r_{k-1})=0$ for some $i$, but no strict subset has this property. It suffices to show that such a set cannot exist.

Now we perform a bit of further expansion
\[e_i(r_1,r_2,\dotsc,r_{k-2})+r_{k-1}e_{i-1}(r_1,r_2,\dotsc,r_{k-2})=e_{i-1}(r_1,r_2,\dotsc,r_{k-2})+r_{k-1}e_{i-2}(r_1,r_2,\dotsc,r_{k-2})=0,\]from which we may deduce that
\[e_i(r_1,r_2,\dotsc,r_{k-2})e_{i-2}(r_1,r_2,\dotsc,r_{k-2})=e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2.\]
By minimality, none of $e_i$, $e_{i-2}$, or $e_{i-1}$ can vanish, for then two consecutive elementary symmetric polynomials must vanish. By "On the nonnegativity of generalized discriminants of quadratics",
\[e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2\ge \frac{i(k-i)}{(i-1)(k-i-1)}e_i(r_1,r_2,\dotsc,r_{k-2})e_{i-2}(r_1,r_2,\dotsc,r_{k-2}).\]Since $e_{i-1}(r_1,r_2,\dotsc,r_{k-2})^2$ is strictly positive, we have a contradiction.

Wow. I spent around two hours in contest looking for an inequality-based approach to proving this but failed. Honestly, given the premise of the problem, I'm slightly surprised one exists. Is there any clean way to prove the inequality you cite? It looks like the terms could line up with some Cauchy-Schwartz / Sum of Squares approach, but I don't immediately see how to do it.
(I ultimately found the Descartes' Rule of Signs approach, so fortunately I avoided a crisis. :P)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mathandski
757 posts
#47 • 1 Y
Y by KevinYang2.71
We show the converse that, if all polynomials dividing $P$ have some zero coefficient, $P(x)$ has a nonzero root. Clearly, it suffices to show $k = n - 1$. WLOG let $a_k = 1$ and WLOG only consider monic degree-$k$ polynomials dividing $P$. Let, $P(x) = (x -  r_1) \dots (x - r_n)$. Its divisors with degree $n - 1$ have,
\[\frac{P(x)}{x - r_1}, \frac{P(x)}{x - r_2}, \dots, \frac{P(x)}{x - r_n}\]
FTSOC, let each of these divisors have a zero coefficient. By Pigeonhole, since they are monic degree $n-1$ polynomials, two of them share a zero coefficient in the same spot. WLOG, let $\frac{P(x)}{x - r_1}, \frac{P(x)}{x - r_2}$ both have a coefficient of zero for $x^{n-1-e}$.

Let $\sigma (S, k)$ be the $k$-th symmetric sum of a set $S$. Let $R = \{r_1, \dots, r_n\}$. By Vieta's,
\[\sigma (R - r_1, e) = \sigma (R - r_2, e) = 0\]\[\iff r_2 \sigma (R - r_1 - r_2, e - 1) + \sigma (R - r_1 - r_2, e) = r_1 \sigma (R - r_1 - r_2, e - 1) + \sigma (R - r_1 - r_2, e) = 0\]
As a result, $r_2 \sigma (R - r_1 - r_2, e - 1) = r_1 \sigma (R - r_1 - r_2, e - 1)$. Since roots are distinct, we must have $\sigma (R - r_1 - r_2, e - 1) = 0$. Plugging this back in, $\sigma (R - r_1 - r_2, e) = 0$ as well.

(Above is the extent to which I wrote up my solution on the test)

By Vieta's, $Q(x) = \frac{P(x)}{(x - r_1)(x - r_2)}$ has two consecutive zero coefficients for $x^m, x^{m-1}$ for some $m$. However, it also has distinct real roots. We show that this is impossible on real polynomials.

(In red is what I couldn't figure out during the test :wallbash_red:)
Whenever we take the derivative of a polynomial with distinct real roots, which we label $r_1 < r_2 < \dots < r_p$, by the Mean value theorem, there exists a point on each of $(r_1, r_2), (r_2, r_3), \dots, (r_{p-1}, r_p)$ where the derivative is zero meaning there are $p-1$ instances where the derivative is zero. Therefore, since there cannot be additional roots by FToA, the derivative of the polynomial also has distinct real roots.

