document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1120495>Question 1120495</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given regular hexagon ABCDEF,, with Centre O and sides of length 12.  Let G be the midpoint of BC. Let H be the  midpoint of DE.  AH intersects EB at J and FG intersects EB at K..Find JK.<BR>\n" );
document.write( "Hint given in question ::: draw auxiliary lines HG and DA.\r<BR>\n" );
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document.write( "Thanks for any help .  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #737154 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=737154\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "I didn't see any use in the suggested auxiliary line segments HG and DA....<br><BR>\n" );
document.write( "Perhaps there is a clever way to solve the problem using them; my solution is not very difficult.<br><BR>\n" );
document.write( "Consider the regular hexagon with the center O at (0,0), with sides AB and DE horizontal; A in quadrant III, B in quadrant IV, D in quadrant I, and E in quadrant II.  Then the coordinates of the vertices of the hexagon are<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A%28-6%2C-6%2Asqrt%283%29%29 ALIGN=MIDDLE  ALT=\"A%28-6%2C-6%2Asqrt%283%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=B%286%2C6%2Asqrt%283%29%29 ALIGN=MIDDLE  ALT=\"B%286%2C6%2Asqrt%283%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=C%2812%2C0%29 ALIGN=MIDDLE  ALT=\"C%2812%2C0%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=D%286%2C6%2Asqrt%283%29%29 ALIGN=MIDDLE  ALT=\"D%286%2C6%2Asqrt%283%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=E%28-6%2C6%2Asqrt%283%29%29 ALIGN=MIDDLE  ALT=\"E%28-6%2C6%2Asqrt%283%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=F%28-12%2C0%29 ALIGN=MIDDLE  ALT=\"F%28-12%2C0%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Then the two midpoints in the problem have coordinates<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=G%289%2C-3%2Asqrt%283%29%29 ALIGN=MIDDLE  ALT=\"G%289%2C-3%2Asqrt%283%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=H%280%2C6%2Asqrt%283%29%29 ALIGN=MIDDLE  ALT=\"H%280%2C6%2Asqrt%283%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Now triangles HJE and AJB are similar, with the ratio of similarity 1:2 because of the lengths of the bases EH (length 6) and AB (length 12).  Since the height of the hexagon is 12*sqrt(3), we can determine that the y coordinate of point J is 2*sqrt(3).  (It is 1/3 of the way from side DE -- y value 6*sqrt(3) -- and side AB -- y value -6*sqrt(3).)<br><BR>\n" );
document.write( "So we can determine that the coordinates of J are <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=J%28-2%2C2%2Asqrt%283%29%29 ALIGN=MIDDLE  ALT=\"J%28-2%2C2%2Asqrt%283%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<br><BR>\n" );
document.write( "The exact same similarity exists between triangles BKG and EKF, leading us to the coordinates of K as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=K%282%2C-2%2Asqrt%283%29%29 ALIGN=MIDDLE  ALT=\"K%282%2C-2%2Asqrt%283%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<br><BR>\n" );
document.write( "Then, making JK the hypotenuse of a right triangle, the Pythagorean Theorem (or the observation that we have a 30-60-90 right triangle, with legs 4 and 4*sqrt(3)) gives us the length of JK as 8.<BR>\n" );
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