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

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
Two circles tangents
Eeightqx   11
N 4 minutes ago by awesomeming327.
Source: 2024 GPO P2
Let $\triangle ABC$ be a acute triangle with altitudes $AD,BE,CF$. $H$ is the orthocenter of $\triangle ABC$. The angle bisectors of $\angle BEC,\angle BFC$ intersect $BC$ at $X,Y$, respectively. The circle $\odot(BC)$ with diameter $BC$ intersects segment $AD$ at $Z$. Show that $\odot(BC)$ is tangent to $\odot(XYZ)$.
11 replies
Eeightqx
Feb 14, 2024
awesomeming327.
4 minutes ago
Polynomial...
Sadigly   0
15 minutes ago
Source: Azerbaijan Senior NMO 2020
Find all nonzero polynomials $P(x)$ with real coefficents, that satisfies $$P(x)^3+3P(x)^2=P(x^3)-3P(-x)$$for all real numbers $x$
0 replies
Sadigly
15 minutes ago
0 replies
Regular 2021-gon
Sadigly   0
17 minutes ago
Source: Azerbaijan Senior NMO 2020
A regular 2021-gon is divided into 2019 triangles,such that no diagonals intersect. Prove that at least 3 of the 2019 triangles are isoscoles
0 replies
Sadigly
17 minutes ago
0 replies
Circumcircle of ADM
v_Enhance   67
N 20 minutes ago by Ilikeminecraft
Source: USA TSTST 2012, Problem 7
Triangle $ABC$ is inscribed in circle $\Omega$. The interior angle bisector of angle $A$ intersects side $BC$ and $\Omega$ at $D$ and $L$ (other than $A$), respectively. Let $M$ be the midpoint of side $BC$. The circumcircle of triangle $ADM$ intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. Let $N$ be the midpoint of segment $PQ$, and let $H$ be the foot of the perpendicular from $L$ to line $ND$. Prove that line $ML$ is tangent to the circumcircle of triangle $HMN$.
67 replies
v_Enhance
Jul 19, 2012
Ilikeminecraft
20 minutes ago
Nice geometry...
Sadigly   0
20 minutes ago
Source: Azerbaijan Senior NMO 2020
Let $ABC$ be a scalene triangle, and let $I$ be its incenter. A point $D$ is chosen on line $BC$, such that the circumcircle of triangle $BID$ intersects $AB$ at $E\neq B$, and the circumcircle of triangle $CID$ intersects $AC$ at $F\neq C$. Circumcircle of triangle $EDF$ intersects $AB$ and $AC$ at $M$ and $N$, respectively. Lines $FD$ and $IC$ intersect at $Q$, and lines $ED$ and $BI$ intersect at $P$. Prove that $EN\parallel MF\parallel PQ$.
0 replies
Sadigly
20 minutes ago
0 replies
Diophantine involving cube
Sadigly   0
25 minutes ago
Source: Azerbaijan Senior NMO 2020
$a;b;c;d\in\mathbb{Z^+}$. Solve the equation: $$2^{a!}+2^{b!}+2^{c!}=d^3$$
0 replies
Sadigly
25 minutes ago
0 replies
Easy combinatorics
Sadigly   5
N 30 minutes ago by GreekIdiot
Source: Azerbaijan Junior NMO 2018
First $20$ positive integers are written on a board. It is known that, after you erase a number from the board, there exists a number that is equal to the arithmetic mean of the rest of the numbers left on the board. Find all the numbers that could've been erased.
5 replies
1 viewing
Sadigly
an hour ago
GreekIdiot
30 minutes ago
Easy algebra
Sadigly   1
N 35 minutes ago by Vivaandax
Source: Azerbaijan Junior NMO 2018
$x^{11}+x^7+x^3=1$. $$x^{\alpha}=x^4+x^3-1.\hspace{4mm} \alpha=?$$
1 reply
Sadigly
an hour ago
Vivaandax
35 minutes ago