Therefore, by induction, if we take the $m-1$th derivative of $Q(x)$, we still have a polynomial with distinct roots. However, the coefficient of the $x$ term and the constant term is zero so the polynomial has a double root at zero. (!)
This post has been edited 1 time. Last edited by Mathandski, Mar 26, 2025, 4:11 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Tintarn
9042 posts
#48 • 1 Y
Y by ItalianZebra
Math4Life2020 wrote:
Wow. I spent around two hours in contest looking for an inequality-based approach to proving this but failed. Honestly, given the premise of the problem, I'm slightly surprised one exists. Is there any clean way to prove the inequality you cite? It looks like the terms could line up with some Cauchy-Schwartz / Sum of Squares approach, but I don't immediately see how to do it.
Well, the inequality we need is just the famous Newton inequality
\[S_k^2>S_{k-1}S_{k+1}\]for non-zero $x_1,\dots,x_n$, which are not all equal. Here, $S_k=\frac{\sigma_k}{\binom{n}{k}}$ is the normalized version of the elementary symmetric polynomials of $x_1,\dots,x_n$. This of course finishes off the problem immediately, so I am slightly surprised that no one has mentioned it before... :maybe:
This post has been edited 1 time. Last edited by Tintarn, Apr 12, 2025, 12:11 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SimogmH1
2 posts
#49
Y by
KevinYang2.71 wrote:
Let $n$ and $k$ be positive integers with $k<n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0,\,a_1,\,\ldots,\,a_k$ such that the polynomial $a_kx^k+\cdots+a_1x+a_0$ divides $P(x)$, the product $a_0a_1\cdots a_k$ is zero. Prove that $P(x)$ has a nonreal root.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megarnie
5607 posts
#50 • 2 Y
Y by KevinYang2.71, brainfertilzer
Assume FTSOC $P$ has $n$ real roots. WLOG $P$ has leading coefficient $1$ so let $P(x)=:(x-r_1)\cdots(x-r_n)$. WLOG $k=n-1$, as otherwise we can take $Q:=(x-r_1)\cdots(x-r_{k+1})$ and deal with $Q$ instead of $P$.

By the assertion, each $\frac{P(x)}{x-r_i}$ for $1\leq i\leq n$ has a coefficient equal to $0$. Since $r_i\neq 0$ for all $i$, this coefficient is not the leading or constant coefficient, so it has $k-1=n-2$ choices. By pigeonhole, there exists $1\leq i<j\leq n$ such that $Q(x):=\frac{P(x)}{x-r_i}$ and $R(x):=\frac{P(x)}{x-r_j}$ have no $x^\alpha$ term, for some $\alpha\geq 1$. WLOG $i=n-1$ and $j=n$. Hence
\begin{align*}
P(x)&=(x-r_1)\cdots(x-r_{n-2})(x-r_n)\\
Q(x)&=(x-r_1)\cdots(x-r_{n-2})(x-r_{n-1}).
\end{align*}Then $P(x)-Q(x)=(r_{n-1}-r_n)(x-r_1)\cdots(x-r_{n-2})$ has no $x^\alpha$ term, so $A(x):=(x-r_1)\cdots(x-r_{n-2})$ has no $x^\alpha$ term since $r_{n-1}\neq r_n$. Also, $r_{n-1}P(x)-r_nQ(x)=x(r_{n-1}-r_n)(x-r_1)\cdots(x-r_{n-2})$ has no $x^\alpha$ term so $A(x)$ has no $x^{\alpha-1}$ term.

Claim. $A(x)$ cannot have $n-2$ distinct real roots.
Proof. Suppose FTSOC $A(x)$ has $n-2$ distinct real roots. We claim that $A^{(m)}(x)$, the $m$th derivative of $A(x)$, has $n-2-m$ distinct real roots for all $m\leq n-2$. We proceed by induction on $m$ with the base case $m=0$ trivial.

Assume $A^{(m)}(x)$ has $n-2-m$ distinct real roots. Between every $2$ consecutive real roots there exists a local extremum, where $A^{(m+1)}(x)$ is $0$. $A^{(m)}(x)$ has $n-2-m-1$ pairs of consecutive real roots so $A^{(m+1)}(x)$ has $n-2-(m+1)$ distinct real roots, completing the induction step.

$A^{(\alpha-1)}$ has no $x^{(\alpha-1)-(\alpha-1)}=x^0$ or $x^{\alpha-(\alpha-1)}=x$ term, so $0$ is a double root of $A^{(\alpha-1)}$. However, all the roots of $A^{(\alpha-1)}$ are distinct and real, a contradiction. $\square$

This is a contradiction since $r_1,\,\ldots,\,r_{n-2}$ are distinct real numbers. Thus $P$ has a nonreal root. $\square$
Z K Y
N Quick Reply
G
H
=
a