由 aaddfg 於 星期六 十二月 09, 2006 11:34 pm
#ed_op#DIV#ed_cl#9a/11應該是對的...我也是這個答案#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#設AP=k,BQ=2k,CR=3k#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#ΔAPR : ΔABC=ak-k#ed_op#SUP#ed_cl#2 #ed_op#/SUP#ed_cl#: a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#ΔBQR : ΔBCA=2ak-2k#ed_op#SUP#ed_cl#2 #ed_op#/SUP#ed_cl#: a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#ΔCRQ : ΔCAB=3ak-3k#ed_op#SUP#ed_cl#2 #ed_op#/SUP#ed_cl#: a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#(頂角相同,三角形面積比=夾邊乘積比)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#ΔPQR : ΔABC=(√3)/4*(a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-6ak+11k#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#) : (√3)/4*a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#ΔPQR=(11√3)/4*(k#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#-6a/11+9a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#/121)-9a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#/121*(11√3)/4+(√3)/4*a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SUP#ed_cl# #ed_op#/SUP#ed_cl#=(11√3)/4*(k-3a/11)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+√3a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#/22≥√3a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#/22#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#意謂當k=3a/11時有最小值√3a#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#/22#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#此時CR=9a/11#ed_op#/DIV#ed_cl#