For an advanced method, search Lagrange Interpolation (those who knows)
Sadigly   2
N 36 minutes ago by Sadigly
Source: Azerbaijan Senior NMO 2018
$P(x)$ is a fifth degree polynomial. $P(2018)=1$, $P(2019)=2$ $P(2020)=3$, $P(2021)=4$, $P(2022)=5$. $P(2017)=?$
2 replies
Sadigly
an hour ago
Sadigly
36 minutes ago
Easy geometry...
Sadigly   1
N 39 minutes ago by Nuran2010
Source: Azerbaijan NMO 2018
A circle $\omega$ and a point $T$ outside the circle is given. Let a tangent from $T$ to $\omega$ touch $\omega$ at $A$, and take points $B,C$ lying on $\omega$ such that $T,B,C$ are colinear. The bisector of $\angle ATC$ intersects $AB$ and $AC$ at $P$ and $Q$,respectively. Prove that $PA=\sqrt{PB\cdot QC}$
1 reply
Sadigly
an hour ago
Nuran2010
39 minutes ago
Easy inequality...
Sadigly   0
41 minutes ago
Source: Azerbaijan Senior NMO 2020
$x,y,z\in\mathbb{R^+}$. If $xyz=1$, then prove the following: $$\sum\frac{x^6+2}{x^3}\geq3(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})$$
0 replies
Sadigly
41 minutes ago
0 replies
Easy inequality...
Sadigly   2
N 41 minutes ago by ehuseyinyigit
Source: Azerbaijan Junior NMO 2018
$a;b\in\mathbb{R^+}$. Prove the following inequality: $$\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\leq\sqrt[3]{2(a+b)(\frac1{a}+\frac1{b})}$$
2 replies
Sadigly
an hour ago
ehuseyinyigit
41 minutes ago
reals associated with 1024 points
bin_sherlo   1
N an hour ago by AnSoLiN
Source: Türkiye 2025 JBMO TST P8
Pairwise distinct points $P_1,\dots,P_{1024}$, which lie on a circle, are marked by distinct reals $a_1,\dots,a_{1024}$. Let $P_i$ be $Q-$good for a $Q$ on the circle different than $P_1,\dots,P_{1024}$, if and only if $a_i$ is the greatest number on at least one of the two arcs $P_iQ$. Let the score of $Q$ be the number of $Q-$good points on the circle. Determine the greatest $k$ such that regardless of the values of $a_1,\dots,a_{1024}$, there exists a point $Q$ with score at least $k$.
1 reply
bin_sherlo
3 hours ago
AnSoLiN
an hour ago
JBMO Combinatorics vibes
Sadigly   0
an hour ago
Source: Azerbaijan Senior NMO 2018
Numbers $1,2,3...,100$ are written on a board. $A$ and $B$ plays the following game: They take turns choosing a number from the board and deleting them. $A$ starts first. They sum all the deleted numbers. If after a player's turn (after he deletes a number on the board) the sum of the deleted numbers can't be expressed as difference of two perfect squares,then he loses, if not, then the game continues as usual. Which player got a winning strategy?
0 replies
Sadigly
an hour ago
0 replies
Game of Polynomials
anantmudgal09   13
N Apr 28, 2025 by Mathandski
Source: Tournament of Towns 2016 Fall Tour, A Senior, Problem #6
Petya and Vasya play the following game. Petya conceives a polynomial $P(x)$ having integer coefficients. On each move, Vasya pays him a ruble, and calls an integer $a$ of his choice, which has not yet been called by him. Petya has to reply with the number of distinct integer solutions of the equation $P(x)=a$. The game continues until Petya is forced to repeat an answer. What minimal amount of rubles must Vasya pay in order to win?

