2025 AMC 12A Problems/Problem 5

The following problem is from both the 2025 AMC 10A #13 and 2025 AMC 12A #5, so both problems redirect to this page.

Problem

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).

$[asy] unitsize(1cm); int n = 25; real s = 6; real ratio = 0.767; real a = s; for (int i = 0; i < n; ++i) { real b = a * ratio; // Draw current square draw(box((-a/2,-a/2),(a/2,a/2))); if (i % 2 == 1) { fill(box((-a/2,-a/2),(a/2,a/2)), gray(1)); } else { fill(box((-a/2,-a/2),(a/2,a/2)), lightred); } a = b; } draw(box((-a/2,-a/2),(a/2,a/2))); [/asy]$

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

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

Video Solution

https://youtu.be/CCYoHk2Af34

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, meaning that there wouldn't be a loss of generality if we let the side length equal some value, getting the same answer $\boxed{D}$ . This was done in solution 3.

~vgarg (minor clarifications by ~Logibyte)

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 $p = k^2$ . 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 (Minor LaTeX edits by ~roblmin235)

Solution 4 (The fastest, no sequences/series)

Since this is an infinite fixed pattern, we can actually take a huge shortcut. Suppose we take off the outer "layer" and then swap the colors. We have essentially created a smaller copy of the original diagram, specifically with a $k:1$ ratio of side lengths. Assume the original diagram had 100 units of area. Then, based on the givens, the original diagram had 64 units of shaded area and 36 units of unshaded area, meaning our new smaller diagram has 36 units of shaded area. Since the ratio of corresponding side lengths between our new diagram and the original diagram is $k$ , the ratio of corresponding areas is $k^2$ . Thus, we get $k^2 = \frac{36}{64}$ , so $k = \boxed{\text{(D) }\dfrac{3}{4}}$ .

Note: the explanation seems complicated to explain, but the thought process while doing it is extremely fast, much faster than any of the methods involving a series.

~dg6665

~remark: At least personally, the final $k^2 = \frac{36}{64}$ seemed to have a nonobvious origin, so here as an explanation of it. In the original figure, the nonshaded part makes up 36/100 units. In the new diagram, the nonshaded part makes up 64/100 units, but since this new diagram has area k^2, the nonshaded part has area $\frac{64k^2}{100}$ setting these two equal gives $k^2 = \frac{36}{64}$ ~Neber100

Solution 5 (based very closely on solution 4)

Similarly to solution 4, we can imagine "peeling" off the outer layer to get an inner figure that is similar to the outer figure. Letting the large square have side 1, we know the outer shaded layer has area $1-k^2$ . The shaded part of the inner figure (with the outer shaded part removed now must make up $\frac{36}{100} = \frac{9}{25}$ of the inner area of $k^2$ . Together, the inner shaded part and the outer shaded part must add up to $\frac{64}{100}$ , as they make up 64% of the square. Thus, we get the equation $1-k^2 + \frac{9k}{25} = 16/25$ - Solving this gives k = 3/4 ~Neber100

Solution 6 (The longest way)

