2025 AMC 10A Problem 13

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

Every ring is a scaled copy of the one two steps out, by a factor $k^2$ in length and $k^4$ in area, which is exactly what makes the shaded total a geometric series. The factor $1 - k^2$ then cancels, leaving the memorable formula: shaded fraction $= 1/(1 + k^2)$, always between $1/2$ and $1$. The cancellation is also a sanity check: as $k \to 1$ the rings become thin and shading alternates evenly, giving $1/2$; as $k \to 0$ the outer ring swallows everything and the fraction tends to $1$.

Use self-similarity. Let $S$ be the shaded fraction of any such figure. The picture inside the second square is a copy of the whole figure scaled by $k$ with the colours swapped, and it occupies area $k^2$. Hence
$$S = \left(1 - k^2\right) + k^2 (1 - S) \implies S\left(1 + k^2\right) = 1 \implies S = \frac{1}{1 + k^2},$$
with no series summation at all. Then $S = \frac{16}{25}$ gives $k = \frac34$ as before.

• Setting the outermost ring alone equal to 64%, i.e. 1 - k^2 = 0.64, which gives k = 3/5, choice (A).
• Reading $64\%$ as the value of $k$ or of $k^2$ and picking $\frac{16}{25}$, choice (B).
• Using ratio $k^2$ for the shaded series instead of $k^4$, which forgets that the unshaded rings sit in between.