Consider a square of side length 5. Place four equal circles in the corners of the square, each circle touching the two adjacent sides of the square. Then, add a fifth circle inside the square such that it touches all four corner circles but does not overlap with them. What is the area of the inner circle?

Let's break down the problem step by step.

### 1. Square and Circles Configuration
You are given a square with side length \( 5 \). Four equal circles are placed in the corners of the square, each touching two adjacent sides. Additionally, a fifth circle is placed inside the square, touching all four corner circles but not overlapping with them.

### 2. Radius of Corner Circles
Let \( r \) be the radius of each of the four corner circles. Since each circle touches two adjacent sides of the square, the distance from the center of a corner circle to the corner of the square is equal to \( r \). Thus, the center of each corner circle is at a distance \( r \) from both adjacent sides.

The diagonal distance between the centers of two corner circles along a side of the square is equal to \( 5 - 2r \), because the total side length is \( 5 \), and each circle reduces this length by \( r \) on either end.

### 3. Fifth Circle Configuration
Let \( R \) be the radius of the inner circle. The inner circle touches all four corner circles. This means the distance between the center of the inner circle and the center of any corner circle is \( r + R \).

### 4. Geometric Relationship
The diagonal distance between the centers of two corner circles that are along the diagonal of the square is equal to \( 5\sqrt{2} - 2r\sqrt{2} \), which simplifies to:

\[
\text{Diagonal distance} = 5\sqrt{2} - 2r
\]

Now, the distance between the centers of two diagonally opposite corner circles is also equal to \( 2(r + R) \) because they are each separated by two radii of the circles.

Equating these two expressions gives:

\[
2(r + R) = 5\sqrt{2} - 2r
\]

### 5. Solving for \( R \)
Rearrange the equation to isolate \( R \):

\[
2R = 5\sqrt{2} - 4r
\]

Now we need to find the value of \( r \). From the arrangement of the circles along the side of the square:

\[
2r = 5 - 2r
\]

Thus, \( r = 1 \).

Substitute \( r = 1 \) into the equation for \( R \):

\[
2R = 5\sqrt{2} - 4
\]

\[
R = \frac{5\sqrt{2} - 4}{2}
\]

### 6. Area of the Inner Circle
The area of the inner circle is given by \( \pi R^2 \). Substituting \( R = \frac{5\sqrt{2} - 4}{2} \):

\[
A = \pi \left( \frac{5\sqrt{2} - 4}{2} \right)^2
\]

This simplifies to the exact area of the inner circle.

Would you like more detailed steps on how to simplify this expression, or any further clarifications?

Here are 5 related questions to deepen your understanding:
1. How would the arrangement change if the square had a different side length?
2. What happens to the area of the inner circle if the radius of the corner circles increases?
3. How could this configuration be extended to a cube with spherical objects?
4. How does the geometry change if the circles are not touching the sides of the square?
5. What would happen if a sixth circle is added inside the square, touching the inner circle?

**Tip:** In geometry problems involving circles inside polygons, symmetries often help simplify the relationships between different components.

Solve the geometric problem of finding the area of an inner circle that touches four corner circles within a square of side length 5. This problem involves concepts like the Pythagorean theorem, circle properties, and algebraic relationships. Ideal for students in Grades 10-12.

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