document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1089830>Question 1089830</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Let a, b and c be sides of a triangle in arithmetic progression in this order. Show that cos((A-C)/2)/cos((A+C)/2)=2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #704388 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=704388\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> There's almost certainly a more elegant way than this, but here goes:\r<BR>\n" );
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document.write( "WLOG, assume a = 1-x, b = x, and c = 1+x. Let I be the incenter, and r be the inradius. We have semiperimeter 3/2. Additionally, if we let D, E, and F be the points of tangency with the incircle and sides BC, CA, AB respectively, we have that <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large BD = BF = \frac{1}{2}\">, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large CD = CE = \frac{1}{2} - x\">, and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large AE = AF = \frac{1}{2} + x\">. Finding the area of triangle ABC using A = rs and Heron's formula we have:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large [ABC] = \frac{3}{2}r = \sqrt{\frac{3}{2} \cdot \frac{1}{2} \cdot (\frac{1}{2} - x)(\frac{1}{2} + x)}\">\r<BR>\n" );
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document.write( "Squaring both sides and simplifying gives <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large 3r^2 = \frac{1}{4} - x^2\">. (*)\r<BR>\n" );
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document.write( "Now we are trying to show <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large \cos \frac{A-C}{2} = 2 \cos \frac{A+C}{2}\">. Using sum and difference formulas we have\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large \cos \frac{A-C}{2} = \cos \frac{A}{2} \cos \frac{C}{2} + \sin \frac{A}{2} \sin \frac{C}{2}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large = \frac{\frac{1}{4} - x^2}{AI \cdot IC} + \frac{r^2}{AI \cdot IC}\">\r<BR>\n" );
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document.write( "and also:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large \cos \frac{A+C}{2} = \cos \frac{A}{2} \cos \frac{C}{2} - \sin \frac{A}{2} \sin \frac{C}{2}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large = \frac{\frac{1}{4} - x^2}{AI \cdot IC} - \frac{r^2}{AI \cdot IC}\">\r<BR>\n" );
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document.write( "It follows that <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large \cos \frac{A-C}{2} = 2 \cos \frac{A+C}{2}\"> if and only if <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large \frac{1}{4} - x^2 + r^2 = 2(\frac{1}{4} - x^2 - r^2)\">. This simplifies to <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large 3r^2 = \frac{1}{4} - x^2\"> which we showed is true in (*). Therefore <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\large \frac{\cos \frac{A-C}{2}}{\cos \frac{A+C}{2}} = 2\">. </SPAN>\n" );
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