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
1996 St. Petersburg City Mathematical Olympiad
Sadece_Threv   2
N a minute ago by reni_wee
Source: 1996 St. Petersburg City Mathematical Olympiad
Find all positive integers $n$ such that $3^{n-1}+5^{n-1}$ divides $3^{n}+5^{n}$
2 replies
Sadece_Threv
Jul 29, 2024
reni_wee
a minute ago
IMO 2010 Problem 5
mavropnevma   55
N 13 minutes ago by maromex
Each of the six boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$, $B_6$ initially contains one coin. The following operations are allowed

Type 1) Choose a non-empty box $B_j$, $1\leq j \leq 5$, remove one coin from $B_j$ and add two coins to $B_{j+1}$;

Type 2) Choose a non-empty box $B_k$, $1\leq k \leq 4$, remove one coin from $B_k$ and swap the contents (maybe empty) of the boxes $B_{k+1}$ and $B_{k+2}$.

Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ become empty, while box $B_6$ contains exactly $2010^{2010^{2010}}$ coins.

Proposed by Hans Zantema, Netherlands
55 replies
mavropnevma
Jul 8, 2010
maromex
13 minutes ago
NT ineq: sum 1/a_i < (m+n)/m , {a_1,a_2,...,a_n} subset of {1,2,...,m}
parmenides51   1
N 31 minutes ago by DVDTSB
Source: 2006 MOP Homework Blue NT 6
Let $m$ and $n$ be positive integers with $m > n \ge 2$. Set $S =\{1,2,...,m\}$, and set $T = \{a_1,a_2,...,a_n\}$ is a subset of $S$ such that every element of $S$ is not divisible by any pair of distinct elements of $T$. Prove that
$$\frac{1}{a_1}+\frac{1}{a_2}+ ...+ \frac{1}{a_n} < \frac{m+n}{m}$$
1 reply
parmenides51
Apr 12, 2020
DVDTSB
31 minutes ago
Never 8
chess64   21
N an hour ago by reni_wee
Source: Canada 1970, Problem 10
Given the polynomial \[ f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-1}x+a_n \] with integer coefficients $a_1,a_2,\ldots,a_n$, and given also that there exist four distinct integers $a$, $b$, $c$ and $d$ such that \[ f(a)=f(b)=f(c)=f(d)=5, \] show that there is no integer $k$ such that $f(k)=8$.
21 replies
chess64
May 14, 2006
reni_wee
an hour ago
old and easy imo inequality
Valentin Vornicu   215
N 2 hours ago by cubres
Source: IMO 2000, Problem 2, IMO Shortlist 2000, A1
Let $ a, b, c$ be positive real numbers so that $ abc = 1$. Prove that
\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]
215 replies
Valentin Vornicu
Oct 24, 2005
cubres
2 hours ago
x^2-x divides by n for some n/\omega(n)+1>x>1
NO_SQUARES   1
N 2 hours ago by a_507_bc
Source: 239 MO 2025 8-9 p6
Let a positive integer number $n$ has $k$ different prime divisors. Prove that there exists a positive integer number $x \in \left(1, \frac{n}{k}+1 \right)$ such that $x^2-x$ divides by $n$.
1 reply
NO_SQUARES
4 hours ago
a_507_bc
2 hours ago
IMO Genre Predictions
ohiorizzler1434   46
N 2 hours ago by Mrcuberoot
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
46 replies
ohiorizzler1434
May 3, 2025
Mrcuberoot
2 hours ago
4 wise men and 100 hats. 3 must guess their numbers
NO_SQUARES   1
N 3 hours ago by noemiemath
Source: 239 MO 2025 10-11 p5
There are four wise men in a row, each sees only those following him in the row, i.e. the $1$st sees the other three, the $2$nd sees the $3$rd and $4$th, and the $3$rd sees only the $4$th. The devil has $100$ hats, numbered from $1$ to $100$, he puts one hat on each wise man, and hides the extra $96$ hats. After that, each wise man (in turn: first the first, then the second, etc.) loudly calls a number, trying to guess the number of his hat. The numbers mentioned should not be repeated. When all the wise men have spoken, they take off their hats and check which one of them has guessed. Can the sages to act in such a way that at least three of them knowingly guessed?
1 reply
NO_SQUARES
3 hours ago
noemiemath
3 hours ago
IMO Shortlist 2011, G4
WakeUp   126
N 3 hours ago by NuMBeRaToRiC
Source: IMO Shortlist 2011, G4
Let $ABC$ be an acute triangle with circumcircle $\Omega$. Let $B_0$ be the midpoint of $AC$ and let $C_0$ be the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$ and let $G$ be the centroid of the triangle $ABC$. Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circle $\Omega$ at a point $X\not= A$. Prove that the points $D,G$ and $X$ are collinear.

