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Author Topic: ?square peg in round hole?  (Read 4351 times)

Offline CZARCAR

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?square peg in round hole?
« on: 23/05/2011 18:03:58 »
or round peg in a square hole. Which has the most area for the internal piece?


 

Offline imatfaal

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?square peg in round hole?
« Reply #1 on: 23/05/2011 19:11:27 »
If we take a square which is 2x units to a side

Area of square is   4 x^2 square units

Areas of circle inside is π x^2 square units  (radius is 1x unit)

Area of circle outside is 2π  x^2 square units (radius (√8)x/2 ie √2x units)

take ratios of filled area

circle inside a square is π/4

square inside a circle is 4/(2π)

 π/4      = 0.785398163

 4/(π.2) =  0.636619772

I am sure to have slipped up somewhere
 

Offline CliffordK

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?square peg in round hole?
« Reply #2 on: 23/05/2011 19:29:42 »
If you are thinking of a square that precisely fits into a circle.
Or a circle that precisely fits into a square.



Then you can calculate the areas of each.

Using r as the radius of the circle.

Square in Circle:
Area of circle:
πr2

Area of the square.  =  4 x triangle with base&height equal to r
4 x r2/2 = 2r2

Difference is:
πr2 - 2r2 = (π-2)r2 = (3.14-2)r2 = 1.14r2

Circle in Square:
Area of circle:
πr2

Area of square = 4 x squares with base&height equal to r
4r2

Difference is:
4r2-πr2 = (4-π)r2 = (4-3.14)r2 = 0.86r2

But..
As the previous poster (imatfaal) mentioned, taking differences isn't adequate as the areas are different.

So...
Taking a ratio of the inner to the outer is probably a better way to look at it.

Square in Circle:
2r2/πr2 = 2/π = 2/3.14 = 0.64

Circle in Square:
πr2/4r2 = π/4 = 3.14/4 = 0.785

So the circle in the square actually wins with more area in the square vs the area in the outer container.

And, it looks like imatfaal and I came up with the same answer   :)
 

Offline imatfaal

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?square peg in round hole?
« Reply #3 on: 23/05/2011 19:35:18 »

And, it looks like imatfaal and I came up with the same answer   :)

Phew!
 

Offline CZARCAR

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?square peg in round hole?
« Reply #4 on: 23/05/2011 20:44:50 »
If you are thinking of a square that precisely fits into a circle.
Or a circle that precisely fits into a square.



Then you can calculate the areas of each.

Using r as the radius of the circle.

Square in Circle:
Area of circle:
πr2

Area of the square.  =  4 x triangle with base&height equal to r
4 x r2/2 = 2r2

Difference is:
πr2 - 2r2 = (π-2)r2 = (3.14-2)r2 = 1.14r2

Circle in Square:
Area of circle:
πr2

Area of square = 4 x squares with base&height equal to r
4r2

Difference is:
4r2-πr2 = (4-π)r2 = (4-3.14)r2 = 0.86r2

But..
As the previous poster (imatfaal) mentioned, taking differences isn't adequate as the areas are different.

So...
Taking a ratio of the inner to the outer is probably a better way to look at it.

Square in Circle:
2r2/πr2 = 2/π = 2/3.14 = 0.64

Circle in Square:
πr2/4r2 = π/4 = 3.14/4 = 0.785

So the circle in the square actually wins with more area in the square vs the area in the outer container.

And, it looks like imatfaal and I came up with the same answer   :)
great piece i gotta read again but more urgently , i should try to beat a square peg in a round hole instead of a round peg in a square hole?
 

Offline imatfaal

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?square peg in round hole?
« Reply #5 on: 23/05/2011 21:28:36 »

great piece i gotta read again but more urgently , i should try to beat a square peg in a round hole instead of a round peg in a square hole?

in the time you spent to quote and write above - why not read either of the answers given? 

edited to fix quotes
« Last Edit: 23/05/2011 22:15:55 by imatfaal »
 

Offline CliffordK

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?square peg in round hole?
« Reply #6 on: 23/05/2011 22:04:36 »
in the time you spent to quote and write above - why not read either of the answers given? 
Reading back, it is interesting that we had two different approaches to the same problem.

I used r=radius of circle being the same in both problems...  essentially:



You, on the other hand used the 2x = side of the square being the same, essentially calculating:

 

Offline imatfaal

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?square peg in round hole?
« Reply #7 on: 23/05/2011 22:22:34 »
Cliff - I was about to respond, yes but mine is easier; but it isn't really is it?  The only tricky part is that I work out the hypotenuse from two sides and you work out the two sides from the hypotenuse (but you even managed to avoid that - which was nice).  And you went to the trouble of diagrams
 

Offline Geezer

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?square peg in round hole?
« Reply #8 on: 24/05/2011 00:55:53 »
This is a pictorial method of comparing the two ratios.



 

Offline CZARCAR

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?square peg in round hole?
« Reply #9 on: 24/05/2011 13:06:10 »

great piece i gotta read again but more urgently , i should try to beat a square peg in a round hole instead of a round peg in a square hole?

in the time you spent to quote and write above - why not read either of the answers given? 

edited to fix quotes

hard to follow the math & diagrams but simply= both the circle & square's area will be defined by the formation of 2 inner triangles inside both the inner circle & the inner square. The 2 triangles will define the total area of the square. The 2 triangles within the inner circle will define the area of the inner circle but will have extra area that lies outside the 2 triangles.
« Last Edit: 24/05/2011 13:07:48 by CZARCAR »
 

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?square peg in round hole?
« Reply #9 on: 24/05/2011 13:06:10 »

 

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