2025 AMC 10A Problems/Problem 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 $k,$ where $0 < k < 1.$ The spaces between squares are alternately shaded, as shown in the figure (which is not necessarily drawn to scale).

$[asy] unitsize(1cm); int n = 25; // number of squares real s = 6; // side length of largest square real ratio = 0.8; // shrink factor for each square real a = s; // current square side for (int i = 0; i < n; ++i) { real b = a * ratio; // next smaller square // Draw current square draw(box((-a/2,-a/2),(a/2,a/2))); if (i % 2 == 1) { fill(box((-a/2,-a/2),(a/2,a/2)), gray(1)); // light shading on the outside } else { fill(box((-a/2,-a/2),(a/2,a/2)), gray(0.4)); // dark shading on the outside } a = b; // move to the next square } // Draw innermost square outline draw(box((-a/2,-a/2),(a/2,a/2))); [/asy]$

The area of the shaded portion of the figure is $64\%$ of the area of the original square. What is $k?$

$\textbf{(A) } \frac 35 \qquad \textbf{(B) } \frac {16}{25} \qquad \textbf{(C) } \frac 23 \qquad \textbf{(D) } \frac 34 \qquad \textbf{(E) } \frac 45$

Solution 1

Let the side length of the largest square be $a,$ so it has area $a^2.$ Hence, the second-largest square has area $a^2k^2,$ the third-largest has $a^2k^4,$ and so on.

It follows that the total shaded area is $a^2-a^2k^2+a^2k^4-a^2k^6+...=a^2(1-k^2+k^4-k^6+...)=a^2\dfrac{1}{1+k^2}.$

The ratio of the area of the shaded region to that of the original square is then $\dfrac{a^2\frac{1}{1+k^2}}{a^2}=\dfrac{1}{1+k^2}=\dfrac{64}{100}$ $\implies 64+64k^2=100\implies k^2=\dfrac{36}{64}\implies k=\boxed{\text{(D) }\dfrac{3}{4}}.$

~Tacos_are_yummy_1

Remark

We can just let $a=1$ because the question deals with ratios to make calculations easier and have a less chance of messing up, getting the same answer $\boxed{D}$ .

~vgarg

Solution 2

Let the side length of the first square be $a$ and the second square be $b$ . The area of the original square is $a^2.$ The area of the outermost shaded region is $a^2 - b^2.$ Let $\frac{a}{b} = r.$ We have $b = \frac{a}{r}.$ So the outermost shaded region area becomes:

$a^2 - b^2 =$ $a^2 - \frac{a^2}{r^2}$

Now let the next square's side lengths be $c$ and $d.$ Similarly, $c = \frac{b}{r} = \frac{a}{r^2}$ and $d = \frac{c}{r} = \frac{a}{r^3}$ and the area of the next shaded region becomes: $c^2 - d^2 = \frac{a^2}{r^4} - \frac{a^2}{r^6}$

Notice the pattern of adding $4$ to the exponent. If this sequence continues infinitely, we ultimately get:

$(a^2 + \frac{a^2}{r^4} + \frac{a^2}{r^8} + ...) - (\frac{a^2}{r^2} + \frac{a^2}{r^6} + ...)$

Which can be simplified using the infinite geometric sequence formula. The problem also tells us that all of this equals $64\% a^2,$ or $\frac{16}{25}a^2:$

$\frac{a^2}{1 - \frac{1}{r^4}} - \frac{\frac{a^2}{r^2}}{1 - \frac{1}{r^4}} = \frac{16}{25}a^2$

$\frac{1}{1 - \frac{1}{r^4}} - \frac{\frac{1}{r^2}}{1 - \frac{1}{r^4}} = \frac{16}{25}$

$r^2 = \frac{16}{9} \implies r = \frac{4}{3}$ Since the problem tells us the ratio must be less than one, we take the reciprocal to finally get $k = \frac{1}{r} = \boxed{\text{(D) }\dfrac{3}{4}}.$

~grogg007

Solution 3 (Faster)

Let the outside square be of side length $1$ , and let $p = k²$ . Then the area of the square is $1-p+p^{2}-p^{3}+p^{4}-p^{5} \dots$ or $\left(1-p\right)\left(1+p^{2}+p^{4}+p^{6}+\dots\right)$ . Then, using geometric series, we get $\left(1-p\right)\left(\frac{1}{1-p^{2}}\right)$ = $\frac{1}{1+p}$ . This is equal to $\frac{64}{100}$ . Therefore, $1+p=\frac{25}{16}$ , so $p = 9/16$ and $k = \frac{3}{4}$ . Therefore, the answer is $\boxed{(D)\frac{3}{4}}.$

~Moonwatcher22 (Latex fixed by ~roblmin235)

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2025 AMC 10A Problems/Problem 13

Contents

Solution 1

Remark

Solution 2

Solution 3 (Faster)