Let \(P\) be a point inside triangle \(ABC\). Let \({G}_{1}\)\({G}_{2}\), and \({G}_{3}\) be the centroids of triangles \(PBC\), \(PCA\),and \(PAB\), respectively. If the area of triangle \(ABC\) is 18, then find the area of triangle \({G}_{1}{G}_{2}{G}_{3}\).

Diagram to help with problem

TheMathCoder  Apr 20, 2018

Anyone here??? Please help me quickly!

TheMathCoder  Apr 21, 2018

I'm working on it, but this is hard!

GYanggg  Apr 21, 2018

Same here! I'll try my best!

tertre  Apr 21, 2018

I am still struggling to find the values...

TheMathCoder  Apr 21, 2018

X^2: Did you see this question, which was posted 2-3 days ago and hasn't been answered? I'm just curious for myself to know how to solve this seemingly very difficult Geometry-Trig problem. I don't even know how to approach the problem since I don't fully understand the properties of "centroids" of triangles. Thanks.

Guest Apr 23, 2018


TheMathCoder  Apr 24, 2018

Let the height of the triangle ABC (the vertical height of A above BC), be h,

and the height of P be \(\alpha h \text{ where } 0<\alpha<1 .\)


The centroid of the triangle BPC, \(G_{1},\)

lies on the median of that triangle from P, (that's the line from P to the mid-point of BC), and is 2/3 of the way down that line from P. It follows that the height of G1 is \(\alpha h/3.\)


The centroid of the triangle BPA, G3, can be defined in a similar way to that of G1.

The mid-point of AB will be at a height of h/2, so the vertical displacement between this point and P will be, (using the given diagram),

\(\displaystyle \frac{h}{2}-\alpha h,\)

so the height of G3 will be

\(\displaystyle \alpha h + \frac{2}{3}\left(\frac{h}{2}-\alpha h\right)=\frac{h}{3}+\frac{\alpha h}{3}.\)

(I've used the diagram as given. If P lies above the mid-point of AB heightwise, the calculation above will be slightly different, but leads to the same result.)


Calculation of the height of G2 leads to exactly the same result, so the line G3-G2 will parallel to BC, and it will follow that the triangle made up of the centroids will be similar to the triangle ABC.


The height of this triangle will be \(\displaystyle \frac{h}{3}+\frac{\alpha h}{3}-\frac{\alpha h}{3}=\frac{h}{3}.\)

The same result will follow for each of the other two 'heights'.


Since the linear dimensions ABC  to G1.G2.G3 are reduced by a factor of 3, it follows that the area will be reduced by a factor of 9 meaning that the area of G1.G2.G3 will be 18/9 = 2.



Guest Apr 24, 2018
edited by Guest  Apr 24, 2018

Bravo Tiggsy!! Thank you very much.

Guest Apr 24, 2018

Wowowowowowowow, you are a great person tiggsy! Thank you a lot!

TheMathCoder  Apr 24, 2018

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