2025 AMC 12A Problem 5

Below is the professionally curated solution for Problem 5 of the 2025 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12A solutions, or check the answer key.

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Concepts:geometric sequencearea ratio

Difficulty rating: 1270

5.

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,k, where 0<k<1.0 \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%64\% of the area of the original square. What is k?k?

35\dfrac{3}{5}

1625\dfrac{16}{25}

23\dfrac{2}{3}

34\dfrac{3}{4}

45\dfrac{4}{5}

Solution:

Let the outer square have area 1.1. The nested squares have areas 1,k2,k4,,1, k^2, k^4, \ldots, so the ring between the nnth and (n+1)(n+1)th squares has area k2n(1k2).k^{2n}(1-k^2).

The shaded rings are the alternate ones n=0,2,4,,n = 0, 2, 4, \ldots, with total area j=0k4j(1k2)=1k21k4=11+k2.\sum_{j=0}^{\infty} k^{4j}(1-k^2) = \frac{1-k^2}{1-k^4} = \frac{1}{1+k^2}.

Setting 11+k2=1625\dfrac{1}{1+k^2} = \dfrac{16}{25} gives 1+k2=2516,1 + k^2 = \dfrac{25}{16}, so k2=916k^2 = \dfrac{9}{16} and k=34.k = \dfrac{3}{4}.

Thus, the correct answer is D.

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