If X, Y, Z are feet of the perpendiculars from the centroid G of $ABC$ upon the sides of AB,BC, CA. Prove $[XYZ] = 4\Delta^2(a^2+b^2+c^2)/9a^2b^2c^2$

Question

If $X$, $Y$, $Z$ are feet of the perpendiculars from the centroid $G$ of $\Delta{ABC}$ upon the sides $BC$, $CA$, $AB$, prove that $[XYZ] = \frac{4\Delta^2(a^2+b^2+c^2)}{9a^2b^2c^2}$ where $\Delta$ is the area of $\Delta{ABC}$

I noticed XCGY is cyclic with GC diameter, so I thought of ptolemy's theorem to find XY, YZ and ZX but that requires stuff like GX, XC etc. I tried basic angle chasing but nothing seemed to help. Any hints would be appreciated.

score 1 · Accepted Answer · answered Jul 19 '23 at 08:30

There is obviously something wrong in the formula, since its dimension is "adimensional", a constant, a number, but we expect something behaving like an area. (So it should be measured in m${}^2$, it should scale with $\lambda^2$ if we scale $a,b,c$ simultaneously with $\lambda$ to $\lambda a,\lambda b,\lambda $c, et caetera.)

This will become clear at the end of the story, we get the corrected version looking "almost like" the given formula.

Let us draw the height $AA_H$ of length $h_A$, and the median $AA_G$ from $A$. So $A_H$ is the projection of $A$ on the opposite side $BC$, and $A_G$ is the mid point of this side. Then in the triangle $AA_HA_G$ the point $G$ is on the side $AA_G$ at one third of it distance from $A_G$. And $GX\|AA_H$. This implies $GX=\frac13 h_A$. And similarly, the similar relations using similar notations.

I will use below $S$ for the area (surface) of $\Delta ABC$, since it is easier to type, and since $\Delta$ is always a letter when spelling $\Delta ABC$. Let $R$ be the circumradius of the given triangle with sides $a,b,c$. Then the area of $\Delta GYZ$ is: $$ \begin{aligned}{} [GYZ] & = \frac 12 GY\cdot GZ\cdot \sin\widehat{YGZ} = \frac 12 \cdot \frac 19h_B h_C\sin\widehat{YGZ} =\frac 12\cdot \frac 19\cdot\frac {2S}b\cdot\frac {2S}c\cdot \sin \hat A \\ &= \frac 12\cdot \frac 19\cdot\frac {2S}b\cdot\frac {2S}c\cdot \frac a{2R} =\frac 19\cdot \frac {S^2}{abc}\cdot\frac 1R\cdot a^2 =\frac 19\cdot \frac {S^2}{abc}\cdot\frac 1{abc/4S}\cdot a^2 \\ &=\frac 19\cdot\frac{4S^{\color{red}3}}{a^2b^2c^2}\cdot a^2\ . \end{aligned} $$ Now write the similar formulas for $[GZX]$ and $[GXY]$ and add to get: $$ [XYZ]= \frac 19\cdot\frac{4S^{\color{red}3}}{a^2b^2c^2}\cdot (a^2 + b^2+c^2) \ . $$

If X, Y, Z are feet of the perpendiculars from the centroid G of $ABC$ upon the sides of AB,BC, CA. Prove $[XYZ] = 4\Delta^2(a^2+b^2+c^2)/9a^2b^2c^2$

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