Consider the triangle whose vertices are $(0,0),(0,3),$ and $(4,0)$. Let $\Gamma_{1}$ be the greatest circle that can be drawn inside the triangle. For $n>1$, let $\Gamma_{n}$ be the greatest circle that can be drawn inside the triangle and does not overlap the $(n-1)$ greater circles, and is not in the region bounded by $\Gamma_{1}$ and the coordinate axes. Which of the following points is closest to the center of $\Gamma_{1992}?$
$\text{(A)}\space(0,1)\space \space \space \space \space\text{(B)}\space(0,3)\space \space \space \space \space\text{(C)}\space(1,0)\space \space \space \space \space\text{(D)}\space(\frac{8}{5},\frac{9}{5})\space \space \space \space \space\text{(E)}\space(4,0)$
This problem appeared in one of the math competitions in my city. The average time to solve this problem is $15$ minutes.
I tried to find a pattern between $n$ and the given points. (of course $n \ne 1$ because it is equidistance from $\text{(A),(C)}$, and $\text{(D)}$). For $n=2$, the correct option is $\text{(E)}$. For $n = 3$, the correct option is $\text{(B)}$. Then for $n \ge 4$ is is tedious!
Any help would be appreciated. THANKS!
I used Mathematica to construct iteratively 2723 circles (with a cutoff of $r=1/700$) and then reordered the list to find n° 1992. It turns out that circles n° 1991 and 1992 have the same radius ($0.00181207$) and their centers are located at: $$ C_{1991}=(2.45007, 1.05146); \quad C_{1992}=(2.12919, 0.0887912). $$ Hence $C_{1991}$ is nearer to point $D$, while $C_{1992}$ is nearer to point $C$ and no definite answer can be given, unless some other criteria are adopted for ordering circles.
Below is a picture of the result, but the above circles are too small to be seen at this scale.