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发表于 2014-3-22 22:49:04 |只看该作者

superacid 发表于 2014-3-21 20:02
你发一个解析几何的解法出来，我就公开道歉，如何？
先做出实事再说话

哪里有规定说一定要发解法
我的方法能不能用，LZ用不用这种方法都是他自己的事，关你P事

以B为原点建立坐标系(B,BC;y)
令BC=a (a>0)
假设AB不垂直AD(BC),那么直线AB的斜率k存在,m≠0且k≠0若DB⊥BC,则AB=AD<sqr(AD^2+BD^2),所以m≠-a
A(m,km)
B(0,0)
C(a,0)
D(a+m,km)
BA: y=kx
CD: y=k(x-a)
BD: y=kmx/(a+m)
F(a+n,kn)
E(b,0)

则
AE: y=km(x-b)/(m-b)
AF: y=(kn-km)(x-m)/(a+n-m)+km

又∵∠EAF=1/2∠BAD
cos^2(∠EAF)=(1+cos(∠BAD))/2
余弦定理
((a+n-m)^2+(kn-km)^2+(b-m)^2+(km)^2-(b-a-n)^2-(kn)^2))/(2*sqr(((a+n-m)^2+(kn-km)^2))*((b-m)^2+(km)^2)))=(1+(m^2+(km)^2+a^2-(a+m)^2-(km)^2)/(2*sqr(m^2+(km)^2)*a))/2
化简得
(-2nm-2am-2kkmn-2bb-2bm+2ab+2bn)/(sqr(((a+n-m)^2+(kn-km)^2))*((b-m)^2+(km)^2)))=1-1/sqr(k^2+1)

AE、BD相交于点G:
y=km(x-b)/(m-b)
y=kmx/(a+m)
得
G(b/(1-(m-b)/(a+m)),kmb/(a-b))

AF、BD相交于点H:
y=(kn-km)(x-m)/(a+n-m)+km
y=kmx/(a+m)
得
H((-m(kn-km)(a+m)/(km(a+n-m))+a+m)/(1-(kn-km)(a+m)/(km(a+n-m))),km(-m(kn-km)(a+m)/(km(a+n-m))+a+m)/((1-(kn-km)(a+m)/(km(a+n-m)))(a+m)))
GF、EH相交于点I:
GF:
(y-kn)/(kmb/(a-b)-kn)=(x-a-n)/(b/(1-(m-b)/(a+m))-a-n)
EH:
y/(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m)/((1-(kn-km)(a+m)/(km(a+n-m)))(a+m)))=(x-b)/((-m(kn-km)(a+m)/(km(a+n-m))+a+m)/(1-(kn-km)(a+m)/(km(a+n-m)))-b)
得
I(([-b/((-m(kn-km)(a+m)/(km(a+n-m))+a+m)/(1-(kn-km)(a+m)/(km(a+n-m)))-b) + ((a+n)(kmb/(a-b)+kn)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m)))/(b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))] (b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))/(kmb/(a-b)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m))))/(1-[(b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))/((-m(kn-km)(a+m)/(km(a+n-m))+a+m)/(1-(kn-km)(a+m)/(km(a+n-m)))-b)/(kmb/(a-b)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m)))]),(kmb/(a-b)-kn)(([-b/((-m(kn-km)(a+m)/(km(a+n-m))+a+m)/(1-(kn-km)(a+m)/(km(a+n-m)))-b) + ((a+n)(kmb/(a-b)+kn)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m)))/(b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))] (b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))/(kmb/(a-b)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m))))/(1-[(b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))/((-m(kn-km)(a+m)/(km(a+n-m))+a+m)/(1-(kn-km)(a+m)/(km(a+n-m)))-b)/(kmb/(a-b)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m)))])-a-n)/(b/(1-(m-b)/(a+m))-a-n)+kn)

∵AI⊥EF
有AI·EF=
(a+n-b)(([-b/((-m(kn-km)(a+m)/(km(a+n-m))+a+m)/(1-(kn-km)(a+m)/(km(a+n-m)))-b) + ((a+n)(kmb/(a-b)+kn)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m)))/(b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))] (b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))/(kmb/(a-b)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m))))/(1-[(b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))/((-m(kn-km)(a+m)/(km(a+n-m))+a+m)/(1-(kn-km)(a+m)/(km(a+n-m)))-b)/(kmb/(a-b)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m)))]))+kn((kmb/(a-b)-kn)(([-b/((-m(kn-km)(a+m)/(km(a+n-m))+a+m)/(1-(kn-km)(a+m)/(km(a+n-m)))-b)+((a+n)(kmb/(a-b)+kn)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m)))/(b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))] (b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))/(kmb/(a-b)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m))))/(1-[(b/(1-(m-b)/(a+m))-a-n)+kn)(km(-m(kn-km)(a+m)/(km(a+n-m))+a+m))/((-m(kn-km)(a+m)/(km(a+n-m))+a+m)/(1-(kn-km)(a+m)/(km(a+n-m)))-b)/(kmb/(a-b)((a+m)-(kn-km)(a+m)(a+m)/(km(a+n-m)))])-a-n)/(b/(1-(m-b)/(a+m))-a-n)+kn)
=0

联立两式得

-km(m-n)(a+n-m)/(a+m)^2=0

当且仅当mk=0时等式成立，与假设矛盾
所以AB⊥AD

和你用程序参加最少步周赛一样，自欺欺人罢了

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