Question

Let $H$ be the orthocenter of $△ABC$ where $A\equiv (0,0)$, $B\equiv (8,0)$ and $C\equiv (4,4\sqrt{3}).$ If $G_1,G_2,G_3,G_4,\dots .$ are centroids of triangles $\Delta AHB,\Delta A{G}_{1}B,\Delta A{G}_{2}B,\Delta A{G}_{3}B,\dots ..$ respectively then $\underset{n\to \infty }{\lim }\sum _{i=1}^n\text{Ar}\left(\Delta A{G}_{i}B\right)$ equals

• A. $8\sqrt{3}$ sq. units
• B. $16\sqrt{3}$ sq. units
• C. $\frac{8}{\sqrt{3}}$ sq. units
• D. $\frac{16}{\sqrt{3}}$ sq. units

Answer 1

Answer: (C)

$△ABC$ is equilateral
$H\to$ orthocenter or centroid
$H=\left(4,\frac{4}{\sqrt{3}}\right);G_1=\left(4,\frac{4}{3\sqrt{3}}\right);G_2=\left(4,\frac{4}{9\sqrt{3}}\right)$
$::$
$::$
$G_n=\left(4,\frac{4}{3\sqrt[n]{3}}\right)$
$\therefore \underset{n\to \infty }{\lim }\left(A_1+A_2+\dots A_n\right)$
$[$ All ${\Delta }^{\prime }$ s are isosceles $]$
$=\frac{1}{2}\times 8\left[\frac{4}{3\sqrt{3}}+\frac{4}{9\sqrt{3}}+\frac{4}{3\sqrt[n]{3}}\right]$
$=4\times \frac{4}{3\times \sqrt{3}}\underset{n\to \infty }{\lim }\left[1+\frac{1}{3}+\frac{1}{3^2}+\dots .+\frac{1}{3^{n-1}}\right]$
$=\frac{16}{3\sqrt{3}}\times \frac{1}{1-\frac{1}{3}}=\frac{8}{\sqrt{3}}$