(Anant Mudgal)

(Translated from here.)
13 replies
anantmudgal09
Apr 22, 2017
Mathandski
Apr 28, 2025
Game of Polynomials
G H J
Source: Tournament of Towns 2016 Fall Tour, A Senior, Problem #6
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
anantmudgal09
1980 posts
#1 • 7 Y
Y by Wave-Particle, Ankoganit, dgrozev, Adventure10, Mango247, TheHimMan, ohiorizzler1434
Petya and Vasya play the following game. Petya conceives a polynomial $P(x)$ having integer coefficients. On each move, Vasya pays him a ruble, and calls an integer $a$ of his choice, which has not yet been called by him. Petya has to reply with the number of distinct integer solutions of the equation $P(x)=a$. The game continues until Petya is forced to repeat an answer. What minimal amount of rubles must Vasya pay in order to win?

(Anant Mudgal)

(Translated from here.)
This post has been edited 2 times. Last edited by anantmudgal09, Feb 5, 2018, 1:11 PM
Reason: Titular reference to "Game of Thrones"
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ankoganit
3070 posts
#2 • 9 Y
Y by anantmudgal09, Vietnamisalwaysinmyheart, Vrangr, hansu, Wizard_32, kamatadu, Adventure10, Ru83n05, EvansGressfield
This is a very cute problem, so let's hope I got this right for Anant's sake :P

We claim that Vasya could do with $4$ rubles. We'll use the following notation: for an integer $a$, $n(a)$ is the number of distinct integer roots of $P(x)=a$. We'll first prove a lemma:

Lemma: If $P(x)=a$ has $3$ or more distinct integer root for some $P(x)\in\mathbb Z[x]$ and $a\in \mathbb Z$, then both the equations $P(x)=a+1$ and $P(x)=a-1$ have no integer root.
Proof: We can shift the origin appropriately and assume WLOG $a=0$. Then $$P(x)=(x-a_1)^{b_1}(x-a_2)^{b_2}(x-a_3)^{b_3}Q(x)$$for integers $a_i$'s , $b_i$'s and $Q\in\mathbb Z[x]$. Here $a_i$'s are all distinct and $b_i$'s are all positive. Then to have $|P(x)|=1$, we'll need $(x-a_i)=\pm 1$ for each $i$; but by PHP, this is impossible with distinct $a_i$'s. This proves our lemma. $\square$

Now for the main problem. We'll first show that Vasya can win in at most $4$ moves. Let Vasya call the integers $0,-1,+1$ in the first three moves. If $n(0)\ge 3$, then $n(1)=n(-1)=0$ by the lemma and Vasya wins. If one of $n(1)$ and $n(-1)$ is $\ge 3$, say WLOG $n(1)=3$; then Vasya can call $2$ in the next move and win, because then $n(0)=n(2)=0$ by our lemma. So suppose $n(0), n(1),n(-1)$ are all $\le 2$. If some of two these were equal, Vasya would've won already, so let's say $n(0),n(-1),n(1)$ are $0,1,2$ in some order. Then at least one of $n(1)$ and $n(-1)$ is nonzero; say WLOG $n(1)\ne 0$. Then let Vasya call $2$ in the fourth move. $n(2)$ can't be $\ge 3$, or we would have $n(1)=0$; so $n(2)\in\{0,1,2\}$, so it has to collide with one of $n(0),n(1),n(-1)$, letting Vasya win.

Now it remains to show that Vasya can't be sure of winning in $3$ moves or less. Indeed, we may WLOG assume the first integer called by Vasya is $0$. Then if he calls the integers $a$ and $b$ in the next two moves and hopes to win, then Petya could smash his hopes by conceiving the polynomial $$P(x)\equiv -a(x-1)(x-3)\left( (x-2)^2\left(x^{2016}+|b|+2017^{2017}\right)+1\right).$$Indeed, $P(x)=0$ has two roots $1$ and $3$; $P(x)=a$ has only one root $2$, because $$P(x)=a\implies |(x-1)(x-3)|=1\implies x=2;$$and $P(x)=b$ has no roots at all, because for $x\not\in \{1,2,3\}$, we have $|P(x)|>b$. Thus we're done. $\blacksquare$
This post has been edited 2 times. Last edited by Ankoganit, Apr 23, 2017, 11:43 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dgrozev
2463 posts
#3 • 3 Y
Y by anantmudgal09, Adventure10, Mango247
@Anant Mudgal: Not that it matters, but Petya and Vasya are russian male names. So- him/his. :)
I suppose your original statement was different, but the russians change it in their Petya/Vasya game style ?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
anantmudgal09
1980 posts
#4 • 3 Y
Y by Ankoganit, Adventure10, Mango247
dgrozev wrote:
@Anant Mudgal: Not that it matters, but Petya and Vasya are russian male names. So- him/his. :)
I suppose your original statement was different, but the russians change it in their Petya/Vasya game style ?

