Question

锐角△ABC中,AD,BE,CF相交于点P,若AP+BP+CP=6,设PD=x,PE=y,PF=z,且xy+yz+xz=

Answer 1

应该是AP=BP=CP=6吧如图：∵S△PBC=1 /2 PM•BC,S△ABC=1 /2 AN•BC,

∴S△PBC/ S△ABC =PM /AN =PD/ AD =x/ x+6 ,

同理：S△PAC/ S△ABC =y /y+6 ,S△PAB /S△ABC =z/ z+6 ,

∵S△ABC=S△PBC+S△PCA+S△PAB,

∴x /(x+6) +y/( y+6) +z /(z+6) =1．

即1-6 /(x+6) +1-6 /(y+6) +1-6 /(z+6) =1,

∴3 /(x+6 )+3 /(y+6) +3/( z+6) =1,

∴3（yz+zx+xy）+36（x+y+z）+324

=xyz+6（xy+yz+zx）+36（x+y+z）+216,

∴xy+yz+zx=28．

∴xyz=108-3（xy+yz+zx）=24．