document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1189710>Question 1189710</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In the diagram attached below, Circles M and N are tangent to each other, and to Line AB and Line BC. If Angle ABC=120°, what is the ratio of the radius of Circle M to the radius of circle N\r<BR>\n" );
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document.write( "Diagram: https://imgur.com/a/4dfUycL </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #821267 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=821267\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "The tutor greenestamps has a great answer. I'll provide a (slightly) different viewpoint.\r<BR>\n" );
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document.write( "I'll use the letter notation he has set up.<BR>\n" );
document.write( "Here's what the drawing looks like with those points mentioned.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" width = \"35%\" src = \"https://i.imgur.com/Gyt7sPs.png\">\r<BR>\n" );
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document.write( "Let circle N have a radius of 1. This means EN = ND = 1.<BR>\n" );
document.write( "<font color=blue>Goal: Find the length of FM</font><BR>\n" );
document.write( "This will effectively give the ratio FM/EN, aka the ratio of the two radii<BR>\n" );
document.write( "Note how FM/EN = FM/1 = FM\r<BR>\n" );
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document.write( "As mentioned by the other tutor, the 120 degree angle ABC is bisected into two smaller 60 degree angles<BR>\n" );
document.write( "angle ABM = angle MBC = 60<BR>\n" );
document.write( "This leads to triangles BEN and BFM being 30-60-90 triangles<BR>\n" );
document.write( "The points of tangency E and F have the 90 degree angles at those locations.\r<BR>\n" );
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document.write( "The useful formulas for any 30-60-90 triangle are that<BR>\n" );
document.write( "hypotenuse = 2*(short leg)<BR>\n" );
document.write( "long leg = sqrt(3)*(short leg)<BR>\n" );
document.write( "That second formula rearranges to<BR>\n" );
document.write( "short leg = (long leg)/sqrt(3) = (long leg)*sqrt(3)/3\r<BR>\n" );
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document.write( "For triangle BEN we have EB as the short leg and EN as the long leg<BR>\n" );
document.write( "short leg = (long leg)*sqrt(3)/3<BR>\n" );
document.write( "EB = (EN)*sqrt(3)/3<BR>\n" );
document.write( "EB = (1)*sqrt(3)/3<BR>\n" );
document.write( "EB = sqrt(3)/3\r<BR>\n" );
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document.write( "And,<BR>\n" );
document.write( "hypotenuse = 2*(short leg)<BR>\n" );
document.write( "BN = 2*EB<BR>\n" );
document.write( "BN = 2*sqrt(3)/3 is the hypotenuse of triangle BEN \r<BR>\n" );
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document.write( "Let x be the radius of circle M<BR>\n" );
document.write( "FM = x<BR>\n" );
document.write( "DM = x as well since they're both radii of the same circle M<BR>\n" );
document.write( "Now focus on triangle BFM \r<BR>\n" );
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document.write( "Since it is also a 30-60-90 triangle we can say<BR>\n" );
document.write( "short leg = (long leg)*sqrt(3)/3<BR>\n" );
document.write( "BF = (FM)*sqrt(3)/3<BR>\n" );
document.write( "BF = x*sqrt(3)/3<BR>\n" );
document.write( "and also<BR>\n" );
document.write( "hypotenuse = 2*(short leg)<BR>\n" );
document.write( "BM = 2*(BF)<BR>\n" );
document.write( "BM = 2x*sqrt(3)/3\r<BR>\n" );
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document.write( "The key things to recap is that we found<BR>\n" );
document.write( "BN = 2*sqrt(3)/3<BR>\n" );
document.write( "ND = 1<BR>\n" );
document.write( "DM = x\r<BR>\n" );
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document.write( "Now break segment BM into smaller pieces<BR>\n" );
document.write( "BM = BN + ND + DM<BR>\n" );
document.write( "BM = 2*sqrt(3)/3 + 1 + x\r<BR>\n" );
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document.write( "We have<BR>\n" );
document.write( "BM = 2x*sqrt(3)/3<BR>\n" );
document.write( "and<BR>\n" );
document.write( "BM = 2*sqrt(3)/3 + 1 + x\r<BR>\n" );
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document.write( "Set those two right hand sides equal to one another. Solve for x.\r<BR>\n" );
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document.write( "2*sqrt(3)/3 + 1 + x = 2x*sqrt(3)/3<BR>\n" );
document.write( "2*sqrt(3) + 3 + 3x = 2x*sqrt(3) ............. multiply both sides by 3 to clear out the fractions<BR>\n" );
document.write( "2*sqrt(3) + 3 = 2x*sqrt(3)-3x<BR>\n" );
document.write( "2*sqrt(3) + 3 = x(2*sqrt(3)-3)<BR>\n" );
document.write( "x = (2*sqrt(3) + 3)/(2*sqrt(3)-3)\r<BR>\n" );
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document.write( "Let's multiply top and bottom of that expression by 2*sqrt(3)+3 to rationalize the denominator\r<BR>\n" );
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document.write( "I recommend using a tool like WolframAlpha to check your work.\r<BR>\n" );
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document.write( "If EN = 1, then FM = x = 7+4*sqrt(3)\r<BR>\n" );
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document.write( "Therefore, the ratio of the radius of circle M to circle N is exactly 7+4*sqrt(3)<BR>\n" );
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