Let R 
 be the set of all rectangles centered at the origin and with perimeter 1 
 (the center of a rectangle is the intersection point of its two diagonals). Let S 
 be a region that contains all of the rectangles in R 
 (region A 
 contains another region B 
 if B is completely inside of  A). find the area of 
S which has the form pi multiplied by a , where a 
 is a real number. Find a to the power of minus 1 .

Let \( R \) be the set of all rectangles centered at the origin with perimeter 1, and \( S \) be the region that contains all the rectangles in \( R \). We are tasked with finding the area of \( S \), which is of the form \( \pi \cdot a \), and ultimately computing \( a^{-1} \).

### Step 1: Parametrizing the rectangles
The perimeter of a rectangle is given by:
\[
2 \cdot (\text{length} + \text{width}) = 1
\]
Let the length of the rectangle be \( 2x \) and the width be \( 2y \). The factor of 2 is introduced to simplify calculations later, as the rectangle is symmetric about the origin.

Thus, the perimeter constraint becomes:
\[
2 \cdot (2x + 2y) = 1 \quad \Rightarrow \quad x + y = \frac{1}{2}
\]
This gives us a relation between \( x \) and \( y \).

### Step 2: Equation of the boundary of \( S \)
Each rectangle is centered at the origin, so we need to determine the maximum extents of these rectangles. Since the center is at the origin, the vertices of any rectangle are symmetric about both the x-axis and y-axis.

The coordinates of the vertices of the rectangle would be \( (\pm x, \pm y) \). The set of all possible rectangles is bounded by the largest possible values of \( x \) and \( y \) that satisfy the perimeter condition.

Since \( x + y = \frac{1}{2} \), the maximum value of \( x \) is \( \frac{1}{2} \) when \( y = 0 \), and the maximum value of \( y \) is \( \frac{1}{2} \) when \( x = 0 \). Therefore, the largest rectangle is aligned along the axes and forms a square with vertices at \( \left( \pm \frac{1}{2}, 0 \right) \) and \( \left( 0, \pm \frac{1}{2} \right) \).

### Step 3: Shape of the region \( S \)
The region \( S \) that contains all the rectangles is actually an ellipse. This is because, as the rectangle rotates and changes shape, its vertices trace out an elliptical region centered at the origin.

The equation of this ellipse can be determined by considering the maximum extents of the rectangles. The semi-major axis of the ellipse corresponds to the maximum value of \( x \), which is \( \frac{1}{2} \), and the semi-minor axis corresponds to the maximum value of \( y \), which is also \( \frac{1}{2} \). Thus, the equation of the ellipse is:
\[
\frac{x^2}{\left( \frac{1}{2} \right)^2} + \frac{y^2}{\left( \frac{1}{2} \right)^2} = 1 \quad \Rightarrow \quad 4x^2 + 4y^2 = 1 \quad \Rightarrow \quad x^2 + y^2 = \frac{1}{4}
\]
This represents a circle with radius \( \frac{1}{2} \).

### Step 4: Area of the region \( S \)
The area of the region \( S \), which is a circle of radius \( \frac{1}{2} \), is given by:
\[
\text{Area of } S = \pi \cdot \left( \frac{1}{2} \right)^2 = \pi \cdot \frac{1}{4}
\]
Thus, the area is of the form \( \pi \cdot a \), where \( a = \frac{1}{4} \).

### Step 5: Finding \( a^{-1} \)
Now, we compute \( a^{-1} \):
\[
a^{-1} = \left( \frac{1}{4} \right)^{-1} = 4
\]

Thus, \( a^{-1} = 4 \).

---

Would you like further clarification on any part of the solution?

### 5 Related Questions:
1. What is the general formula for the area of an ellipse?
2. How does the orientation of the rectangle affect the shape of the region \( S \)?
3. Why does the set of rectangles form an elliptical region?
4. How would the problem change if the perimeter constraint was different?
5. Can we generalize this to rectangles centered at points other than the origin?

### Tip:
When solving geometry problems involving shapes, sketching a diagram can help visualize the relationships between the different parts of the problem.

Let R be the set of all rectangles centered at the origin and with perimeter 1. Let S be a region that contains all of the rectangles in R. Find the area of S, which has the form pi multiplied by a, where a is a real number. Find a to the power of minus 1.

Explore how to find the area of a region containing all rectangles with a perimeter of 1 centered at the origin. Using geometric principles, this problem leads to the calculation of an elliptical area, providing insight into key concepts like perimeter and area of shapes. Suitable for students in Grades 10-12.

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