+295 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 \({G}_{1}{G}_{2}{G}_{3}\).
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.
Tiggsy
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.
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.
Tiggsy
+118673 Hi Tiggsy,
This question has been referred to recently and I have tried to understand you logic.
This statement confuses me. and is 2/3 of the way down that line from P.
Where did the 2/3 come from? Is this just some fact that I am supposed to know?
Tiggsy is not here a lot so if someone else can explain this I would appreciate it.
(Please do not respond if you do not know what you are talking about)
+118673 This site looks very relevant but I have not managed to work through it properly yet.
http://jwilson.coe.uga.edu/EMAT6680Fa06/Chitsonga/MEDIAN/THE%20MEDIANS%20OF%20A%20TRIANGLE.htm