Notes that: The sum $S$ can be written as: $\begin{align*} S &= 1^2 - k^2 + (k^2)^2 - (k^3)^2 + (k^4)^2 - (k^5)^2 + (k^6)^2 - \dots \\ &= k^0 - k^2 + k^4 - k^6 + k^8 - k^{10} + \dots \\ &= (k^0 + k^4 + k^8 + k^{12} + \dots) - (k^2 + k^6 + k^{10} + k^{14} + \dots) \\ \end{align*}$ So we have: $\begin{align*} S &= \lim_{n\rightarrow\infty}\left[1\times\dfrac{1-(k^4)^n}{1-k^4}-k^2\times\dfrac{1-(k^4)^n}{1-k^4}\right]\\ &= \dfrac{1}{1-k^4}-\dfrac{k^2}{1-k^4} \\ &= \dfrac{1-k^2}{1-k^4} \\ &= \dfrac{1-k^2}{(1+k^2)(1-k^2)} \\ &= \dfrac{1}{1+k^2} \end{align*}$ We are given the condition that: $\because \dfrac{1}{1+k^2}=64\%=\dfrac{64}{100}=\dfrac{16}{25}$ We can now solve for $k$ : $\begin{align*} \therefore \dfrac{1}{1+k^2} &= \dfrac{16}{25} \\ 1+k^2 &= \dfrac{25}{16} \\ k^2 &= \dfrac{25}{16} - 1 \\ k^2 &= \dfrac{25}{16} - \dfrac{16}{16} \\ k^2 &= \dfrac{9}{16} \\ k &= \pm\sqrt{\dfrac{9}{16}} \\ k &= \pm\dfrac{3}{4} \end{align*}$ Therefore, ignore the negative value (Because $0<k<1$), so the final answer is $\color{red}\boxed{\color{black}\dfrac{3}{4}}\color{black}$ . We are done ~~ : )

~ funkCCP

Solution 7

Assume that the side length for the largest square in the diagram is $1.$ Since the ratio of the next inner square to the largest square is $k,$ we get that the side length for the second largest square is $k,$ $k^2$ for the third largest square, and $k^3$ for the fourth largest square. So, we can write that $(1-k^2)+(k^4-k^6)+...=\frac{64}{100}$ by using the diagram since the area of a square with side length $s$ is $s^2.$ This is a geometric sequence with first term $1-k^2$ and ratio $k^4.$ So, we can sum the geometric series for $\frac{1-k^2}{1-k^4}=\frac{64}{100}.$ Cross multiplying, we have that $100-100k^2=64-64k^4.$ Moving the terms to one side, we have $64k^4-100k^2+36=0,$ and we can factor as $(64k^2-36)(k^2-1)=0.$ Since $k<1,$ $k^2-1=0$ is invalid. So, we have $64k^2-36=0,$ and simplifying we get $k^2=\frac{36}{64}.$ Since $k>0,$ $k=\frac{6}{8}=\boxed{\frac{3}{4}}.$

~ gogogo2022

Solution 8 (Cheese 🧀)

Measure the side length of the first square to be 2 inches. Measure the side length of the second square to be 1.5 inches. $\frac{1.5}{2} = \frac{3}{4} = \boxed{\text{(D) } \frac{3}{4}}$

~Aeioujyot

~remark: this is probably a bad idea since MAA could intentionally put a not-to-scale diagram on it. They make it clear diagrams are not necessarily to scale ~dg6665

~🧀

~🧀🧀🧀 ~🧀

~remark-This is REALLY risky, NEVER do this thing on the actual test. Solution 4 is more accurate and logical and explains basically the same idea. - Mathhman645

Chinese Video Solution

https://www.bilibili.com/video/BV1nhkUByEV4/

~metrixgo

Video Solution (Intuitive and Easy)

https://youtu.be/1GzN3-H9BRo

~Education, the Study of Everything

Video Solution by Power Solve

https://www.youtube.com/watch?v=Vd_kvodRjNQ

Video Solution (In 1 Min)

https://youtu.be/5H8L7_M9BWc?si=ZDOYb3ZKllXsWFyn ~ Pi Academy

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Video Solution by Daily Dose of Math

https://youtu.be/5Fjos1vBt0A

~Thesmartgreekmathdude

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2025 AMC 12A (Problems • Answer Key • Resources)
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2025 AMC 12A Problems/Problem 5

Contents

Problem

Video Solution

Solution 1

Remark

Solution 2

Solution 3 (Faster)

Solution 4 (The fastest, no sequences/series)

Solution 5 (based very closely on solution 4)

Solution 6 (The longest way)

Solution 7

Solution 8 (Cheese 🧀)

Chinese Video Solution

Video Solution (Intuitive and Easy)

Video Solution by Power Solve

Video Solution (In 1 Min)

Video Solution by SpreadTheMathLove

Video Solution

Video Solution by Daily Dose of Math

See Also