推 testy :太感謝了!! 這個方法好 :P 06/05 14:30
※ 編輯: JohnMash 來自: 112.104.141.206 (06/05 18:52)
※ 引述《testy (塔提)》之銘言：
: 取三角形ABC重心G,於原本大三角形可得三塊小三角形ABG、BCG、CAG
: 現在在大三角形ABC內任取一點D,
: 如何得之它屬於哪一個小三角形
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AG=(AB+AC)/3
if D in GBC
then there exists a point K in BC such that
GD= a GK, where 0<a<1
and K in BC iff AK=u AB +(1-u) AC, where 0<u<1
hence, AD=AG+GD=AG+a GK = AG + a (AK-AG)
=(1-a)AG + a AK
=(1-a)(AB+AC)/3 + a (u AB +(1-u) AC)
=(1/3 + a(u-1/3)) AB + (1/3 + a(2/3 -u)) AC
Hence D in GBC iff AD=pAB+qAC
p+q=2/3+a/3, p-q=2au-a=a(2u-1)
3p+3q-2=a, ((p-q)/a+1)/2=u
then
0<3p+3q-2<1 , 0<(2p+q-1)/(3p+3q-2)<1
that is
0<2p+q-1<3p+3q-2<1
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if D in GAB
then there exists a point K in AB such that
GD= a GK, where 0<a<1
and K in AB iff AK=u AB , where 0<u<1
hence, AD=AG+GD=AG+a GK = AG + a (AK-AG)
=(1-a)AG + a AK
=(1-a)(AB+AC)/3 + au AB
=(1/3 + a(u-1/3)) AB + (1/3 - a/3) AC
Hence, D in GAB iff AD=pAB+qAC
p+q=2/3+a(u-2/3), p-q=au
p+q=2/3+p-q-2a/3
a=1-3q
then
0<(p-q)<1-3q<1
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Similarly
Hence, D in GCA iff AD=pAB+qAC
0<(q-p)<1-3p<1
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