Re: [中學] 大三角形內的點屬於哪一塊小三角形的問題

看板Math作者 (Paul)時間14年前 (2011/06/05 13:12), 編輯推噓1(100)
留言1則, 1人參與, 最新討論串2/2 (看更多)
※ 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 ---------------------------------------------------------- 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 --------------------------------------------------------------- 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 -------------------------------------------------------- Similarly Hence, D in GCA iff AD=pAB+qAC 0<(q-p)<1-3p<1 ------------------------------------------------------------ -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 112.104.141.72

06/05 14:30, , 1F
太感謝了!! 這個方法好 :P
06/05 14:30, 1F
※ 編輯: JohnMash 來自: 112.104.141.206 (06/05 18:52)
文章代碼(AID): #1Dwn17r3 (Math)
文章代碼(AID): #1Dwn17r3 (Math)