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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.

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[*]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]
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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
2025 KMO Inequality
Jackson0423   1
N 4 minutes ago by Quantum-Phantom
Source: 2025 KMO Round 1 Problem 20

Let \(x_1, x_2, \ldots, x_6\) be real numbers satisfying
\[
x_1 + x_2 + \cdots + x_6 = 6,
\]\[
x_1^2 + x_2^2 + \cdots + x_6^2 = 18.
\]Find the maximum possible value of the product
\[
x_1 x_2 x_3 x_4 x_5 x_6.
\]
1 reply
+1 w
Jackson0423
Yesterday at 4:32 PM
Quantum-Phantom
4 minutes ago
Simple Geometry
AbdulWaheed   2
N 11 minutes ago by Sadigly
Source: EGMO
Try to avoid Directed angles
Let ABC be an acute triangle inscribed in circle $\Omega$. Let $X$ be the midpoint of the arc $\overarc{BC}$ not containing $A$ and define $Y, Z$ similarly. Show that the orthocenter of $XYZ$ is the incenter $I$ of $ABC$.
2 replies
AbdulWaheed
Today at 5:15 AM
Sadigly
11 minutes ago
inequality thing
BinariouslyRandom   2
N 41 minutes ago by BinariouslyRandom
Source: Philippine MO 2025 P5
Find the largest real constant $k$ for which the inequality \[ (a^2+3)(b^2+3)(c^2+3)(d^2+3) + k(a-1)(b-1)(c-1)(d-1) \ge 0 \]holds for all real numbers $a$, $b$, $c$, and $d$.

answer
2 replies
BinariouslyRandom
3 hours ago
BinariouslyRandom
41 minutes ago
Geometry Concurrence
KHOMNYO2   0
an hour ago
Given triangle $XYZ$ such that $XY \neq XZ$. Let excircle-$X$ be tangent with $YZ, ZX, XY$ at points $U, V, W$ respectively. Let $R$ and $S$ be points on the segment $XZ, XY$ respectively such that $RS$ is parallel to $YZ$. Lastly, let $\gamma$ be the circle that is externally tangent with the excircle-$X$ on point $T$. Prove that $VW, UT$, and $RS$ concur at a point.
0 replies
KHOMNYO2
an hour ago
0 replies
inequality
mathematical-forest   1
N an hour ago by lbh_qys
Positive real numbers $x_{1} ,x_{2} \cdots ,x_{n}$,satisfied $\sum_{i=1}^{n}x_{i} =1$
Proof:$$\sum_{i=1}^{n} \frac{\min  \left \{  x_{i-1},x_{i}\right \}\max \left \{  x_{i},x_{i+1}\right \}  }{x_{i}} \le 1$$
1 reply
mathematical-forest
2 hours ago
lbh_qys
an hour ago
Some number theory
EeEeRUT   5
N an hour ago by shafikbara48593762
Source: Thailand MO 2025 P9
Let $p$ be an odd prime and $S = \{1,2,3,\dots, p\}$
Assume that $U: S \rightarrow S$ is a bijection and $B$ is an integer such that $$B\cdot U(U(a)) - a \: \text{ is a multiple of} \: p \: \text{for all} \: a \in S$$Show that $B^{\frac{p-1}{2}} -1$ is a multiple of $p$.
5 replies
EeEeRUT
May 14, 2025
shafikbara48593762
an hour ago
Sharygin 2025 CR P7
Gengar_in_Galar   5
N an hour ago by Blackbeam999
Source: Sharygin 2025
Let $I$, $I_{a}$ be the incenter and the $A$-excenter of a triangle $ABC$; $E$, $F$ be the touching points of the incircle with $AC$, $AB$ respectively; $G$ be the common point of $BE$ and $CF$. The perpendicular to $BC$ from $G$ meets $AI$ at point $J$. Prove that $E$, $F$, $J$, $I_{a}$ are concyclic.
Proposed by:Y.Shcherbatov
5 replies
Gengar_in_Galar
Mar 10, 2025
Blackbeam999
an hour ago
60^o angle wanted, equilateral on a square
parmenides51   5
N 2 hours ago by sunken rock
Source: 2019 Austrian Mathematical Olympiad Junior Regional Competition , Problem 2
A square $ABCD$ is given. Over the side $BC$ draw an equilateral triangle $BCS$ on the outside. The midpoint of the segment $AS$ is $N$ and the midpoint of the side $CD$ is $H$. Prove that $\angle NHC = 60^o$.
.
(Karl Czakler)
5 replies
parmenides51
Dec 18, 2020
sunken rock
2 hours ago
Random walk
EthanWYX2009   0
2 hours ago
As shown in the graph, an ant starts from $4$ and walks randomly. The probability of any point reaching all adjacent points is equal. Find the probability of the ant reaching $1$ without passing through $6.$
0 replies
EthanWYX2009
2 hours ago
0 replies
Lemma on tangency involving a parallelogram with orthocenter
Gimbrint   0
2 hours ago
Source: Own
Let $ABC$ be an acute triangle ($AB<BC$) with circumcircle $\omega$ and orthocenter $H$. Let $M$ be the midpoint of $AC$. Line $BH$ intersects $\omega$ again at $L\neq B$, and line $ML$ intersects $\omega$ again at $P\neq L$. Points $D$ and $E$ lie on $AB$ and $BC$ respectively, such that $BEHD$ is a parallelogram.