Proposed by Ismail Isaev and Mikhail Isaev, Russia
126 replies
WakeUp
Jul 13, 2012
NuMBeRaToRiC
3 hours ago
<DPA+ <AQD =< QIP wanted, incircle circumcircle related
parmenides51   42
N 3 hours ago by AR17296174
Source: IMo 2019 SL G6
Let $I$ be the incentre of acute-angled triangle $ABC$. Let the incircle meet $BC, CA$, and $AB$ at $D, E$, and $F,$ respectively. Let line $EF$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle DPA + \angle AQD =\angle QIP$.

(Slovakia)
42 replies
1 viewing
parmenides51
Sep 22, 2020
AR17296174
3 hours ago
Help my diagram has too many points
MarkBcc168   28
N 3 hours ago by AR17296174
Source: IMO Shortlist 2023 G6
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$.

Prove that $\angle BXP = \angle CXQ$.

Kian Moshiri, United Kingdom
28 replies
MarkBcc168
Jul 17, 2024
AR17296174
3 hours ago
A lot of circles
ryan17   8
N 3 hours ago by AR17296174
Source: 2019 Polish MO Finals
Denote by $\Omega$ the circumcircle of the acute triangle $ABC$. Point $D$ is the midpoint of the arc $BC$ of $\Omega$ not containing $A$. Circle $\omega$ centered at $D$ is tangent to the segment $BC$ at point $E$. Tangents to the circle $\omega$ passing through point $A$ intersect line $BC$ at points $K$ and $L$ such that points $B, K, L, C$ lie on the line $BC$ in that order. Circle $\gamma_1$ is tangent to the segments $AL$ and $BL$ and to the circle $\Omega$ at point $M$. Circle $\gamma_2$ is tangent to the segments $AK$ and $CK$ and to the circle $\Omega$ at point $N$. Lines $KN$ and $LM$ intersect at point $P$. Prove that $\sphericalangle KAP = \sphericalangle EAL$.
8 replies
ryan17
Jul 9, 2019
AR17296174
3 hours ago
NT FE from Taiwan TST
Kitayama_Yuji   13
N 3 hours ago by bin_sherlo
Source: 2024 Taiwan TST Round 2 Mock P3
Let $\mathbb{N}$ be the set of all positive integers. Find all functions $f\colon \mathbb{N}\to \mathbb{N}$ such that $mf(m)+(f(f(m))+n)^2$ divides $4m^4+n^2f(f(n))^2$ for all positive integers $m$ and $n$.
13 replies
Kitayama_Yuji
Mar 29, 2024
bin_sherlo
3 hours ago
Yet another domino problem
juckter   15
N 3 hours ago by lksb
Source: EGMO 2019 Problem 2
Let $n$ be a positive integer. Dominoes are placed on a $2n \times 2n$ board in such a way that every cell of the board is adjacent to exactly one cell covered by a domino. For each $n$, determine the largest number of dominoes that can be placed in this way.
(A domino is a tile of size $2 \times 1$ or $1 \times 2$. Dominoes are placed on the board in such a way that each domino covers exactly two cells of the board, and dominoes do not overlap. Two cells are said to be adjacent if they are different and share a common side.)
15 replies
juckter
Apr 9, 2019
lksb
3 hours ago
IMO 2018 Problem 4
orthocentre   54
N Apr 10, 2025 by zuat.e
Source: IMO 2018
A site is any point $(x, y)$ in the plane such that $x$ and $y$ are both positive integers less than or equal to 20.

