document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=200809>Question 200809</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many distinct triangles have a perimeter of 15 units and integral length sides? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #151053 by </B><A HREF=/tutors/aboutme.mpl?userid=vleith><B>vleith(2983)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=vleith><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=151053\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I'm sure there must be an algebraic way to do this, but for this one, I would just grind through them.\r<BR>\n" );
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document.write( "Start with an equilateral triangle 5,5,5<BR>\n" );
document.write( "Now take one unit from one side and add it another side, 4,6,5 or 4.5,6. which leaves unique one of 4,6,5<BR>\n" );
document.write( "Again 3,7,5 or 3,6,6<BR>\n" );
document.write( "Again 2,8,5 or 2,7,6 or 2,7,6 or 2,7,6 which leaves just 2,8,5 and 2,7,6<BR>\n" );
document.write( "Again 1,9,5 or 1,8,6 or 1,7,7\r<BR>\n" );
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document.write( "So we have 7 possibilites. Look at the possible answers and make sure that for each, the sum of any two sides is greater than the remaining side.\r<BR>\n" );
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document.write( "1,9,5 won't work since 1+5 <9<BR>\n" );
document.write( "Likewise 2,8,5 wont work since 2+5<8\r<BR>\n" );
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document.write( "So I say we have 5 that work </SPAN>\n" );
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