AMC 10 · 2024 · #16

Grade 8 geometry-2dalgebra

Archetype Multiplicative Ratio with Integrality Constraints

similar-figuresarea-rectanglesratio-proportionexponents convert-to-algebraidentify-subproblems ↑ Prerequisites: area-rectanglessimilar-figuresexponents

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

All of the rectangles in the figure below, which is drawn to scale, are similar to the enclosing rectangle. Each number represents the area of the rectangle. What is length $AB$ ?

$\textbf{(A) }4+4\sqrt5\qquad\textbf{(B) }10\sqrt2\qquad\textbf{(C) }5+5\sqrt5\qquad\textbf{(D) }10\sqrt[4]{8}\qquad\textbf{(E) }20$

Pick an answer.

(A)

$4+4\sqrt5$

(B)

$10\sqrt2$

(C)

$5+5\sqrt5$

(D)

$10\sqrt[4]{8}$

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: An outer rectangle is tiled by smaller rectangles that are all similar to the outer one. Each small rectangle is labeled with its area. Find the length $AB$ of the outer rectangle's top side.

Givens: Every rectangle in the figure is similar to the enclosing one; Labels show areas: $1, 8, 16, 18, 25, 32, 36, 49$ (and the enclosing rectangle); From the figure, the top edge $AB$ is covered by the rectangles of area $32, 25, 49$; The bottom edge is covered by the rectangles of area $36, 16, 8, 18$; Answer choices: (A) $4+4\sqrt5$, (B) $10\sqrt2$, (C) $5+5\sqrt5$, (D) $10\sqrt[4]{8}$, (E) $20$

Unknowns: The length of the top side $AB$

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #7 Identify Subproblems

Eight different areas with one shared shape is a textbook trigger for Tool #13 (Convert to Algebra): name the short and long side of a unit-area rectangle $x$ and $y$, and every other rectangle's sides become $x\sqrt{A}$ and $y\sqrt{A}$. Once everything is in $(x, y)$, the figure becomes one equation (top edge = bottom edge), which fixes the ratio $y/x$. Tool #7 (Identify Subproblems) then splits the work into clean stages — (A) get the ratio from the side-equality, (B) get $x$ and $y$ from the area constraint $xy=1$, (C) plug back into the top-edge sum for $AB$. Each stage is short on its own.

Execute — Answer: D

#13 Convert to Algebra 6.EE.A.2 Step 1

Name the sides of the unit-area rectangle.
Let $x$ be the short side and $y$ the long side of the rectangle with area $1$, so $xy = 1$.
Because all rectangles are similar, the rectangle with area $A$ has short side $x\sqrt{A}$ and long side $y\sqrt{A}$.

$$xy = 1, \qquad \text{sides of area-}A \text{ rectangle} = (x\sqrt{A},\; y\sqrt{A})$$

💡 Areas scale with the square of length, so square roots of areas are the right length unit. Naming just two letters $x, y$ captures every side in the figure.

#7 Identify Subproblems 7.G.A.1 Step 2

Write the top edge $AB$ as a sum.
From the figure, $AB$ is the horizontal run across the rectangles of area $32, 25, 49$.
The area-$32$ rectangle lies with its long side horizontal, contributing $y\sqrt{32} = 4\sqrt{2}\,y$.
The area-$25$ and area-$49$ rectangles lie with their short side horizontal, contributing $x\sqrt{25} = 5x$ and $x\sqrt{49} = 7x$.

$$AB = y\sqrt{32} + x\sqrt{25} + x\sqrt{49} = 4\sqrt{2}\,y + 12x$$

💡 Reading orientations off a scale drawing is Grade 7 work: the picture itself tells you which side of each similar rectangle is horizontal.

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

Write the bottom edge the same way.
It crosses the rectangles of area $36, 16, 8, 18$.
The area-$36$ and area-$16$ rectangles are oriented short-side horizontal ($6x$ and $4x$); the area-$8$ and area-$18$ rectangles are oriented long-side horizontal ($2\sqrt{2}\,y$ and $3\sqrt{2}\,y$).

$$\text{bottom} = 6x + 4x + 2\sqrt{2}\,y + 3\sqrt{2}\,y = 10x + 5\sqrt{2}\,y$$

💡 Same idea on the opposite edge — the four bottom-row rectangles give four length contributions, two short and two long.

#13 Convert to Algebra 6.EE.B.7 Step 4

Equate the top and bottom edges to solve for the ratio $y/x$.
Both expressions describe the same horizontal length of the outer rectangle, so set them equal and isolate.

$$4\sqrt{2}\,y + 12x = 10x + 5\sqrt{2}\,y \;\Rightarrow\; 2x = \sqrt{2}\,y \;\Rightarrow\; \dfrac{y}{x} = \sqrt{2}$$

💡 Collect the $x$ terms on one side, the $y$ terms on the other, divide. Grade 6 one-variable equation — the only new thing is that the answer is irrational.

