Intersection of concentric, unit area sphere and square

  • #1
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Main Question or Discussion Point

Where do a concentric sphere and square, both of area 1, intersect?
 

Answers and Replies

  • #2
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The shortest distance from the center on this square is 1/2 units. The radius of the sphere with a surface area of 1 is inferior to 1/2, meaning that the square and the sphere meat at every point on the circle of the same radius which is concentric and coplanar to the square.
 
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  • #3
Integral
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Do you mean square and circle? Or Sphere and cube? Area Or Volumn?
 
  • #4
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Sorry, it was getting late. I meant to say

Where do a concentric circle and square, both of area 1, intersect?
 
  • #5
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Well, where in terms of what? Interval of points on the circle? If so, assuming the sides of the square are parallel to the axes, we have to calculate on which interval does the distance between the center (here the origin) and the edges of the square is equal or greater than [tex]\frac{1}{\sqrt{\pi}}[/tex].
 
  • #6
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To recap and revise:

A square of 4 units area is concentric to a circle of 4 units area. Given that their common center falls on an origin of Cartesian coordinates and the square's corners are located at (1, 1) (1, -1) (-1,-1) (-1, 1), what coordinates describe the points at which the square and circle overlap?
 
  • #7
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Well, first, what's the radius of the circle? You can get that from the area.

Now, draw yourself a picture where you estimate where the intersections are (you can do this now, since you know the radius of the circle). Each intersection point will be on a side of the square, so that will give you one of the coordinates of the point. To get the other one, draw the line from the origin to the point and the line from the point to the other axis (the one whose coordinate you're missing). You should now be able to get the second coordinate for the point. The other seven points' coordinates should then be obvious by inspection, given the symmetry of the problem.
 
  • #8
HallsofIvy
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A circle of area 1 has radius [itex]1/(2\pi)[/itex] and so a circle of area 1 with center at the origin of a coordinate system can be represented by the equation [itex]x^2+ y^2= 1/4\pi^2[/itex].
A square of area 1 has sides of length 1 and so a square of area 1 with center ata the origin of a coordinate system can be represented as the are contained within the lines x= 1/2, y= 1/2, x= -1/2, y= -1/2.

Where does [itex]x^2+ y^2= 1/4\pi^2[/itex] intersect each of those?

Loren, I would have thought you would have been able to do a problem like this easily!
 
  • #9
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Thanks for your confidence, Walls of Ivy. Indeed I must think twice before assuming something significant.

Arrgh! Next time I will refer to paper and pencil before jumping to confusions.
 
  • #10
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HallsofIvy is clearly correct, but I find a geometric solution easier than an algebraic one. I would simply draw the right triangle with the origin at one end of the hypotenuse and the intersection point in question at the other end. One leg of the triangle has length 1 and the hypotenuse has length R, the radius of the circle. Good ol' Pythagoras gives us the other leg, which is the other coordinate of the intersection point.

Different strokes for different folks - there's nothing better about this method than the one HallsofIvy offered; it's just personal preference. Of course, you end up solving the same equations, you just get there a different way.
 
  • #11
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A circle of area 1 has radius [itex]1/(2\pi)[/itex] and so a circle of area 1 with center at the origin of a coordinate system can be represented by the equation [itex]x^2+ y^2= 1/4\pi^2[/itex].
A square of area 1 has sides of length 1 and so a square of area 1 with center ata the origin of a coordinate system can be represented as the are contained within the lines x= 1/2, y= 1/2, x= -1/2, y= -1/2.

Where does [itex]x^2+ y^2= 1/4\pi^2[/itex] intersect each of those?

Loren, I would have thought you would have been able to do a problem like this easily!
You got distracted a bit there, you were thinking of circumference instead of area. The radius is [tex]\frac{1}{\sqrt{\pi}}[/tex].
 

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