Comparing a rectangle to a circle

[BMAD]

Assume we have a circle and a rectangle.  The circumference of the circle is the same as the perimeter of the rectangle and the areas of each shape is also the same.  Write the ratio of the circle's circumference to the diagonal of the rectangle as a function in terms of the width of the rectangle.

Edited April 9, 2018 by BMAD

[bonanova]

Spoiler

Circle: Area = pi r²  Circumference = 2 pi r  
Rectangle: Area = ab = pi r²  Perimeter = 2 (a + b) = 2 pi r  Diagonal = d = sqrt (a² + b²)^1/2  Width = b.

2 (a + b) = 2 pi r   =>   (a + b) = pi r   =>   (a + b)² = pi² r².

[1]  (a² + 2ab +  b²) = pi² r².

ab = pi r²   =>

[2]  2ab = 2 pi r²

[2] - [1]   (a² + b²) = pi² r² - 2 pi r² = pi r² (pi - 2) = ab (pi - 2)  =>

[3]  a² - (pi - 2) b a +  b² = 0.

a = { (pi - 2) b +/- [ (pi - 2)² b² - 4b⁴ ]^1/2 } / 2 = f (b)

Ratio of circumference to diagonal = g.

Write g in terms of rectangle width b.

g = 2 pi r / d = 2 (a + b) / ( a² + b² )^1/2

g(b) = 2 (  f (b) + b ) / (  f² (b) + b² )^1/2

 

 

[BMAD]

On 4/14/2018 at 3:11 AM, bonanova said:

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Circle: Area = pi r²  Circumference = 2 pi r  
Rectangle: Area = ab = pi r²  Perimeter = 2 (a + b) = 2 pi r  Diagonal = d = sqrt (a² + b²)^1/2  Width = b.

2 (a + b) = 2 pi r   =>   (a + b) = pi r   =>   (a + b)² = pi² r².

[1]  (a² + 2ab +  b²) = pi² r².

ab = pi r²   =>

[2]  2ab = 2 pi r²

[2] - [1]   (a² + b²) = pi² r² - 2 pi r² = pi r² (pi - 2) = ab (pi - 2)  =>

[3]  a² - (pi - 2) b a +  b² = 0.

a = { (pi - 2) b +/- [ (pi - 2)² b² - 4b⁴ ]^1/2 } / 2 = f (b)

Ratio of circumference to diagonal = g.

Write g in terms of rectangle width b.

g = 2 pi r / d = 2 (a + b) / ( a² + b² )^1/2

g(b) = 2 (  f (b) + b ) / (  f² (b) + b² )^1/2

 

 

excellent work as always bonanova but should 

Spoiler

{ (pi - 2) b +/- [ (pi - 2)² b² - 4b⁴ ]^1/2 } / 2 actually be { (pi - 2) b +/- [ (pi - 2)² b² - 4b² ]^1/2 } / 2

 

[bonanova]

7 minutes ago, BMAD said:

excellent work as always bonanova but should 

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{ (pi - 2) b +/- [ (pi - 2)² b² - 4b⁴ ]^1/2 } / 2 actually be { (pi - 2) b +/- [ (pi - 2)² b² - 4b² ]^1/2 } / 2

 

It certainly should. It clearly fails a units check. Good catch.

[bonanova]

But actually ...

Spoiler

If Rectangle (a,b) and Circle (r=1) have the same area, then ab = pi.
If they also have the same perimeter, a+b = pi.
Substituting a=pi/b into the second equation leads to

b² - pi b + pi = 0

which has no real solutions.
So the premise of the OP is not true.

That's because circles maximize area for a given perimeter.
An equal-perimeter rectangle must have less area.

The most efficient rectangle is a square, with side pi/2 and area pi²/4.
That's (pi/4), or about 78%, of the circle's area.

Good one. You had me going for a few days.

 

[BMAD]