YLL討論網

#ed_op#DIV#ed_cl#抱歉!英文不好#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#還是用中文寫吧!#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2006.7#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#先證f是1-1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#若取相異兩點A,B而它們有相同的像f(A)=f(B)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#再選取C,D使得p(A,C,D)≠p(B,C,D)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#但是顯然p(f(A),f(C),f(D))=p(f(B),f(C),f(D))得到矛盾#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#故f是1-1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#接下來我有兩種寫法,我都寫出來聽聽打大家的意見#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#<法一>#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#a) claim f(A)f(B)≤AB#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#若 f(A)f(B)>AB#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#在線段AB上選一點C#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#AB+BC+AC=f(A)f(B)+f(B)f(C)+f(A)f(C)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(B)f(C)+f(A)f(C)<BC+AC=AB<f(A)f(B)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#這與三角不等式矛盾#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#b) claim f(A)f(B)≥AB#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#既然f是1-1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f就有反函數f#ed_op#SUP#ed_cl#-1#ed_op#/SUP#ed_cl#且與f有相同的性質#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#故仿a可得證#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#於是f(A)f(B)=AB#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f是保距變換#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#<法二>#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#若f(A)f(B)≠AB#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#對於線段AB之間的點C#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f(C)就不會在線段f(A)f(B)上#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#而在以f(A),f(B)為焦點的橢圓上#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#於是任取C,D在線段AB上#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#對於線段CD上的點會在以f(C),f(D)為焦點的橢圓上#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#但兩個橢圓最多只有四個交點#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#而CD之間有無限多個點#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f又是1-1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#這不可能#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#故f(A)f(B)=AB#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#f是保距變換#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#*************************************#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#之前兩題也是2006的#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1.2.3.4比較簡單就不寫過程了#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(1) 1/2 (2) (2,2,2,3) (3) 型如k#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#,k#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+k,k#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+2k (4) 1/216#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#至於8.9還沒想出來#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#唉#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#考這麼難#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#分明是要挖優秀的人才嗎!#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#今年我們學校還鼓勵大家報名#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#去接受挫折嗎?#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#SUP#ed_cl##ed_op#/SUP#ed_cl# #ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl#5.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#if some one of a,b,c is greater then 1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#it is trivial#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#then consider a,b,c all less then 1#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#since  a+b+c+(1-a)+(1-b)+(1-c)=3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(3/6)^6≥a(1-b)b(1-c)c(1-a)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#i.e. a(1-b)b(1-c)c(1-a)≤1/64#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#so at least one of  a(1-b),b(1-c),c(1-a) is lower or equal then 1/4#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#6.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#let the vertices are (0,0),(a,b),(c,d) where a,b,c,d are integral numbers#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#(c+di)/(a+bi)=(1+i√3)/2 or (1-i√3)/2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#c=(a-b√3)/2,d=(b+a√3)/2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#it is possible only at a=b=c=d=0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl#

#ed_op#DIV#ed_cl##ed_op#A href="http://www.tcfsh.tc.edu.tw/cou/France/test_2005.pdf"#ed_cl#http://www.tcfsh.tc.edu.tw/cou/France/test_2005.pdf#ed_op#/A#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#A href="http://www.tcfsh.tc.edu.tw/cou/France/MPSI.pdf"#ed_cl#http://www.tcfsh.tc.edu.tw/cou/France/MPSI.pdf#ed_op#/A#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#誰來挑戰看看呢？我想了好久只會三四題，還有些題目看不懂的－　－#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#可以英文回答#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#