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Triangle Medians and Altitudes

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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 $${G}_{1}{G}_{2}{G}_{3}$$. Apr 20, 2018

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Apr 21, 2018
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I'm working on it, but this is hard!

GYanggg  Apr 21, 2018
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Same here! I'll try my best!

Apr 21, 2018
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I am still struggling to find the values...

Apr 21, 2018
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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.

Apr 23, 2018
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HEEEEELP!

TheMathCoder  Apr 24, 2018
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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

Apr 24, 2018
edited by Guest  Apr 24, 2018
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Bravo Tiggsy!! Thank you very much.

Apr 24, 2018
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Wowowowowowowow, you are a great person tiggsy! Thank you a lot!

Apr 24, 2018