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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
2-var inequality
sqing   0
a few seconds ago
Source: Own
Let $ a,b > 0 ,   a+b+a^2+ b^2 +ab  = 3.$ Prove that
$$  \frac{1}{a}+ \frac{ 1}{ b }-\frac{6}{ 5ab } \geq\frac{2(\sqrt{10}-2)}{9}$$$$\frac{1}{a}+\frac{1}{b}-\frac{1}{ab } \geq\frac{4\sqrt{10}-5}{9}$$$$\frac{1}{a}+\frac{1}{b}+ \frac{1}{ab } \geq\frac{17+8\sqrt{10}}{9}$$
0 replies
1 viewing
sqing
a few seconds ago
0 replies
Rational Points in n-Dimensional Space
steven_zhang123   0
23 minutes ago
Let \( T = (x_1, x_2, \ldots, x_n) \), where \( x_i \) is rational for \( i = 1, 2, \ldots, n \). A vector \( T \) is called a rational point in \( n \)-dimensional space. Denote the set of all such vectors \( T \) as \( S \). For \( A = (x_1, x_2, \ldots, x_n) \) and \( B = (y_1, y_2, \ldots, y_n) \) in \( S \), define the distance between points \( A \) and \( B \) as \( d(A, B) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2 + \cdots + (x_n - y_n)^2} \). We say that point \( A \) can move to point \( B \) if and only if there is a unit distance between two points in \( S \).

Prove:
(1) If \( n \leq 4 \), there exists a point that cannot be reached from the origin via a finite number of moves.
(2) If \( n \geq 5 \), any point in \( S \) can be reached from any other point via moves.
0 replies
1 viewing
steven_zhang123
23 minutes ago
0 replies
Inspired by old results
sqing   5
N 30 minutes ago by sqing
Source: Own
Let $a,b,c $ be reals such that $a^2+b^2+c^2=3$ .Prove that
$$(1-a)(k-b)(1-c)+abc\ge -k$$Where $ k\geq 1.$
$$(1-a)(1-b)(1-c)+abc\ge -1$$$$(1-a)(1-b)(1-c)-abc\ge -\frac{1}{2}-\sqrt 2$$
5 replies
sqing
Yesterday at 7:36 AM
sqing
30 minutes ago
equation in integers
Pirkuliyev Rovsen   2
N an hour ago by ytChen
Solve in $Z$ the equation $a^2+b=b^{2022}$
2 replies
1 viewing
Pirkuliyev Rovsen
Feb 10, 2025
ytChen
an hour ago
No more topics!
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
164 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
3190 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
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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
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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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