#7 Identify Subproblems 8.EE.A.2 Step 5

Use $y = x\sqrt{2}$ together with $xy = 1$ to solve for $x$ and $y$.

$$x(x\sqrt{2}) = 1 \;\Rightarrow\; x^2 = \dfrac{1}{\sqrt{2}} \;\Rightarrow\; x = \dfrac{1}{\sqrt[4]{2}}, \quad y = x\sqrt{2} = \sqrt[4]{2}$$

💡 Taking a square root of $1/\sqrt{2}$ is the same as taking the fourth root of $1/2$ — Grade 8 exponent rules turn radicals into clean powers of $2$.

#13 Convert to Algebra 8.EE.A.1 Step 6

Plug $x = 2^{-1/4}$ and $y = 2^{1/4}$ back into the top-edge expression for $AB$.

$$AB = 4\sqrt{2}\,(2^{1/4}) + 12\,(2^{-1/4}) = 4 \cdot 2^{3/4} + 6 \cdot 2^{3/4} = 10 \cdot 2^{3/4} = 10\sqrt[4]{8} \;\Rightarrow\; \textbf{(D)}$$

💡 Rewrite $12/2^{1/4}$ by multiplying top and bottom by $2^{3/4}$ to clear the radical; both terms collapse to a common $2^{3/4}$ factor.

[1] #13 6.EE.A.2 Name the sides of the unit-area rectangle. Let $x$ be the short side and $y$ the

[2] #7 7.G.A.1 Write the top edge $AB$ as a sum. From the figure, $AB$ is the horizontal run ac

[3] #7 7.G.A.1 Write the bottom edge the same way. It crosses the rectangles of area $36, 16, 8

[4] #13 6.EE.B.7 Equate the top and bottom edges to solve for the ratio $y/x$. Both expressions d

[5] #7 8.EE.A.2 Use $y = x\sqrt{2}$ together with $xy = 1$ to solve for $x$ and $y$.

[6] #13 8.EE.A.1 Plug $x = 2^{-1/4}$ and $y = 2^{1/4}$ back into the top-edge expression for $AB$

Review

Reasonableness: The ratio $k = \sqrt{2}$ matches the famous $\text{A4}$-paper ratio, which is exactly the ratio that lets a rectangle tile copies of itself similarly — a strong sanity signal. Numerically $10\sqrt[4]{8} \approx 10 \cdot 1.6818 \approx 16.82$, which sits between the other choices (B) $10\sqrt{2} \approx 14.1$ and (E) $20$. Also, computing the bottom edge with the same $x, y$ gives $10x + 5\sqrt{2}\,y = 10 \cdot 2^{-1/4} + 5\sqrt{2}\cdot 2^{1/4} = 6 \cdot 2^{3/4} + 4 \cdot 2^{3/4} = 10 \cdot 2^{3/4}$ — the same number, confirming the equation was consistent.

Alternative: Tool #1 (Draw a Diagram) plus measurement: since the figure is drawn to scale, measure $AB$ and the side of (say) the area-$36$ rectangle in the printed figure, take their ratio, and multiply by $\sqrt{36} = 6$ scaled by the appropriate side. With a ruler this lands near $16.8$, matching $10\sqrt[4]{8}$ but missing the other choices' values ($\approx 12.9, 14.1, 16.2, 20$). Useful for elimination on the test even without the full algebra.

CCSS standards used (min grade 8)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Naming the unit-area rectangle's sides $x$ and $y$, so every other rectangle's sides become $x\sqrt{A}$ and $y\sqrt{A}$ — a uniform way to write every horizontal contribution to the outer edge.)
7.G.A.1 Solve problems involving scale drawings of geometric figures (Reading each small rectangle's orientation (short side horizontal vs. long side horizontal) directly from the scale figure to write the top- and bottom-edge sums.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form x+p=q and px=q (Setting top edge $= $ bottom edge gives a one-variable equation in $x$ and $y$, which collapses to $\dfrac{y}{x} = \sqrt{2}$.)
8.EE.A.2 Use square root and cube root symbols to represent solutions to equations of the form x^2 = p and x^3 = p (Solving $x^2 = 1/\sqrt{2}$ yields $x = 2^{-1/4}$ and $y = 2^{1/4}$ — Grade 8 square-root manipulation with fractional powers.)
8.EE.A.1 Know and apply the properties of integer exponents (Combining $2^{1/2} \cdot 2^{1/4} = 2^{3/4}$ and $1/2^{1/4} = 2^{-1/4}$ to fold both terms of $AB$ into the single form $10 \cdot 2^{3/4} = 10\sqrt[4]{8}$.)

⭐ Similar rectangles all share one side ratio $k$. Naming the unit rectangle's sides $x$ and $y$ turns every length in the figure into a multiple of $x$ or $y$, and the equality of opposite edges of the outer rectangle gives one clean equation that pins down $k = \sqrt{2}$ — the same ratio as A4 paper.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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