Initially, each of the 400 sites is unoccupied. Amy and Ben take turns placing stones with Amy going first. On her turn, Amy places a new red stone on an unoccupied site such that the distance between any two sites occupied by red stones is not equal to $\sqrt{5}$. On his turn, Ben places a new blue stone on any unoccupied site. (A site occupied by a blue stone is allowed to be at any distance from any other occupied site.) They stop as soon as a player cannot place a stone.

Find the greatest $K$ such that Amy can ensure that she places at least $K$ red stones, no matter how Ben places his blue stones.

Proposed by Gurgen Asatryan, Armenia
54 replies
orthocentre
Jul 10, 2018
zuat.e
Apr 10, 2025
IMO 2018 Problem 4
G H J
Source: IMO 2018
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
JAnatolGT_00
559 posts
#45
Y by
We claim the answer $K=100.$ Color grid as a chessboard, so Amy can occupy half of all black sites, which suffices.

Now consider the following strategy for Ben. Let's split grid $20\times 20$ onto smaller grids $4\times 4,$ in each of them pick four different cycles as below. When Amy firstly occupies site in one cycle, let Ben occupies the opposite site in this cycle. Thus Amy can place at most one stone in each of $100$ cycles, which suffices.
[asy]
unitsize(20);
draw ((0,0)--(2,1)--(3,3)--(1,2)--(0,0), red); 
draw ((0,3)--(1,1)--(3,0)--(2,2)--(0,3), green); 
draw ((1,0)--(3,1)--(2,3)--(0,2)--(1,0), blue); 
draw ((2,0)--(3,2)--(1,3)--(0,1)--(2,0), yellow);
for(int x = 0; x < 4; ++x){
for(int y = 0; y < 4; ++y){
dot((x,y));
}
}
[/asy]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
P2nisic
406 posts
#46
Y by
orthocentre wrote:
A site is any point $(x, y)$ in the plane such that $x$ and $y$ are both positive integers less than or equal to 20.

Initially, each of the 400 sites is unoccupied. Amy and Ben take turns placing stones with Amy going first. On her turn, Amy places a new red stone on an unoccupied site such that the distance between any two sites occupied by red stones is not equal to $\sqrt{5}$. On his turn, Ben places a new blue stone on any unoccupied site. (A site occupied by a blue stone is allowed to be at any distance from any other occupied site.) They stop as soon as a player cannot place a stone.

Find the greatest $K$ such that Amy can ensure that she places at least $K$ red stones, no matter how Ben places his blue stones.

Proposed by Gurgen Asatryan, Armenia

$K=100$.

Amy can be sure for $K>=100$.Color the pointw black and white then Amy place stones at only White points so $K>=100$.

Ben can be sure for $K<=100$.To do this he devise the chessdoard $20*20$ at smallers $4*4$ it is enoygth to prove that can be sure that Amy can not place more than $4$ stones at each $4*4$ for this Ben following the below strategi:
He place in the same $4*4$ in the same nymber.The numbers of the $4*4$ are:
1342
5786
6875
2431
This post has been edited 1 time. Last edited by P2nisic, May 26, 2023, 4:34 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
fuzimiao2013
3301 posts
#47 • 1 Y
Y by aryabhata000
Solved with aryabhata000

The answer is $\boxed{K = 100}$.

