Let be P the centroid of triangle ABC. If triangle ABP is an equilateral triangle with a side length of 2, then find the perimeter of triangle ABC.
+1486 Let be P the centroid of triangle ABC. If triangle ABP is an equilateral triangle with a side length of 2, then find the perimeter of triangle ABC.
AB = 2
PF = sqrt(3)
CF = PF * 3
AC = BC = sqrt[(AF)² + (CF)²]
Perimeter of triangle ABC = AB + AC + BC 
Dragan, could you please give your explanation and format your answer as a root? Please and thank you!
\(4\sqrt7+2\text{ is the answer}\)(I assume this is an AoPS challenge problem and thus you don't need any information. If you do, then remember how 1:2 ratios work with centroids)
+128406 Using Dragan's illustration.....
AP = 2
And
sin (60°) = PF / AP
sin(60°) = PF / 2 ⇒ PF = √3
But CF is a median of triangle ABC and the PF is 1/3 of the altitude FC = 3√3 = √27
And using the Pythagorean Theorem
CB =√ [ FC^2 + FB^2 ]= √ [27 + 1 ] = √28 = 2√7 = CA
So the perimeter = CB + CA + AB = 2√7 + 2√7 + 2 = 4√7 + 2
