Question

Consider ABC an equilateral triangle of sides 4cm.

(a) determine the position of the center of gravity of the equilateral triangle

(b) determine and construct the set of points C such that IMA + MB + MCII = 12/√3

(c) name the set of points C

Answer 1

a.

The position of the center of mass is $(\frac{l}{2},x).$ Wher l is the length of the sides of the triangle and x is gotten as follows.

$\frac{2}{2\sqrt3}=\frac{x}{2}\\ x=\frac{2}{3}\sqrt3$

Therefore, the position of the center of mass is $(2,\frac{2}{3}\sqrt3)$ .

b.

$|MA|=2\sqrt3-\frac{2}{3}\sqrt3=\frac{4}{3}\sqrt3\\ |MB|=2\sqrt3-\frac{2}{3}\sqrt3=\frac{4}{3}\sqrt3\\ |MA+MB+MC|=\frac{12}{\sqrt3}\\ \frac{4}{3}\sqrt3+\frac{4}{3}\sqrt3+|MC|=4\sqrt3\\ |MC|=\frac{4}{3}\sqrt3$

Therefore, the set of points of C is $(4,0)$ . Since

$MA=\sqrt{(2-4)^2+(\frac{2}{3}\sqrt3-0)^2}=\frac{4}{3}\sqrt3$

c.

the set of point of C is $(4,0)$