Thanks for pointing, I had this misconception for a while :). As you said, my original formulation involved Alice and Bob, which was taken as it is in the English version. Interestingly, I didn't pose it in the form of a game involving money, but it looks like Russian paper setters are fond of it.


@Ankoganit, that's almost the same idea I had in mind (except my phrasing of the key lemma wasn't as tidy). I had a different construction though; and it turns out that the contestants gave many other constructions.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dgrozev
2463 posts
#5 • 17 Y
Y by BartSimpsons, anantmudgal09, Ankoganit, Tintarn, rafayaashary1, 62861, claserken, InCtrl, tarzanjunior, Wizard_32, AlastorMoody, starchan, Assassino9931, Adventure10, Mango247, Supercali, kiyoras_2001
anantmudgal09 wrote:
...Interestingly, I didn't pose it in the form of a game involving money, but it looks like Russian paper setters are fond of it.
Don't worry, it could have been... glasses of vodka. :)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
anantmudgal09
1980 posts
#6 • 2 Y
Y by Adventure10, Mango247
...........
This post has been edited 1 time. Last edited by anantmudgal09, Feb 2, 2020, 3:47 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ankoganit
3070 posts
#7 • 9 Y
Y by anantmudgal09, rafayaashary1, biomathematics, Wizard_32, AlastorMoody, kamatadu, anurag27826, Adventure10, Mango247
Who said he's gonna drink it?
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
anantmudgal09
1980 posts
#8 • 4 Y
Y by rafayaashary1, Ankoganit, Wizard_32, Adventure10
I'll post my solution for the part about construction. Proof for sufficiency of $4$ moves is the same as in post #2.

WLOG, assume Vasya is done in $3$ moves (for lesser number, he simply asks random questions to reach $3$). Petra replies with $1,2,0$ in this order. To see his answers are consistent, let $m, n, l$ be the numbers asked by Vasya in this order. Then, for $$P(X) \overset{\text{def}}{:=} tX^4+(n-m-t)X^2+m,$$where $t$ is a positive integer, so that $P$ is increasing over positive integers. We have $P(0)=m$ and $P(\pm 1)=n$ and $P(2)>\operatorname{max}\{m, n, l \}$ so we are done.
This post has been edited 1 time. Last edited by anantmudgal09, Feb 2, 2020, 3:48 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
primesarespecial
364 posts
#9
Y by
The minimum number of rubles Vasya needs to pay is $4$.
We divide into two cases.Firstly Vasya asks $0$.
If $P$ does not have an integer root,then let Vasya ask integers $d>1$ and $d+1$.We must have solutions to $P(x)=d,d+1$,otherwise we are done.
Let $a_1<a_2,..<a_r$ and $b_1<b_2...<b_k$ be such that $P(a_i)=d,P(b_i)=d+1$.WLOG,assume $r>k>0$,since $b_i -a_j|1$,for any $i,j$,we can conclude that for this to be true,$r=2,k=1$ and we have $a_2=a_1+2,b_1=a_1+1$.Now,if Vasya asks $d-1$,we can conclude by the same logic for $d-1,d$ that $P(x)=d-1$ has no solutions and we are done in at most $4$ moves.

Now,if $P$ has an integer root,then we can be done by Vasya asking $0,1,-1,-2$ or $0,1,-1,2$(depending on whether $P(x)-1$ or $P(x)+1$ has no integer roots),by the same logic in the first case.

Now,we give a construction where $3$ fails.Let Vasya ask $a<b<c$.
Then Petya conceives $P(x)=(a-b)(x+1)(x+3)+a$,for which $P(x)=a,b,c$ has $2,1,0$ solutions respectively.
This post has been edited 2 times. Last edited by primesarespecial, Jan 21, 2022, 7:24 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IAmTheHazard
5001 posts
#10
Y by
The answer is $4$ rubles.

