Animated Solution for Mathematics - Circles: Comprehension Passage

Imagine standing at the origin of a coordinate plane, looking out at a vast circular region $M$ with radius $r$. Inside this region, we place a sequence of circles, $C_n$, that shrink and shift in a mesmerizing pattern.

We start with a geometric progression $a_n=\frac{1}{2^{n-1}}$. The terms are $1,\frac{1}{2},\frac{1}{4},\dots$, and their sum $S_n={\sum }_{i=1}^n{a}_i$ is given by:

This sum acts as the heartbeat of our problem, defining the spatial progression of our circles.

For the circles $C_n$, the center is $(S_{n-1},0)$ and the radius is $a_n$. As $n$ increases, the centers march steadily along the x-axis toward the value $2$, while the circles themselves shrink rapidly.

For a circle $C_n$ to be contained within $M$, its farthest point from the origin must not exceed $r$. The farthest point is at a distance of $S_{n-1}+a_n$ from the origin. Because $S_{n-1}+a_n=S_n$, our condition for containment is simply $S_n\le r$.

With $r=\frac{1025}{513}$, we solve:

This simplifies to:

Thus, $2^{n-1}\le 513$. Since $2^9=512$, we find $n-1\le 9$, or $n\le 10$. We have $k=10$ circles.

We now ask: how many of these $10$ circles can we pick such that no two intersect? We checked adjacent circles $C_n$ and $C_{n+1}$ and found they always intersect because the distance between their centers is $a_n$, which is less than the sum of their radii $a_n+a_{n+1}$.

However, if we skip one, the geometry changes. For $C_n$ and $C_{n+2}$, the distance between centers is $a_n+a_{n+1}$, and the sum of radii is $a_n+a_{n+2}$. Since $a_{n+1}>a_{n+2}$, the distance between centers is greater than the sum of radii, meaning they do not intersect.

To maximize our count $l$, we pick the alternate circles: $C_1,C_3,C_5,C_7,C_9$. That gives us $l=5$. The final calculation is:

Finally, we turn to $D_n$, where the centers are $(S_{n-1},S_{n-1})$. These centers lie on the line $y=x$. The distance from the origin to the center is $\sqrt{S_{n-1}^2+S_{n-1}^2}=\sqrt{2}{S}_{n-1}$.

The condition for $D_n$ to be inside $M$ is $\sqrt{2}{S}_{n-1}+a_n\le r$. With $r=\frac{(2^{199}-1)\sqrt{2}}{2^{198}}=2\sqrt{2}-\frac{\sqrt{2}}{2^{198}}$, we substitute and simplify.

The $2\sqrt{2}$ terms cancel out, leaving us with an inequality that, after careful manipulation, reveals:

Since $2-\frac{1}{\sqrt{2}}\approx 1.293$, which is less than $2$, we conclude $n-1\le 198$, so $n\le 199$. There are $199$ such circles.