document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1181730>Question 1181730</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 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.<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #849973 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=849973\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Here's how to solve this problem:\r<BR>\n" );
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document.write( "1. **Visualize the Setup:** Imagine three spheres, each with its center on the surface of the other two.  The plane containing the centers cuts each sphere along a great circle (a circle with radius R).  We're looking for the area where all three great circles overlap.\r<BR>\n" );
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document.write( "2. **The Triangle:** The centers of the three spheres form an equilateral triangle.  Since each center is on the surface of the other two spheres, and all spheres have the same radius R, the side length of this equilateral triangle is 2R.\r<BR>\n" );
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document.write( "3. **The Great Circles:** The plane cutting through the centers creates three great circles. Because the centers form an equilateral triangle, these great circles intersect each other at 60-degree angles.\r<BR>\n" );
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document.write( "4. **The Common Area:** The area common to all three great circles forms an equilateral triangle in the center. To find the area of this triangle, we need to know its side length.\r<BR>\n" );
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document.write( "5. **Finding the Side Length:** Let's call the side length of the central equilateral triangle 's'.  Consider one of the 60-degree sectors of a great circle. The central equilateral triangle cuts off 1/3 of each 60-degree angle, or 20 degrees. The side length of the equilateral triangle is related to the radius R by the law of cosines.<BR>\n" );
document.write( "  s² = R² + R² - 2R²cos(60°)<BR>\n" );
document.write( "  s² = 2R² - 2R²(1/2) = R²<BR>\n" );
document.write( "  s = R\r<BR>\n" );
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document.write( "6. **Area of the Central Triangle:** The area of an equilateral triangle with side length 's' is given by (s²√3)/4.  In our case, s = R.\r<BR>\n" );
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document.write( "Area = (R²√3)/4\r<BR>\n" );
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document.write( "Therefore, the area common to the three great circles is (R²√3)/4.<BR>\n" );
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