[/quote]
,
Prove that
Where 
be odd integers such that 
and 
.
and 
for some integers 
and 
,
.
such that 
Prove that
Where 
Prove that
parallel to one of the coordinate axes the difference (in absolute value) between the numbers of white and red points on 
?
and 
.
be reals such that 
and 
Prove that
Let 
and ![$$ a^2+b^2+c^2\geq \frac{3}{ \sqrt[3]{2}}$$](http://latex.artofproblemsolving.com/6/2/2/62211222c148c86d65526d86dde603cb6598925f.png)
![$$ a^2+2b^2+c^2\geq 2\sqrt[3]{4} $$](http://latex.artofproblemsolving.com/3/8/3/383d38cc87c1b1ea1e17e10e4bb95bed79329786.png)
such that: 
.
such that 
then find minimal value of 
be the set of positive integers. Find all functions 
,
and taking values in 
for every positive integer 
.
Where 
be a nonconstant polynomial of degree 
of degree less than 
be positive real numbers so that 
.![\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1.
\]](http://latex.artofproblemsolving.com/7/6/e/76e794a62090c8c4b5f3a70ad4d036669418ca69.png)
Y by Adventure10, Mango247, ehuseyinyigit
Some checkers placed on an 
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
checkers have been placed on the board.
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
checkers have been placed on the board.
units available, where 
,
checkers, we have turned on at most 
checkers. Since there are 
squares on the board, we need 
or 
.
,
.
(but slightly larger due to edge effects).
checkers on the board (and 
empty squares), so from (b), the checkers form a connected graph 
.
denote the number of pairs of neighboring squares both containing checkers and 
denote the number of pairs of neighboring squares with exactly one containing a checker. Then 
.
,
.![\[n^2-f(n) \le Y\le4f(n)-2X,\]](http://latex.artofproblemsolving.com/9/c/9/9c909775e04d0a06c9508bf69e338d9faede7b30.png)
so![\[5f(n)\ge n^2+2X\ge n^2+2f(n)-2\implies f(n)\ge\frac{n^2-2}{3},\]](http://latex.artofproblemsolving.com/e/8/6/e8646f5549102e0cd19ade247788912995656147.png)
as desired.
squares, as each checker individually covers three squares, that is the two adjacent squares not covered by another checker as well as the square that the checker itself is on, while the first and last squares additionally cover another square each. Note that no square is covered twice. It follows that 
,
is the area of the shape. We can then calculate that 
.
,
checkers, we have that it will take greater than 
.
and the denominator of three in the fraction. The three suggested looking at how many new squares each new checker in a "snake" of checkers would cover, while the 
of checkers where the 
edges of the tree are given by orthogonal adjacency. By condition (a) we have ![\[ \sum_{v \in V} (4-\deg v) \ge n^2 - |V| \]](http://latex.artofproblemsolving.com/7/5/7/757abf146f7425759345f79a24ca6876bac7a466.png)
and since 
we get ![\[ 4|V| - \left( 2|V|-2 \right) \ge n^2 - |V| \]](http://latex.artofproblemsolving.com/a/7/7/a77a86e75e71be146e7f0a600e670f434bcb9245.png)
which implies 
.
and we are done
squares, and each successive one must add only 
at most
,
(the number of empty squares). On the other hand, the number of pairs of adjacent squares, both filled, is at least 
.
pairs of adjacent squares, at least one of which is filled. Therefore, we have the inequality 
as required.
as each new block after the original one adds at most 
to the perimeter of the filled squares. The minimum perimeter is 
,
![\[1+\frac{n^2-5}{3},\]](http://latex.artofproblemsolving.com/3/2/9/32993f5563d49695dcb6fc04e90ef0a94f339db2.png)
or ![\[ 3k+2 \leq n^2 \implies k \geq \frac{n^2-2}{3} \]](http://latex.artofproblemsolving.com/5/a/a/5aa15265602953636076d6efac2ea40b7d77e8e8.png)
completing the proof. 
squares because of the first condition. Thus,![\[3k+2 \ge n^2 \iff k \ge \frac{n^2-2}{3}. \ \square\]](http://latex.artofproblemsolving.com/c/f/6/cf6c0428fed9892d6cde124c04ed14eea6e629f1.png)
be the number of white cells and let 
be the number of black cells. Then each white cell has at least one black neighbor.
edges. Therefore, the black cells can satisfy at most 
white cells, so 
.
,
be the set of squares with and without a checker, respectively. A upper bound for 
is ![\[\sum_{a \in A} (4-\deg a) = 4|A| - \sum_{a \in A} (\deg a)\]](http://latex.artofproblemsolving.com/4/4/e/44ee2c2b32119a0dcbb221a1fddeddf0b81829d4.png)
.
edges, and thus![\[4|A| - \sum_{a \in A} (\deg a) \ge 4|A| - 2(|A|-1) \ge |B| = n^2 - |A| \implies |A| \ge \frac{n^2-2}{3}. \quad \blacksquare\]](http://latex.artofproblemsolving.com/b/d/b/bdbada60cbe2c0bbcb8e80b2407e32607acd4ad1.png)
be the number of checkers next to square 
,
be the total number of checkers. The checkers must be connected, so the first time we add a checker nothing happens with the checker, but after adding a checker next to it, the count of 
goes up by 
,![\[\sum_{i=1}^{n^2}C_i\geq 2a-2+n^2-a = n^2-2+a.\]](http://latex.artofproblemsolving.com/e/9/2/e92cdc34c0285c1459263edf431385afd95a2a6d.png)
However, notice that each checker is counted at most four times, because it has at most 
surrounding squares. This gives us ![\[4a\geq\sum_{i=1}^{n^2}C_i\geq n^2-2+a,\]](http://latex.artofproblemsolving.com/9/7/3/973228b48b70e04a83dc386505fb20611ce40a87.png)
giving 
,
as the sum of the number of neighbors of every square with a checker. On one hand we have 
,
checkers have at most 
.
.
.
,
.
be the number of black neighbors of 
.
.
be a graph with its vertices as the ![\[\sum_{s\text{ black}} f(s) = \sum_{v \in V} \text{deg}(v) = 2|E| \geq 2k - 2,\]](http://latex.artofproblemsolving.com/0/c/c/0cc9a9977a334ab7f84b9e641a056a16fbb88d65.png)
where the last inequality follows since 
.
is counted 
.![\[4k \geq n^2 + k - 2 \implies k \geq \frac{n^2 - 2}{3}\]](http://latex.artofproblemsolving.com/b/7/5/b7574b0b72819d89a18580af6cb92b9e6b511ac3.png)
as desired. 
,
is never divisible by 
denote the set of edges in this graph.
because each square with a checker on it corresponding to the vertex 
in the graph has 
neighboring squares without checkers on them. Also, a lower-bound for this quality is the number of squares without checkers which is 
because each square without a checker on it must be adjacent to at least one square with a checker on it (condition (a)). Thus, it follows that 
so we can apply the Handshaking Lemma to get
Condition (b) is equivalent to the graph being connected, so we know that 
Hence, it follows that
Therefore, we get
as desired. 
squares with checkers in them. It is clear that all 
squares we need one of their sides to touch a square with a checker. Therefore, the perimeter of our 
(this is from the second condition and that some squares touch sides of the grid). Therefore 
done
If 
we have at least 
empty squares adjacent to a filled square. Thus 
QED
or 
rectangle such that it contains exactly one checker. Count 
.
.
,
.