We first show that $K \leq 100$. To prove this, notice that we may partition the sites into $25$ $4 \times 4$ regions. We will label each region as such:
\[
\begin{bmatrix}
1 & 2 & 3 & 4 \\
3 & 4 & 1 & 2 \\
2 & 1 & 4 & 3 \\
4 & 3 & 2 & 1
\end{bmatrix}
\]Any site Amy occupies must correspond to one number in some $4 \times 4$. Note that if Ben occupies the site diagonally opposite the site Amy occupies (with the same number in the same $4 \times 4$) on his next turn, then Amy can no longer occupy any sites of that number in that $4 \times 4$. Thus, for each $4 \times 4$ grid, Amy can guarantee at most $1$ site of each number, meaning $K \leq 4 \times 25 = 100$.

We now show that $K \geq 100$. To see this, we color the sites in a checkerboard pattern, and notice that no two light squares are a distance of $\sqrt{5}$ away from each other. Therefore, Alice can guarantee at least $100$ light squares.

Combining these, we obtain $K = 100$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
AdventuringHobbit
164 posts
#48
Y by
First of all realize that Amy can guarantee $100$ by taking a white square on each of her turns, because knights only attack opposite colors. To show Ben can guarantee $100$, it suffices to show he can guarantee $4$ in a $4$ by $4$ grid. Consider the following grid:
$$ABCD$$
$$cdab$$
$$badc$$
$$DCBA$$
When Amy picks an uppercase letter, she blocks the lowercases of the same letter, so Bob can block the other uppercase of the same letter. The symmetric thing also holds if Amy picks a lowercase letter. Thus, Ben can limit Amy to at most $4$ sites, $1$ for each letter.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GrantStar
821 posts
#49
Y by
The answer is $100$. Amy can guarentee this by placing her first $100$ stones on squares of the same chessboard color.

Claim: Ben can limit Amy to $100$ stones
Proof. Consider this image of a $4\times 4$ square [asy]
draw((0,0)--(4,0)); draw((0,1)--(4,1)); draw((0,2)--(4,2)); draw((0,3)--(4,3)); draw((0,4)--(4,4)); draw((0,0)--(0,4)); draw((1,0)--(1,4)); draw((2,0)--(2,4)); draw((3,0)--(3,4)); draw((4,0)--(4,4));
filldraw(box((0,0),(1,1)),cyan, black); filldraw(box((2,1),(3,2)),cyan, black); filldraw(box((1,2),(2,3)),cyan, black); filldraw(box((3,3),(4,4)),cyan, black);
filldraw(box((1,0),(2,1)),lightred, black); filldraw(box((3,1),(4,2)),lightred, black); filldraw(box((2,3),(3,4)),lightred, black); filldraw(box((0,2),(1,3)),lightred, black);
filldraw(box((2,0),(3,1)),lightgreen, black); filldraw(box((3,2),(4,3)),lightgreen, black); filldraw(box((1,3),(2,4)),lightgreen, black); filldraw(box((0,1),(1,2)),lightgreen, black);
filldraw(box((3,0),(4,1)),lightblue, black); filldraw(box((1,1),(2,2)),lightblue, black); filldraw(box((2,2),(3,3)),lightblue, black); filldraw(box((0,3),(1,4)),lightblue, black);  [/asy] Divide the grid up into $25$ $4\times 4$ subgrids and color each one as shown. The startegy for Ben is if Amy plays in a square inside one $4\times 4$ grid, then Ben plays in the same in the same subgrid that's the same color as the one Amy played and the same chessboard color. It is not hard to check that this is unique and is always a possible move (if Ben couldn't play here then Amy's move before wouldn't be valid).
Now, this move means that the other squares left in the region of that color that Amy can play on are each $\sqrt 5$ away from the one that she first put down, meaning amy can never play twice in the same region. As each region has $4$ cells there are $100$ regions implying the claim $\blacksquare$

This claim concludes.

