2025 AMC 10A Problem 13

The picture goes on forever, so the rings cannot be added one at a time. The plan is to find a pattern in their areas and add them all at once.

1. Make the outer square have side $1$, so its area is $1$. Then "$64\%$ of the original square" just means the shaded area is $0.64 = \frac{16}{25}$.
2. Each square is $k$ times as wide as the one around it. So the sides are $1, k, k^2, k^3, \ldots$ and the areas (side times side) are $1, k^2, k^4, k^6, \ldots$
3. A ring is the space between two neighboring squares, so its area is the bigger square minus the smaller one:

4. Look at the shaded rows. Each shaded ring is the shaded ring two steps further out, shrunk; its area is $k^4$ times as big. So the shaded total is
$$S = (1 - k^2) + (1 - k^2)k^4 + (1 - k^2)k^8 + \cdots$$
5. This is an infinite geometric series: first term $1 - k^2$, and each term is $k^4$ times the one before. Since $k^4$ is between $0$ and $1$, the sum is "first term divided by ($1$ minus the multiplier)":
$$S = \frac{1 - k^2}{1 - k^4}.$$
6. The bottom is a difference of squares: $1 - k^4 = (1 - k^2)(1 + k^2)$. Cancel the $1 - k^2$ from top and bottom:
$$S = \frac{1}{1 + k^2}.$$
7. Now use $S = \frac{16}{25}$. Flip both sides: $1 + k^2 = \frac{25}{16}$, so $k^2 = \frac{9}{16}$ and $k = \frac34$ (it must be positive).

Check: with $k = \frac34$, $k^2 = \frac{9}{16}$ and $\frac{1}{1 + 9/16} = \frac{16}{25} = 64\%$.

The answer is $\boxed{\textbf{(D)}\ \frac{3}{4}}$.

When a figure repeats forever by shrinking, each piece is a scaled copy of an earlier piece, and that is exactly what makes the areas a geometric series. Remember that areas shrink by the square of the length ratio, and skipping every other ring squares the multiplier again. The final formula $\frac{1}{1 + k^2}$ also makes sense: if $k$ is close to $1$ the rings are thin and about half is shaded; if $k$ is tiny, the outer ring covers almost everything.

Skip the series and use the fact that the figure contains a copy of itself.

Let $S$ be the shaded fraction. Inside the second square is a smaller copy of the whole picture, with area $k^2$, but with the colors swapped (its outer ring is white). So the white fraction of that copy is $1 - S$, and it is shaded in the big picture. Adding the outer ring:
$$S = (1 - k^2) + k^2(1 - S).$$
Multiply out: $S = 1 - k^2 + k^2 - k^2 S = 1 - k^2 S$, so $S(1 + k^2) = 1$ and $S = \frac{1}{1 + k^2}$. Setting this equal to $\frac{16}{25}$ gives $k = \frac34$ again.

Scale so that the outer square has side $1$ and area $1$. The squares then have sides $1, k, k^2, k^3, \ldots$ and areas $1, k^2, k^4, \ldots$, so the ring between the $i$th and $(i+1)$st squares has area
$$k^{2i} - k^{2i+2} = k^{2i}\left(1 - k^2\right), \qquad i = 0, 1, 2, \ldots$$

The shading starts with the outermost ring and takes every other one, so the shaded rings are $i = 0, 2, 4, \ldots$ Their areas form a geometric series with first term $1 - k^2$ and ratio $k^4$:
$$S = \left(1 - k^2\right)\left(1 + k^4 + k^8 + \cdots\right) = \frac{1 - k^2}{1 - k^4} = \frac{1 - k^2}{(1 - k^2)(1 + k^2)} = \frac{1}{1 + k^2}.$$

Setting $S = 0.64 = \frac{16}{25}$:
$$\frac{1}{1 + k^2} = \frac{16}{25} \implies 1 + k^2 = \frac{25}{16} \implies k^2 = \frac{9}{16} \implies k = \frac{3}{4}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{3}{4}}$.