2024 AMC 10A Problems/Problem 16

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 $AB$ ?

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

Solution

Using the rectangle with side length $1$ , let its short side be $x$ and the long side be $y$ . Observe that for every rectangle, since ratios of the side length of the rectangles are directly proportional to the ratios of the square roots of the areas (For example, each side of the rectangle with area $9$ is $\sqrt{9}=3$ times that of the rectangle with area $1$ ), as they are all similar to each other.

The side opposite $AB$ on the large rectangle is hence written as $6x + 4x + 2y\sqrt{2} + 3y\sqrt{2} = 10x+5y\sqrt{2}$ . However, $AB$ can be written as $4y\sqrt{2}+5x+7x = 4y\sqrt{2}+12x$ . Since the two lengths are equal, we can write $10x+5y\sqrt{2} = 4y\sqrt{2}+12x$ , or $y\sqrt{2} = 2x$ . Therefore, we can write $y=x\sqrt{2}$ .

Since $xy=1$ , we have $(x\sqrt{2})(x) = 1$ , which we can evaluate $x$ as $x=\frac{1}{\sqrt[4]{2}}$ . From this, we can plug back in to $xy=1$ to find $y=\sqrt[4]{2}$ . Substituting into $AB$ , we have $AB = 4y\sqrt{2}+12x = 4(\sqrt[4]{2})(\sqrt{2})+\frac{12}{\sqrt[4]{2}}$ which can be evaluated to $\boxed{\textbf{(D) }10\sqrt[4]{8}}$ .

~i_am_suk_at_math_2

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2024 AMC 10A Problems/Problem 16

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