We first outline a strategy for Vasya to win by paying only four rubles. He begins by querying $0$.
If Petya's response is $\geq 2$, then by shifting WLOG let $P(x)=x(x-r)Q(x)$ where $Q$ is an arbitrary integer polynomial (possibly constant), and $r>0$. He then queries $1$ and $-1$. Note that for a response to be nonzero, we must have $x,x-r \in \{-1,1\}$ simultaneously. Since $r>0$ this forces $r=2$ and $x=1$, in which case one answer is zero and the other is one. In this case (since in the other case our responses are 2, 0, 0 and Vasya wins by paying three rubles), query $2$, which forces $x,x-2 \in \{-1,1,-2,2\}$ if the answer is nonzero. $x=1$ fails because $P(1)=1$, and $x \in \{-1,-2.2\}$ fail because then $x-2 \not \in \{-1,1,-2,2\}$, so Petya answers zero and thus repeats a prior answer.
If Petya's response is $0$ or $1$, then Vasya queries $1000$. If the response to that query is $\geq 2$, Vasya then queries $999$ and $1001$. By our previous result (shifted appropriately), if the responses to these two queries are different then they are $0$ and $1$, so there is a repeat. On the other hand, if the response to our $1000$ query is $1$ or $0$ (the opposite of what the first response is), then query $10000$. If the response to that is $0$ or $1$ we're done, and if it's $\geq 2$ then query $10001$ which for similar reasons as before must either be $0$ or $1$. In any case, a response is repeated, and Vasya pays at most four rubles.

Now we prove that Vasya cannot win by paying three rubles. Note that the only information Vasya gains at the conclusion of each query is the response to the previous, so let's say that Petya is feeling nice and decides to tell Vasya beforehand that his answers to the first and second queries will be $2$ and $1$ respectively. However, this action isn't only done out of charity—now that Vasya cannot gain new information based on responses, Petya asks Vasya to write down his three queries beforehand. Despite this, Vasya's task is still made easier. However, Petya still cannot get Vasya to repeat an answer in the first three queries.
By shifting, WLOG let the first query be $0$, the second one be $a$, and the third one be $b$. If Petya then selects the polynomial
$$P(x)=-ax(x-2)(|b|(x(x-1)(x-2))^2+1).$$$P(x)=0$ then has solutions $0,2$, $P(x)=a$ has $1$ as the only solution, since we would require $x,x-2 \in \{-1,1\}$ which forces $x=1$. Finally, $P(x)=b$ has no solutions because the absolute value of the last factor is always greater than $|b|$ unless $x \in \{0,1,2\}$, but these cases have already been handled and don't satisfy $P(x)=b$. Thus the response to the third query will be $0$, and Vasya will have to pay an additional ruble (at least) to get a repeat. We are done. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Dec 19, 2022, 3:25 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HamstPan38825
8863 posts
#11
Y by
The answer is $4$ guesses.

Strategy: The key is the following claim.

Claim. For some integer polynomial $P$, if $P(x) = a$ has at least $3$ solutions for some integer $a$, then $P(x) = a\pm 1$ has no integer solutions.

Proof. Write $$P(x) = (x-x_1)(x-x_2)(x-x_3)R(x) + a$$for integers $x_1, x_2, x_3$ and integer polynomial $R$. For $P(x) = a \pm 1$, $(x-x_1)(x-x_2)(x-x_3) = \pm 1$, but this is impossible as the $x_i$ are distinct. $\blacksquare$

Now Vasya guesses $a = 0$ first. If Petya responds with any integer at least $3$, Vasya can guess $a = 1$ and $a=-1$ and be sure that both answers will be $0$.

If Petya responds with $1$ or $2$, then Vasya may guess $1$ and $-1$, knowing he will receive answers in $\{0, 1, 2\}$. If both guesses reveal $0$, he wins. Otherwise, suppose the answer to $1$ was nonzero; then Vasya can guess $2$ as well. Then by Pigeonhole, two of the four answers will be identical.

If Petya responds with $0$, Vasya can again guess $1$ and $-1$; if one of them is zero, Petya wins; otherwise, the same Pigeonhole argument applies.

