If $X, Y, Z$ are the feet of the perpendiculars from the centroid to the sides $BC, CA, AB$, prove that the area of $\triangle XYZ=\dfrac{4\Delta^2(a^2+b^2+c^2)}{9a^2b^2c^2}$.
Solve only using elementary geometry. ($\Delta$ denotes the area of $\triangle ABC$). a,b and c are the sides of triangle ABC. a is opposite to angle A, b is opposite to angle B and c is opposite to angle C. I tried using some theorems, for example:
- $$\overline{ZY} = \frac{a\overline{AG}}{2R}$$
- $$\overline{XY} = \frac{c\overline{CG}}{2R}$$
- $$\overline{ZX} = \frac{b\overline{BG}}{2R}$$
- Note : G is the centroid, R is the circumradius. I tried to use this but could not proceed in a significant way.
Hints (I will give results for $A$ only, the rest follow by symmetry).
First show that $\angle YGZ +\angle A = 180^\circ$ so that $\sin(\angle YGZ) = \sin(A)$.
Then show that $GX = AD/3$ where $AD$ is the perpendicular from $A$ onto $BC$. By symmetry, $GY = BE/3$ and $GZ = CF/3$.
This is a non-trivial step and requires you to exploit the properties of the Euler line and the fact that $AH = 2OM$. (where $H$ is orthocentre, $O$ is circumcentre, $M$ is mid-point of $BC$)
Now, $$\mathrm {Ar}(\triangle YGZ) = 0.5 \cdot GY\cdot GZ\cdot \sin(\angle YGZ) = 0.5\cdot BE\cdot CF \cdot \sin(A)/9 $$$$= 0.5 \cdot \frac{(2\triangle)(2\triangle)\cdot \sin(A)}{9bc} = \frac{2\triangle^2a}{9bc\cdot 2R} = \frac{\triangle^2}{9Rabc}\cdot a^2$$
Lastly, just sum over $a,b,c$ and use $R = abc/4\triangle$ to get $$\mathrm{Ar}(\triangle XYZ) = \frac{\triangle^2(a^2+b^2+c^2)}{9Rabc} = \boxed{\frac{4\triangle^3(a^2+b^2+c^2)}{9a^2b^2c^2}}$$