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Let $$G$$ be the centroid of $$\triangle ABC$$; that is, the point where the segments joining each vertex to the midpoint of the opposite side all meet. If $$\triangle ABG$$ is equilateral with $$AB=2$$, then compute the perimeter of $$\triangle ABC$$.

Mar 16, 2019

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First, you graph the triangle. It's going to be a very tall isosceles triangle. Using the fact that the centroid of a triangle splits the median into a 2:1 ratio, and the properties of a 30-60-90 triangle, we can see that the altitude of this triangle is $$3\sqrt 3$$. Using the pythagorean theorem, we can see that each of the longer sides is $$\sqrt {82}$$. So adding all three sides up, including the one that was already given, we can see that the perimeter of $$2 + 2\sqrt{82}$$

Sorry! I messed up. All the above steps were correct, I just messed up with the Pythagorean theorem. The answer is actually 2 + 4sqrt7

Hope this helps!

Sorry!

Mar 16, 2019
edited by LagTho  Mar 16, 2019
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Thanks, Lag!!!

Here is a pic :

The height of ABG = sqrt(3)

And because G is a centroid, then CG = 2sqrt(3)

So...the height of ABC = 3sqrt(3)  = sqrt (27)

And using the Pythagorean Theorem

AC^2  =  (height of ABC)^2 + [(1/2)AB]^2

AC^2 =  ( 27) + (1)

AC^2 = 28

AC = sqrt (28)  =  2sqrt(7)   = BC

So.....the perimeter of ABC  =   2 + 2 (2 sqrt(7)   =   2 + 4sqrt(7)

Just  as Lag found   !!!!

Mar 16, 2019
edited by CPhill  Mar 16, 2019