2025 AMC 10A Problem 13

Below is the professionally curated solution for Problem 13 of the 2025 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10A solutions, or check the answer key.

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Concepts:geometric sequencearea ratio

Difficulty rating: 1560

13.

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 side be 1.1. The squares have sides 1,k,k2,,1, k, k^2, \ldots, and the shaded rings alternate, so the shaded area is 1k2+k4k6+=11+k2.1 - k^2 + k^4 - k^6 + \cdots = \frac{1}{1 + k^2}. We're told this equals 64%=1625,64\% = \frac{16}{25}, so 1+k2=2516.1 + k^2 = \frac{25}{16}. Then k2=916,k^2 = \frac{9}{16}, giving k=34.k = \frac{3}{4}. Thus, D is the correct answer.

Problem 13 in Other Years