2025 AMC 12A Problem 5

Question

In the figure below, the outside square contains infinitely many squares, each of them with the same center and sides parallel to the outside square. The ratio of the side length of a square to the side length of the next inner square is

k,

where

0 \lt k \lt 1.

The spaces between squares are alternately shaded, as shown in the figure (which is not necessarily drawn to scale).

The area of the shaded portion of the figure is

64\%

of the area of the original square. What is

k?

\dfrac{3}{5}

\dfrac{16}{25}

\dfrac{2}{3}

\dfrac{3}{4}

\dfrac{4}{5}

Answer: D

Concepts:geometric sequence area ratio

Difficulty rating: 1270

Small hint:

Take the outer square to have area

1

; the squares have areas

1, k^2, k^4, \ldots

Big hint:

The shaded rings form a geometric series that sums to

\dfrac{1}{1+k^2}

Solution:

Let the outer square have area

1.

The nested squares have areas

1, k^2, k^4, \ldots,

so the ring between the

n

th and

(n+1)

th squares has area

k^{2n}(1-k^2).

The shaded rings are the alternate ones

n = 0, 2, 4, \ldots,

with total area

\begin{aligned} \sum_{j=0}^{\infty} k^{4j}(1-k^2) &= \frac{1-k^2}{1-k^4} \\ &= \frac{1}{1+k^2}. \end{aligned}

Setting

\dfrac{1}{1+k^2} = \dfrac{16}{25}

gives

1 + k^2 = \dfrac{25}{16},

so

k^2 = \dfrac{9}{16}

and

k = \dfrac{3}{4}.

Thus, the correct answer is D.