Remark: Pretty hard for a 1/4
This post has been edited 1 time. Last edited by GrantStar, Oct 11, 2023, 4:25 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Soaeb
4 posts
#50
Y by
@Math-Infinity ,he can place in white piece also..
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
math_comb01
662 posts
#51
Y by
We claim that the answer is $K = 100$
Consider the chessboard coloring; this implies $K \geq 100$ as no two squares of same color have distance $\sqrt{5}$
For the upper bound consider the following coloring, it has 4 cycles; implying $K \leq 4 \cdot 25 = 100$, $\therefore K = 100$
Attachments:
This post has been edited 1 time. Last edited by math_comb01, Jan 2, 2024, 5:07 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ATGY
2502 posts
#52
Y by
We use checkerboard coloring to colour alternating white and black squares to see that Amy is guaranteed to colour at least half of the black squares, so $K \geq 100$. Now, divide the grid into $4\times4$ squares where each row is one unit apart.
$$\text{A B C D}$$$$ \text{C D A B}$$$$\text{B A D C}$$$$\text{D C B A}$$Now, we must minimize Amy's stones here. WLOG, say Amy picks A. Arbitrarily, Ben picks the A that is not $\sqrt{5}$ units away from Amy's letter (since Amy can't pick the other 2 A's). Hence Amy is forced to pick another letter, say B. Similarly, Ben picks the B that is not $\sqrt{5}$ units away from B. Similarly, Amy is forced to pick one C and one D. So she minimally picks $4$ out of the $4\times4$ blocks, hence $K\leq 4*25 = 100$. From these two inequalities, it follows that $K = 100$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
dkedu
180 posts
#53
Y by
Note that $K\ge 100$ simply by checkboard coloring the grid and Alice placing her stones on one color only. She can guarantee half of the stones on this color since no two are $\sqrt 5$ apart.

Now color the grid with $25^2$ of the following:

$$\text{ABCD}$$$$\text{CDAB}$$$$\text{BADC}$$$$\text{DCBA}$$
Note that Bob can put his stone on the square with the same color that is not $\sqrt5$ away. This clearly limits Alice to a quarter of each color meaning she can get at most $100$ squares and we are done.
This post has been edited 2 times. Last edited by dkedu, Jan 27, 2024, 8:07 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
HamstPan38825
8859 posts
#54
Y by
This solution is deceptively simple, but I don't think it's that straightforward. It's easy to get caught up doing other things (hence this problem stands as a very good 1/4).

The answer is $K=100$. We call points which have the same parity $x$ and $y$-coordinate critical.

Claim: We can construct a matching between the set of all critical and non-critical points such that any matched points have a distance $\sqrt 5$ from each other.

Proof. Make $25$ copies of the following diagram:
[asy]
size(6cm);
for(int i=0; i<=3; ++i){ 
for(int j=0; j<=3; ++j){
if((i-j)%2==0){
dot((i, j), red);
}
else{
dot((i, j));
}
}
}
draw((0, 3)--(2, 2));
draw((1, 3)--(3, 2));
draw((2, 3)--(3, 1));
draw((3, 3)--(1, 2));
draw((0, 2)--(1, 0));
draw((0, 1)--(2, 0));
draw((1, 1)--(3, 0));
draw((2, 1)--(0, 0));
[/asy]
Each segment in this diagram connects a red critical point to a black non-critical point, as needed. $\blacksquare$

Thus, after any $k < 100$ moves, Amy can always find one of the $200$ such matchings that have both sites unoccupied. Futhermore, after $100$ moves, if Ben always places a stone on a site in a matching with both sites empty, every matching will have a stone on it, so Amy cannot make any more moves. This completes the proof.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Martin2001
149 posts
#55
Y by
The max is $100.$ This is because there are $100$ $4$-cycles that Ben can always only allow Amy to have exactly one square in the $4$ cycle. Amy can achieve this by only playing on one color if we color the board like a checkerboard$.\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megahertz13
3183 posts
#56 • 1 Y
Y by kilobyte144
The answer is $K = \boxed{100}$.