Prove that $BP$ is tangent to the circumcircle of triangle $BDE$.
0 replies
Gimbrint
2 hours ago
0 replies
Consecutive squares are floors
ICE_CNME_4   10
N 2 hours ago by JARP091

Determine how many positive integers \( n \) have the property that both
\[
\left\lfloor \sqrt{2n - 1} \right\rfloor \quad \text{and} \quad \left\lfloor \sqrt{3n + 2} \right\rfloor
\]are consecutive perfect squares.
10 replies
ICE_CNME_4
Yesterday at 1:50 PM
JARP091
2 hours ago
Sharygin 2025 CR P10
Gengar_in_Galar   2
N 2 hours ago by Kappa_Beta_725
Source: Sharygin 2025
An acute-angled triangle with one side equal to the altitude from the opposite vertex is cut from paper. Construct a point inside this triangle such that the square of the distance from it to one of the vertices equals the sum of the squares of distances to to the remaining two vertices. No instruments are available, it is allowed only to fold the paper and to mark the common points of folding lines.
Proposed by: M.Evdokimov
2 replies
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
2 hours ago
Sharygin 2025 CR P12
Gengar_in_Galar   8
N 3 hours ago by Kappa_Beta_725
Source: Sharygin 2025
Circles $\omega_{1}$ and $\omega_{2}$ are given. Let $M$ be the midpoint of the segment joining their centers, $X$, $Y$ be arbitrary points on $\omega_{1}$, $\omega_{2}$ respectively such that $MX=MY$. Find the locus of the midpoints of segments $XY$.
Proposed by: L Shatunov
8 replies
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
3 hours ago
Sharygin 2025 CR P17
Gengar_in_Galar   6
N 3 hours ago by Kappa_Beta_725
Source: Sharygin 2025
Let $O$, $I$ be the circumcenter and the incenter of an acute-angled scalene triangle $ABC$; $D$, $E$, $F$ be the touching points of its excircle with the side $BC$ and the extensions of $AC$, $AB$ respectively. Prove that if the orthocenter of the triangle $DEF$ lies on the circumcircle of $ABC$, then it is symmetric to the midpoint of the arc $BC$ with respect to $OI$.
Proposed by: P.Puchkov,E.Utkin
6 replies
Gengar_in_Galar
Mar 10, 2025
Kappa_Beta_725
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
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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]
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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
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fuzimiao2013
3306 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.
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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.
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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
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Soaeb
4 posts
#50
Y by
@Math-Infinity ,he can place in white piece also..
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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
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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 $K = 100$.
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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
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HamstPan38825
8868 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.
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Martin2001
161 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$.\blacksquare$
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megahertz13
3184 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.
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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
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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.
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zuat.e
66 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.
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