Lesson A problem on three spheres

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A problem on three spheres


Problem 1

The center of each of three spheres of radius  R  lies in the surfaces
of the other two.  Pass a plane containing the centers of the spheres.
Find the area common to the three great circles cut from the spheres by this plane.

Solution 

The centers of the three great circles form an equilateral triangle.
The sides of this triangle are of the length R, since R is the radius of each
of the three spheres.


The area of this triangle is  a = %28sqrt%283%29%2F4%29%2AR%5E2.


But the common intersection area is  highlight%28highlight%28WIDER%29%29 than this triangle.
The common area is formed as the intersection of three  highlight%28highlight%28sectors%29%29 of the great circles,
each sector with the central angle of 60°.


So, the common area can be described as the equilateral triangle with the sides length of R
highlight%28highlight%28PLUS%29%29 three adjacent circular  SEGMENTS  adjacent to each side of this triangle.


The area of each such a segment is  b = %281%2F6%29%2Api%2AR%5E2 - %28sqrt%283%29%2F4%29%2AR%5E2.


The area of the three such segments is  3b = %28pi%2F2%29%2AR%5E2 - %28%283%2Asqrt%283%29%29%2F4%29%2AR%5E2.


Therefore, the total common area of the intersection of the three great circles is


    Area = a + 3b = %28sqrt%283%29%2F4%29%2AR%5E2 + %28pi%2F2%29%2AR%5E2 - %28%283%2Asqrt%283%29%29%2F4%29%2AR%5E2 = 

                  = %28pi%2F2%29%2AR%5E2 - %28%282%2Asqrt%283%29%29%2F4%29%2AR%5E2 = %28pi%2F2%29%2AR%5E2 - %28%28sqrt%283%29%29%2F2%29%2AR%5E2 =

                  = %28%28pi%2F2%29+-+sqrt%283%29%2F2%29%2AR%5E2.


ANSWER.  The common area of intersection of the three great circles is %28%28pi%2F2%29+-+sqrt%283%29%2F2%29%2AR%5E2,

         or about  0.6998*R^2,  approximately.


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    - OVERVIEW of my additional lessons on miscellaneous advanced Geometry problems

To navigate over all topics/lessons of the Online Geometry Textbook use this file/link  GEOMETRY - YOUR ONLINE TEXTBOOK.

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