document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1153218>Question 1153218</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A sphere of radius 1 and a sphere of radius 2 are inscribed in a right circular cone so that the larger sphere is on the base and the smaller sphere is right above it. Find the volume of the cone. <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 #775413 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=775413\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Picture slicing the cone down the middle to get a 2-dimensional picture.  The picture is an isosceles triangle with two inscribed circles of radius 2 and 1.<br><BR>\n" );
document.write( "Let AB be the base of the isosceles triangle and C be the vertex.<BR>\n" );
document.write( "Let O be the center of the base of the cone, P the center of the circle of radius 2, and Q the center of the circle of radius 1.<BR>\n" );
document.write( "Let R be the point of tangency of the two circles, and let S be the \"top\" of the smaller circle.<BR>\n" );
document.write( "Finally, let T be the point of tangency of the larger circle and side BC of the triangle, and let U be the point of tangency of the smaller circle with side BC of the triangle.<br><BR>\n" );
document.write( "The volume of the cone is one-third base times height; we need to determine the measures of OB (radius of the base) and CO (height).<br><BR>\n" );
document.write( "Triangles CUQ, CTP, and COB are all similar, because they are all right triangles that share angle OCB.<br><BR>\n" );
document.write( "Similar triangles CUQ and CTP are similar with ratio 1:2 (the radii of the two spheres, UQ and TP).  That makes CQ half of CP; since PQ=3 (the sum of the two radii), CQ is also 3, and then the height of the cone is 3+3+2 = 8.<br><BR>\n" );
document.write( "To find the radius of the cone, we can use similar triangles CUQ and COB.<br><BR>\n" );
document.write( "QU=1 and CQ=3, so CU = 2*sqrt(2).  Then<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=OB%2FOC+=+QU%2FCU ALIGN=MIDDLE  ALT=\"OB%2FOC+=+QU%2FCU\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=OB%2F8+=+1%2F%282%2Asqrt%282%29%29 ALIGN=MIDDLE  ALT=\"OB%2F8+=+1%2F%282%2Asqrt%282%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=OB+=+8%2F%282%2Asqrt%282%29%29+=+2%2Asqrt%282%29 ALIGN=MIDDLE  ALT=\"OB+=+8%2F%282%2Asqrt%282%29%29+=+2%2Asqrt%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "And then the volume of the cone is<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F3%29%28pi%29%28r%5E2%29%28h%29+=+%281%2F3%29%28pi%29%28%282%2Asqrt%282%29%29%5E2%29%288%29+=+%281%2F3%29%28pi%29%288%29%288%29+=+%2864%2F3%29%28pi%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
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