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由 skywalker 於 星期六 三月 17, 2007 10:18 pm
#ed_op#DIV#ed_cl#考慮三角形ABC與截線PFD，由Menelaus定理#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#有CP/PA*AF/FB*BD/CD=1=>PA/CP=AF/FB*BD/CD=AF/CD=(p-a)/(a-c)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#即PA/CP=(p-a)/(a-c)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#∴PA=b(p-a)/(a-c)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#P#ed_cl#∴PE=PA+AE=b(p-a)/(a-c)+p-a=(p-a)(p-c)/2(a-c)#ed_op#/P#ed_cl##ed_op#P#ed_cl#ME=1/2PE=(p-a)(p-c)/4(a-c)#ed_op#/P#ed_cl##ed_op#P#ed_cl#MA=ME-AE=(p-a)(p-c)/4(a-c)-(p-a)=(p-a)^2/4(a-c)#ed_op#/P#ed_cl##ed_op#P#ed_cl#MC=ME+EC=(p-a)(p-c)/4(a-c)+(p-c)=(p-c)^2/4(a-c)#ed_op#/P#ed_cl##ed_op#P#ed_cl# 於是MA*MC=ME^2#ed_op#/P#ed_cl##ed_op#P#ed_cl#∵ME是點M到三角形ABC內切圓之切線長#ed_op#/P#ed_cl##ed_op#P#ed_cl#∴ME^2是點M到三角形ABC外接圓的冪#ed_op#/P#ed_cl##ed_op#P#ed_cl#MA*MC=ME^2表明點M到三角形ABC外切圓與內切圓的冪相等#ed_op#/P#ed_cl##ed_op#P#ed_cl#從而點M在三角形ABC外接圓與內切圓的根軸上#ed_op#/P#ed_cl##ed_op#P#ed_cl#同理點N也在三角形外接圓與內切圓的根軸上#ed_op#/P#ed_cl##ed_op#P#ed_cl#故OI⊥MN#ed_op#/P#ed_cl#
[問題]2007中國奧林匹克
由 skywalker 於 星期三 一月 31, 2007 7:36 am
#ed_op#DIV#ed_cl##ed_op#FONT face=Verdana#ed_cl#設O,I分別為三角形的外心和內心，三角形ABC的內切圓與邊BC,CA,AB分別相#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT face=Verdana#ed_cl##ed_op#/FONT#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT face=Verdana#ed_cl#交於D,E,F，直線FD與CA相交於點P，直線DE與AB相交於點Q，點M,N分別為#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT face=Verdana#ed_cl##ed_op#/FONT#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#FONT face=Verdana#ed_cl#線段PE,QF的中點，求證OI⊥MN #ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl#