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Difference between revisions of "1999 USAMO Problems/Problem 1"

(→‎Solution 2)
(→‎Solution 2)
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shares a side with a square with a checker on it.
 
shares a side with a square with a checker on it.
  
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Say there are <math>k</math> checkers on the board, sitting on squares <math>s_1, \ldots, s_k</math>. We can assume WLOG that <math>s_1, \ldots, s_k</math> is ordered in such a way that if <math>1 \leq i \leq k</math>, then <math>s_i</math> shares a side with
+
Say there are <math>k</math> checkers on the board. We can assume WLOG that <math>s_1, \ldots, s_k</math> is ordered in such a way that if <math>1 \leq i \leq k</math>, then <math>s_i</math> shares a side with at least one of <math>s_1, \ldots, s_{i - 1}</math>. (If such an ordering were impossible, then some <math>s_i</math> wouldn’t be reachable from <math>s_1</math>, violating condition (b)).
−
at least one of <math>s_1, \ldots, s_{i - 1}</math>. (If such an ordering were impossible, then some <math>s_i</math> wouldn’t be reachable from <math>s_1</math>, violating condition (b)).
 
  
 
Now imagine removing all the checkers and then putting them back onto the squares <math>s_1,
 
Now imagine removing all the checkers and then putting them back onto the squares <math>s_1,
Line 45: Line 44:
 
shares a side with (at most 4) becomes good. When we put a checker onto <math>s_i</math>,
 
shares a side with (at most 4) becomes good. When we put a checker onto <math>s_i</math>,
 
where <math>i > 1</math>, we increase the number of good squares by at most <math>3</math>, since <math>s_i</math> is
 
where <math>i > 1</math>, we increase the number of good squares by at most <math>3</math>, since <math>s_i</math> is
−
already good, and one of its at most <math>4</math> neighbors is already good.  
+
already good, and one of its neighbors already has a checker on it.  
  
 
So the total number of good squares when we’ve put back every checker is at most <math>5 + 3(k - 1) = 3k + 2</math>. Since we know all of the <math>n^2</math> squares are good, <math>3k + 2 \geq n^2</math>, and therefore <cmath>k \geq \frac{n^2 - 2}{3}.</cmath>
 
So the total number of good squares when we’ve put back every checker is at most <math>5 + 3(k - 1) = 3k + 2</math>. Since we know all of the <math>n^2</math> squares are good, <math>3k + 2 \geq n^2</math>, and therefore <cmath>k \geq \frac{n^2 - 2}{3}.</cmath>

Revision as of 13:36, 12 May 2023

Problem

Some checkers placed on an $n\times n$ checkerboard satisfy the following conditions:

(a) every square that does not contain a checker shares a side with one that does;

(b) given any pair of squares that contain checkers, there is a sequence of squares containing checkers, starting and ending with the given squares, such that every two consecutive squares of the sequence share a side.

Prove that at least $(n^{2}-2)/3$ checkers have been placed on the board.

Solution 1

For the proof let's look at the checkers board as a graph $G$, where the checkers are the vertices and the edges are every pair of checkers sharing a side.

Define $R$ as the circuit rank of $G$(the minimum number of edges to remove from a graph to remove all its cycles).

Define $x$ as the number of checkers placed on the board.

1. From (a) for every square containing a checker there can be up to 4 distinct squares without checkers; thus, $4\times x + x = 5\times x \ge n^2$ (this is the most naive upper bound).

2. From (b) this graph has to be connected. which means that no square which contains a checker can share 4 sides with squares that doesn't contain a checker (because it has to share at least one side with a square that contains a checker).

3. Now it is possible to improve the upper bound. Since every edge in G represents a square with a checker that shares a side with a square with another checker, the new upper bound is $5\times x - \text{[sum of degrees in G]} \ge n^2.$

4. Thus, $5x - (2x - 2 + 2R) \ge n^2$ (see lemma below).

5. Thus, $3x + 2 - 2R \ge n^2.$

6. Thus, $x \ge \frac{n^2 - 2 + 2R}3,$ where $R\ge0$ and is equal to 0 if $G$ is a tree.

$\textit{Clarification/Lemma:}$ The sum of degrees of a connected graph $G = (V,E)$ is $2V -2 + 2R = 2E,$ where $R$ is the circuit rank of $G$.

$\textit{Proof.}$ $2 \times V -2$ is the sum of ranks of the spanning tree created by decircuiting the graph $G$. Since the circuit rank of $G$ is $R$, $2R$ is the sum of ranks removed from $G$. Thus, $2V -2 + 2R = 2 E.$ $\blacksquare$

Solution 2

Call a square of the checkerboard “good” if it either has a checker on it or shares a side with a square with a checker on it.

Say there are $k$ checkers on the board. We can assume WLOG that $s_1, \ldots, s_k$ is ordered in such a way that if $1 \leq i \leq k$, then $s_i$ shares a side with at least one of $s_1, \ldots, s_{i - 1}$. (If such an ordering were impossible, then some $s_i$ wouldn’t be reachable from $s_1$, violating condition (b)).

Now imagine removing all the checkers and then putting them back onto the squares $s_1, \ldots, s_k$ in that order, counting the number of good squares after each step. When we put a checker on $s_1$, we increase the number of good squares by at most 5: the square $s_1$ becomes good, and each square it shares a side with (at most 4) becomes good. When we put a checker onto $s_i$, where $i > 1$, we increase the number of good squares by at most $3$, since $s_i$ is already good, and one of its neighbors already has a checker on it.

So the total number of good squares when we’ve put back every checker is at most $5 + 3(k - 1) = 3k + 2$. Since we know all of the $n^2$ squares are good, $3k + 2 \geq n^2$, and therefore \[k \geq \frac{n^2 - 2}{3}.\]

See Also

1999 USAMO (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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