一道数学题.死活弄不出来~Let A(a),B(b),and C(c) be three points in the plane (here a,b,c are complex num-bers).Let G be the gravity center of the triangle ABC.Show that the complexnumber corresponding to G is g =(a + b + c)/3.Hint.Denote by
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一道数学题.死活弄不出来~Let A(a),B(b),and C(c) be three points in the plane (here a,b,c are complex num-bers).Let G be the gravity center of the triangle ABC.Show that the complexnumber corresponding to G is g =(a + b + c)/3.Hint.Denote by
一道数学题.死活弄不出来~
Let A(a),B(b),and C(c) be three points in the plane (here a,b,c are complex num-
bers).Let G be the gravity center of the triangle ABC.Show that the complex
number corresponding to G is g =(a + b + c)/3.
Hint.Denote by G′ the point corresponding to g =( a+b+c)/3 .You will have to show that G′ = G.To this end,denote by M the midpoint of the line segment BC.First show that G′ is on the line AM,i.e.the points G′,A,and M are collinear (use the criterion we proved in class).Then show that AG′ = 2/3AM.This implies G′ = G.
中文的大概意思就是：A,B,C三个点在一个平面里，对应的边是a,b,c（为复杂数）。G是三角形ABC的重心。证明G对应的复杂数是g =(a + b + c)/3。提示：设G′也等于g =(a + b + c)/3，然后证明G‘=G。最后设M为BC的中点。首先证明G’在AM上，即三点共线。然后证明AG′ = 2/3AM。最后得到G'=G。
我主要是不知道怎么证明A,M,G'三点共线。要运用complex number.
一道数学题.死活弄不出来~Let A(a),B(b),and C(c) be three points in the plane (here a,b,c are complex num-bers).Let G be the gravity center of the triangle ABC.Show that the complexnumber corresponding to G is g =(a + b + c)/3.Hint.Denote by
因为G对应的复杂数是g =(a + b + c)/3,
G′也等于g =(a + b + c)/3
M为BC的中点
又因为重心是高线的交点,
所以G'=G
g =(a + b + c)/3，则G点为三角形的重心
实际上可能是证明三角形的重心在中线上
用向量的三角形法则可以证明的啊
这、