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  4. ?square peg in round hole?
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?square peg in round hole?

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Offline CZARCAR (OP)

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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?
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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
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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.

[diagram=627_0]

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   [:)]
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Offline imatfaal

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?square peg in round hole?
« Reply #3 on: 23/05/2011 19:35:18 »
Quote from: CliffordK on 23/05/2011 19:29:42

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

Phew!
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Offline CZARCAR (OP)

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?square peg in round hole?
« Reply #4 on: 23/05/2011 20:44:50 »
Quote from: CliffordK 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.

[diagram=627_0]

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?
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Offline imatfaal

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?square peg in round hole?
« Reply #5 on: 23/05/2011 21:28:36 »
Quote from: CZARCAR on 23/05/2011 20:44:50

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 »
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Offline CliffordK

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?square peg in round hole?
« Reply #6 on: 23/05/2011 22:04:36 »
Quote from: imatfaal on 23/05/2011 21:28:36
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:

[diagram=630_0]

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

[diagram=631_0]
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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
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There’s no sense in being precise when you don’t even know what you’re talking about.  John Von Neumann

At the surface, we may appear as intellects, helpful people, friendly staff or protectors of the interwebs. Deep down inside, we're all trolls. CaptainPanic @ sf.n
 

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.

 [ Invalid Attachment ]

* Square.PDF (2.09 kB - downloaded 1471 times.)
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Offline CZARCAR (OP)

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?square peg in round hole?
« Reply #9 on: 24/05/2011 13:06:10 »
Quote from: imatfaal on 23/05/2011 21:28:36
Quote from: CZARCAR on 23/05/2011 20:44:50

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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