Question

Let G denote the centroid of triangle ABC. If triangle ABG is equilateral with side length 2, then determine the perimeter of triangle ABC.

Answer 1

We know from Euclidean Geometry and the properties of a centroid that GC=2GM. Now GM=sin60*GA=$\sqrt{3}$. Hence CM=GC+GM=3*$\sqrt{3} /$. Now, since GM is normal to AB, we have by the pythagoeran theorem that:
$AC^2=AM^2+GM^2=BM^2+GM^2=BC^2$
Hence, we calculate from this that AC^2= 28, hence AC=2*$\sqrt{7}$=BC. Thus, the perimeter of the triangle is 2+4*$\sqrt{7}$.