AMC 10 · 2025 · #13

Grade 8 geometry-2d

Archetype Geometric Series Share └ Sub-archetype Nested-Similar-Figure Area Sum

sequences-geometricgeometric-series-infinitearea-difference convert-to-algebra ↑ Prerequisites: sequences-geometric

📏 Medium solution 💡 3 insights 📊 Diagram

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).

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

Pick an answer.

(A)

$\frac 35$

(B)

$\frac {16}{25}$

(C)

$\frac 23$

(D)

$\frac 34$

(E)

$\frac 45$

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: A big square holds infinitely many smaller squares, all with the same center and parallel sides. Each square's side is k times the side of the square just outside it, with 0 < k < 1. The frames between neighboring squares are shaded in two alternating colors. One color's frames together cover 64% of the big square's area. Find k.

Givens: The squares are nested, share a center, and have parallel sides.; Each square's side length is k times the next square outside it, with 0 < k < 1.; The gaps (frames) between consecutive squares are shaded in alternating colors.; The shaded color's total area is 64% of the outer square's area.

Unknowns: The side ratio k.

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #16 Change Focus / Count the Complement, #13 Convert to Algebra

The picture repeats forever, so the winning move is to spot the pattern in the areas: each square is a fixed fraction of the one before it. That turns the shaded region into a neat repeating sum. Name the outer area, break the shaded region into frames, notice the endless sum hides a shrunken copy of itself, and one short equation replaces the infinite list. Then the 64% clue pins down k.

Execute — Answer: D

#4 Introduce a Variable 6.EE.B.6 Step 1

Give the outer square area 1

Let the outer square's side be 1, so its area is 1.
Now every area is just a fraction of the whole, which makes the 64% clue easy to use later.

$$\text{outer side}=1,\quad \text{outer area}=1^2=1$$

💡 Calling the whole thing 1 turns every area into a plain fraction of the picture.

#5 Look for a Pattern 6.EE.A.1 Step 2

Each area is k squared of the last

Going one square inward multiplies the side by k, so it multiplies the area by k times k, which is k squared.
Starting from area 1, the areas of the nested squares are 1, then k squared, then k to the fourth, then k to the sixth, and so on.

$$1,\ k^2,\ k^4,\ k^6,\ \dots \qquad(\text{each area}=k^2\times\text{the previous area})$$

💡 Shrinking a side by k shrinks the area by k twice, once for length and once for width.

#7 Identify Subproblems 7.EE.A.1 Step 3

Add up the shaded frames

A frame is one square minus the square just inside it.
The shaded color takes every other frame, starting from the outermost.
So the shaded total is (1 minus k squared), plus (k to the fourth minus k to the sixth), and so on.
Writing it out in a line, the shaded area is 1 minus k squared plus k to the fourth minus k to the sixth, forever alternating.

$$\text{shaded}=(1-k^2)+(k^4-k^6)+\cdots=1-k^2+k^4-k^6+\cdots$$

💡 Each shaded frame is a square minus the one inside it, so lining them up leaves whole areas with plus and minus signs.

#4 Introduce a Variable 8.EE.C.7 Step 4

Fold the endless sum into one equation

Call the shaded total S.
Pull a factor of k squared out of everything after the first 1.
What is left inside the parentheses is exactly the same alternating sum S again.
So S equals 1 minus k squared times S.
Solving that short equation, S plus k squared times S equals 1, so S equals 1 divided by (1 plus k squared).

$$S=1-k^2\big(1-k^2+k^4-\cdots\big)=1-k^2 S \;\Rightarrow\; S(1+k^2)=1 \;\Rightarrow\; S=\dfrac{1}{1+k^2}$$

💡 The infinite sum contains a shrunken copy of itself, so one equation captures the whole endless chain.

#13 Convert to Algebra 8.EE.A.2 Step 5

Use the 64% clue and solve for k

The shaded total S is 64%, which is 64 out of 100, or 16 over 25.
Set 1 divided by (1 plus k squared) equal to 16 over 25.
Flipping both sides, 1 plus k squared equals 25 over 16, so k squared equals 9 over 16.
Taking the positive square root (since k is between 0 and 1), k equals 3 over 4.
That is answer (D).

$$\dfrac{1}{1+k^2}=\dfrac{64}{100}=\dfrac{16}{25} \;\Rightarrow\; 1+k^2=\dfrac{25}{16} \;\Rightarrow\; k^2=\dfrac{9}{16} \;\Rightarrow\; k=\dfrac{3}{4}$$

💡 Once the ratio equals a known fraction, undoing the square root gives k directly.

[1] #4 6.EE.B.6 Let the outer square's side be 1, so its area is 1. Now every area is just a fra

[2] #5 6.EE.A.1 Going one square inward multiplies the side by k, so it multiplies the area by k

[3] #7 7.EE.A.1 A frame is one square minus the square just inside it. The shaded color takes ev

[4] #4 8.EE.C.7 Call the shaded total S. Pull a factor of k squared out of everything after the

[5] #13 8.EE.A.2 The shaded total S is 64%, which is 64 out of 100, or 16 over 25. Set 1 divided

Review

Reasonableness: Put k = 3/4 back in: k squared is 9/16, so 1 plus k squared is 25/16, and 1 divided by 25/16 is 16/25, which is exactly 64%. The value 3/4 sits between 0 and 1 as required, and it is one of the listed choices, so answer (D) holds up.

Alternative: Change focus and use the color swap. Peel off the outermost shaded frame and swap every color in the rest. What remains is an exact copy of the whole picture scaled down (area factor k squared) with the two colors traded. So its shaded part is the original's unshaded 36%. That gives 64 = (1 - k squared)(100) + 36(k squared), which simplifies to 0.36 = 0.64 k squared, again k squared = 9/16 and k = 3/4 — no infinite sum needed.

CCSS standards used (min grade 8)

6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Setting the outer square's side to 1 and letting k stand for the side ratio so every area becomes an expression.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Writing each nested square's area as a power of k (1, k^2, k^4, ...) because area is side squared.)
7.EE.A.1 Add, subtract, factor, and expand linear expressions to produce equivalent expressions (Writing each shaded frame as one square minus the next and combining them into the alternating sum.)
8.EE.C.7 Solve linear equations in one variable (Turning the endless sum into S = 1 - k^2 S and solving for S = 1/(1+k^2).)
8.EE.A.2 Use square root symbols to represent solutions to equations of the form x^2 = p (Solving k^2 = 9/16 by taking the positive square root to get k = 3/4.)

⭐ When a picture repeats itself forever, find the smaller copy hiding inside and let one equation do the endless work.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Give the outer square area 1

Each area is k squared of the last

Add up the shaded frames

Fold the endless sum into one equation

Use the 64% clue and solve for k

Review

CCSS standards used (min grade 8)

More like this