Bound: We show Vasya cannot win in three moves. Indeed, let his first three guesses be $a, b, c$ integers. Without loss of generality assume $a<b<c$ (the order of the guesses doesn't matter). Petya can pick the polynomial $$P(x) = a+(c-a)(x-x_1)(x-(x_1+2))$$for some integer $x_1$. Then $P(x) = a$ has two solutions, $P(x) = c$ has precisely one, and as the second term is at least $c-a$ when it is positive, $P(x) = b$ has no solutions.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
john0512
4187 posts
#12
Y by
The answer is $4$.

To show that $3$ is not enough, suppose Vasya queries $a,b,c$ for $a>b>c$. Then, the polynomial $(a-b)x^2+b$ attains $a$ at $x=\pm 1$, $b$ at $x=0$, and $c$ never. Thus, Vasya does not win.

We claim that querying $0,1,2,3$ is sufficent, that is, no integer polynomial attains $0,1,2,3$ each a different number of times.

The key idea is to use the fact that $a-b|P(a)-P(b)$. Consider the sequence $\dots,P(-2),P(-1),P(0),\dots$. Suppose this sequence has $0,1,2,3$ each a different number of times. All instances of $0$ and all instances of $1$ must be adjacent, and same with all other consecutive pairs.

Some number has to happen at least $3$ times. If $1$ happens 3 times, then no position is adjacent to all three 1s, so $0$ and $2$ can never occur, contradiction. Same thing happens if $2$ happens at least $3$ times. Thus, WLOG assume $0$ happens at least $3$ times (3 is the same). Then $1$ can never occur, since no position is adjacent to all three $0$s.

Therefore, at least one of $2$ and $3$ must appear at least twice, and the other must appear at least once. If $2$ appears at least twice, then the $3$ must be adjacen to both $2$s, which forces $232$. However, since $a-b\mid P(a)-P(b)$, $0$s must be within 2 spots of all $2$s, but there is nowhere for $0$s to go after $232$, contradiction. If we had $323$ instead, we only have two spots that are valid spots for $0$s, but there are $3$ of them, contradiction. Thus we are done.


Remark: The main idea of this problem is to invoke $a-b\mid P(a)-P(b)$ to restrict where outputs can lie. The key observation is that, if the polynomial is only required to have rational coefficents, then Petya can just interpolate through whatever Vasya queries, thus he cannot win. The key feature of integer polynomials that distinguishes it from rational ones is $a-b\mid P(a)-P(b)$ which is why we use it here.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
OronSH
1746 posts
#13
Y by
Answer: $4$. Vasya calls $0$. If it is $\ge 2$, he calls $1,2,-1$. If $1$ returns $\ge 1$, then if $P(a)=P(b)=0$ and $P(c)=1$, then $a-c,b-c\mid 1$ so $a,c,b$ are consecutive and in particular $c$ is unique so $1$ returns $1$. Next if $P(d)=2$, then $a-d,b-d\mid 2$ with $a-b=2$ so $d=c$, contradiction, so $2$ returns $0$, and similarly so does $-1$. If $1$ returns $0$ and $2$ returns $1$, then similarly $-1$ returns $0$. Otherwise $1,2$ both return $0$.

If $0$ returns $\le 1$, then he calls $1,2,3$. If $1$ is $\ge 2$, then similarly either $2,3$ return the same value or $2,3$ return $0,1$, one of which returns the same as $0$. If $1$ is $\le 1$ but $2$ returns $\ge 2$, then similarly $3$ returns either $0,1$ which returns the same as either $0$ or $1$. If $0,1,2$ return $\le 1$ then two are the same.

To show $4$ is minimum, if Vasya calls $a,b,c$, we claim Petya can return $2,1,0$ respectively, since Petya has $P(x)=(a-b)x^2+b$ if $a$ is not between $b$ and $c$, and $P(x)=(a-b)x^{2\left\lfloor\log_4\frac{c-b}{a-b}\right\rfloor+2}+b$ otherwise.
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
#14
Y by
Subjective Difficulty Rating
Attachments:
Z K Y
N Quick Reply
G
H
=
a