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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
 #1
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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
 #2
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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   !!!!

 

 

cool cool cool

 Mar 16, 2019
edited by CPhill  Mar 16, 2019

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