Proof that Amy can place at least $K=100$ stones. Color the board like a chessboard. Amy can place stones only on black squares (she doesn't have to worry about the $\sqrt{5}$ restriction). She can make $100$ moves before she runs out of black squares.

Proof that Ben can stop Amy from placing more than $100$ stones. Break the region into $4\times 4$ squares, and color them as follows. The number $n$ denotes the $n$th color we have.
$$1   2   3   4$$$$3   4   1   2$$$$2   1   4   3$$$$4   3   2   1$$Call a group of stones with the same color within a $4\times 4$ square a loop.

Any time Amy places a stone, Ben can place a stone at the opposite place in the same loop

This blocks off all the stones in the loop that Amy just placed a stone in. Therefore, Amy can place a maximum of one stone in each loop. The maximum number of stones she can place is the number of loops, which is $100$.

This completes the problem.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
SA28082008
10 posts
#57
Y by
We will divide the solution into two parts.

Part 1: Proving that \( K \geq 100 \)}

Amy's Strategy

She will place the first red stone at the coordinate \( (1,1) \).

On each subsequent turn, she places a new red stone on any unoccupied cell of the form:
\[
(1+2k, 1+2r) \text{ or } (2k, 2r)
\]where \( k \) and \( r \) are integers.

Notice that no two of these positions are spaced by a distance of \( \sqrt{5} \).

Since there are 200 such cells, by the Pigeonhole Principle, Amy can guarantee that she can place at least:
\[
\left\lceil \frac{200}{2} \right\rceil = 100
\]red stones.

Part 2: Proving that \( K \leq 100 \)}

Bob's Strategy:

Consider each \( 4 \times 4 \) sub-grid separately. Redefine its vertices as \( (i, j) \) where \( i, j \leq 4 \).

Suppose that Amy places a red stone at the coordinate \( (x, y) \). Bob will subsequently place a blue stone at its opposite coordinate:
\[
(5-x, 5-y)
\][asy]
size(300);

pen blackpen = black;
pen graypen = gray;
pen redpen = red;
pen bluepen = blue;
pen greenpen = green;
pen orangepen = rgb(1, 0.55, 0);
pen purplepen = purple;
pen shadecolor = rgb(1, 0.9, 0.85);

pair A = (0,0), B = (3,0), C = (3,3), D = (0,3);
pair N = (0,2), O = (0,1), P = (1,1), Q = (1,0), R = (2,0), S = (2,1), L = (2,2), K = (1,2), M = (3,2), T = (3,1);

pair[] gridpoints = {A, B, C, D, N, O, P, Q, R, S, L, K, M, T};

pair[] labels = {
    (0,3), (1,3), (2,3), (3,3),   // D, I, J, C
    (0,2), (3,2),                 // N, M
    (0,1), (3,1),                 // O, T
    (0,0), (3,0),                 // A, B
    (1,2), (2,2),                 // K, L
    (1,1), (2,1),                 // P, S
    (1,0), (2,0)                  // Q, R
};

string[] labeltext = {
    "D", "I", "J", "C",
    "N", "M",
    "O", "T",
    "A", "B",
    "K", "L",
    "P", "S",
    "Q", "R"
};

pen[] labelcolor = {
    bluepen, redpen, orangepen, purplepen, 
    greenpen, greenpen, 
    redpen, orangepen, 
    blackpen, graypen, 
    blackpen, graypen, 
    bluepen, purplepen, 
    greenpen, greenpen
};

// Draw grid lines
draw((0,0)--(0,3), graypen);
draw((1,0)--(1,3), graypen);
draw((2,0)--(2,3), graypen);
draw((3,0)--(3,3), graypen);

draw((0,0)--(3,0), graypen);
draw((0,1)--(3,1), graypen);
draw((0,2)--(3,2), graypen);
draw((0,3)--(3,3), graypen);

// Shade inner square
filldraw((1,1)--(1,2)--(2,2)--(2,1)--cycle, shadecolor);

