由 qeypour 於 星期日 八月 05, 2007 5:32 pm
#ed_op#DIV#ed_cl#建立坐標系A(0,0),B(2s,2t),C(2k,0)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#則A'(s+t,t-s),B'(k,k),C'(k+s-t,s-k+t)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#由A'B'C'共線得#ed_op#FONT color=#bf00bf#ed_cl#t^2+k^2-2kt+s^2-sk=0#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#tanA=t/s,tanC=t/(k-s)代入tanA+tanB+tanC=tanA*tanB*tanC#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#可得tanB=kt/(t^2-sk+s^2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#cotA+cotB+cotC=s/t+(k-s)/t+(t^2-sk+s^2)/kt#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#=(#ed_op#FONT color=#ff00ff#ed_cl#k^2+t^2#ed_op#/FONT#ed_cl#-sk+s^2)/kt=(#ed_op#FONT color=#bf00bf#ed_cl#2kt-s^2+sk#ed_op#/FONT#ed_cl##ed_op#FONT color=#000000#ed_cl#-sk+s^2)/kt=2#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl#