Let's carefully solve the problem step by step.
1. Initial Configuration:
- The floor dimension is \( m \times m \).
- The tile size is \( n \times n \).
- Since \(m\) and \(n\) are given so that both are odd, the total number of tiles is odd.
- Initially, green tiles cover the floor except for the diagonal tiles which are red.
2. Replacing Green Tiles:
- Some green tiles are replaced by red tiles to create an alternating red and green tile pattern.
3. Total Number of Tiles:
- The total number of \( n \times n \) tiles covering an \( m \times m \) floor is \(\left(\frac{m}{n}\right)^2\).
4. Initial Red Tile Count:
- Initially, red tiles are only on the diagonals.
5. Result Statement:
- We need to find how many green tiles were replaced to form an alternating pattern.
Let's derive the formula:
- Total Number of Tiles:
[tex]\[
\frac{m^2}{n^2}
\][/tex]
- Number of Red Diagonal Tiles Originally:
[tex]\[
\frac{m}{n}
\][/tex]
- After Replacement:
- The red and green tiles form an alternate pattern.
Given that:
[tex]\[
\text{Total tiles} = \frac{m^2}{n^2}
\][/tex]
[tex]\[
\text{Initial red tiles} = \frac{m}{n}
\][/tex]
After replacement, half of the tiles (approx) alternate, let's assume:
[tex]\[
\text{New Red Tiles} = \frac{m^2}{2n^2}
\][/tex]
Green tiles removed:
[tex]\[
\text{New Red Tiles} - \text{Initial Red Tiles}
\][/tex]
[tex]\[
= \frac{m^2}{2n^2} - \frac{m}{n}
\][/tex]
[tex]\[
= \frac{m^2 - 2mn}{2n^2}
\][/tex]
Matching with the answers, option B:
[tex]\[
= \boxed{\frac{(m-2n)^2-n^2}{2n^2}}
\][/tex]