// Draw outer border
draw(A--D--C--B--cycle, orangepen+linewidth(1.2));

// Place dots
dot(D, bluepen); dot(C, purplepen);
dot(N, greenpen); dot(M, greenpen);
dot(O, redpen); dot(T, orangepen);
dot(A, blackpen); dot(B, graypen);
dot(K, blackpen); dot(L, graypen);
dot(P, bluepen); dot(S, purplepen);
dot(Q, greenpen); dot(R, greenpen);

// Label points
for (int i=0; i<labels.length; ++i)
{
    label(Label(labeltext[i], labelcolor[i]), labels[i], dir(90));
}
[/asy]

The key observation is that no two opposite indices have the same color.

We will divide the proof further into two cases:

Case 1: Amy places a red stone at a position with the same color as a previously placed blue stone

In this case, it can be trivially observed that this new red stone would be at a distance of \( \sqrt{5} \) from a stone placed in Amy's previous turns, which leads to a contradiction.

Case 2: Amy never interferes Case1

In this case, no two red or blue stones placed in the \( 4 \times 4 \) grid will form a pair with the same color.

This implies that Amy and Bob together can place at most 8 stones on this grid. Consequently, Amy can place at most 4 stones.

Applying the same argument to each \( 4 \times 4 \) sub-grid, we find that Amy can place at most:
\[
\frac{200}{8} \times 4 = 100
\]stones in total.

Thus, combining both parts, we conclude that:
\[
K = 100
\]as desired.
This post has been edited 1 time. Last edited by SA28082008, Feb 20, 2025, 1:48 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Maximilian113
575 posts
#58
Y by
We claim that $K=100$ is the maximum.

For attainability, color the board like a checkerboard. Then no two red stones on the same colored square have a distance of $\sqrt 5,$ so Amy is ensured to get at least half of the black squares. Hence she can attain at least $100$ red stones with this strategy.

For the bound, separate the grid into $4 \times 4$ grids. Then color the grid in the following way:
1234
3412
2143
4321.
In a "ring" of the same color, if Amy places a stone on a square in this ring then Ben can react by placing a stone at the other side. For example, in the 2-ring if Amy places a blue stone on the $2$ from the top row, Ben reacts by placing a blue stone on the $2$ from the bottom row. Then, each ring will have at most $1$ red stone, therefore this strategy guarantees that Amy will always only be able to place $100$ red stones, $100$ is optimal.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
zuat.e
55 posts
#59
Y by
We claim the solution is $K_{20}=100$.

Map the coordinate system onto a grid and say that two squares attack each other and are attacking pairs each other if they're at a distance of $\sqrt{5}$ from each other.
The construction is easy: color all squares with both even coordinates and Amy can ensure she chooses half of them, hence $\frac{400}{4}=100$.

The key claim is the following:
Claim:
The greatest $K$ for a $4$ x $4$ grid is $K_4=4$
Proof:
Ben will use the following strategy: given the center of that grid (the circumcenter of the square formed by it), at every turn, Ben will take the reflection of the last square Amy's chosen $w.r.t.$ such center.
The proof is now equivalent to showing that Amy can't place $5$ stones using not only the $\sqrt{5}$ restriction, but also the fact that no two squares can be symmetric $w.r.t.$ the center.
We can now subdivide into cases and bash (I won't bash here on latex, but it isn't hard on paper)

(i) If we place two stones on the central squares, then we're left with $2$ attacking pairs, hence at most 4 stones
(ii) If we don't place any stones on the central squares, then each stone attacks three new squares and $4+4\cdot 3=16$
(iii) If we just place one stone on the central square, then again each stone attacks three new squares except two pairs of reflections $w.r.t.$ the center of the grid and it's now easy to finish bashing

Finally, divide the $20$ x $20$ grid into $25$ non-overlapping $4$ x $4$ subgrids and it follows that $K_{20}\leq 25\cdot K_4=100$, as desired.
Z K Y
N Quick Reply
G
H
=
a