2025 AMC 10A Problem 13

P11-15~4 minPrint

Sequences & Series Quadrilaterals & Polygon Areas Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Cut the figure into known shapes (triangles, rectangles, sectors)

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 $k$ ?

Show solution without answering

Key insight

Each shaded ring is a shrunken copy of the one two rings further out, so the shaded areas add up to 1/(1 + k squared).

Solution

The picture goes on forever, so the rings cannot be added one at a time. The plan is to find a pattern in their areas and add them all at once.

Make the outer square have side $1$ , so its area is $1$ . Then " $64\%$ of the original square" just means the shaded area is $0.64 = \frac{16}{25}$ .
Each square is $k$ times as wide as the one around it. So the sides are $1, k, k^2, k^3, \ldots$ and the areas (side times side) are $1, k^2, k^4, k^6, \ldots$
A ring is the space between two neighboring squares, so its area is the bigger square minus the smaller one:

Ring	Color	Area
1st (outermost)	shaded	$1 - k^2$
2nd	white	$k^2 - k^4$
3rd	shaded	$k^4 - k^6 = k^4(1 - k^2)$
5th	shaded	$k^8 - k^{10} = k^8(1 - k^2)$

4. Look at the shaded rows. Each shaded ring is the shaded ring two steps further out, shrunk; its area is $k^4$ times as big. So the shaded total is
$S = (1 - k^2) + (1 - k^2)k^4 + (1 - k^2)k^8 + \cdots$
5. This is an infinite geometric series: first term $1 - k^2$ , and each term is $k^4$ times the one before. Since $k^4$ is between $0$ and $1$ , the sum is "first term divided by ( $1$ minus the multiplier)":
$S = \frac{1 - k^2}{1 - k^4}.$
6. The bottom is a difference of squares: $1 - k^4 = (1 - k^2)(1 + k^2)$ . Cancel the $1 - k^2$ from top and bottom:
$S = \frac{1}{1 + k^2}.$
7. Now use $S = \frac{16}{25}$ . Flip both sides: $1 + k^2 = \frac{25}{16}$ , so $k^2 = \frac{9}{16}$ and $k = \frac34$ (it must be positive).

Check: with $k = \frac34$ , $k^2 = \frac{9}{16}$ and $\frac{1}{1 + 9/16} = \frac{16}{25} = 64\%$ .

The answer is $\boxed{\textbf{(D)}\ \frac{3}{4}}$ .

Why this works

When a figure repeats forever by shrinking, each piece is a scaled copy of an earlier piece, and that is exactly what makes the areas a geometric series. Remember that areas shrink by the square of the length ratio, and skipping every other ring squares the multiplier again. The final formula $\frac{1}{1 + k^2}$ also makes sense: if $k$ is close to $1$ the rings are thin and about half is shaded; if $k$ is tiny, the outer ring covers almost everything.

Alternative approach

Skip the series and use the fact that the figure contains a copy of itself.

Let $S$ be the shaded fraction. Inside the second square is a smaller copy of the whole picture, with area $k^2$ , but with the colors swapped (its outer ring is white). So the white fraction of that copy is $1 - S$ , and it is shaded in the big picture. Adding the outer ring:
$S = (1 - k^2) + k^2(1 - S).$
Multiply out: $S = 1 - k^2 + k^2 - k^2 S = 1 - k^2 S$ , so $S(1 + k^2) = 1$ and $S = \frac{1}{1 + k^2}$ . Setting this equal to $\frac{16}{25}$ gives $k = \frac34$ again.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Scale so that the outer square has side $1$ and area $1$ . The squares then have sides $1, k, k^2, k^3, \ldots$ and areas $1, k^2, k^4, \ldots$ , so the ring between the $i$ th and $(i+1)$ st squares has area
$k^{2i} - k^{2i+2} = k^{2i}\left(1 - k^2\right), \qquad i = 0, 1, 2, \ldots$

The shading starts with the outermost ring and takes every other one, so the shaded rings are $i = 0, 2, 4, \ldots$ Their areas form a geometric series with first term $1 - k^2$ and ratio $k^4$ :
$S = \left(1 - k^2\right)\left(1 + k^4 + k^8 + \cdots\right) = \frac{1 - k^2}{1 - k^4} = \frac{1 - k^2}{(1 - k^2)(1 + k^2)} = \frac{1}{1 + k^2}.$

Setting $S = 0.64 = \frac{16}{25}$ :
$\frac{1}{1 + k^2} = \frac{16}{25} \implies 1 + k^2 = \frac{25}{16} \implies k^2 = \frac{9}{16} \implies k = \frac{3}{4}.$

The answer is $\boxed{\textbf{(D)}\ \frac{3}{4}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)
Introduction to Algebra, Chapter 11: Special Factorizations