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Difference between revisions of "2024 AMC 10A Problems/Problem 16"

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<math>\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</math>
 
<math>\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</math>
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==Solution==
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Using the rectangle with side length <math>1</math>, let its short side be <math>x</math> and the long side be <math>y</math>. 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 <math>9</math> is <math>\sqrt{9}=3</math> times that of the rectangle with area <math>1</math>), as they are all similar to each other.
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The side opposite <math>AB</math> on the large rectangle is hence written as <math>6x + 4x + 2y\sqrt{2} + 3y\sqrt{2} = 10x+5y\sqrt{2}</math>. However, <math>AB</math> can be written as <math>4y\sqrt{2}+5x+7x = 4y\sqrt{2}+12x</math>. Since the two lengths are equal, we can write <math>10x+5y\sqrt{2} = 4y\sqrt{2}+12x</math>, or <math>y\sqrt{2} = 2x</math>. Therefore, we can write <math>y=x\sqrt{2}</math>.
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Since <math>xy=1</math>, we have <math>(x\sqrt{2})(x) = 1</math>, so <math>x=\frac{1}{\sqrt[4]{2}}</math>, and <math>y=\sqrt[4]{2}</math>. Therefore, <math>AB = 4y\sqrt{2}+12x = 4(\sqrt[4]{2})(\sqrt{2})+\frac{12}{\sqrt[4]{2}} = \boxed{\textbf{(D) }10\sqrt[4]{8}}</math>
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=15|num-a=17}}
 
{{AMC10 box|year=2024|ab=A|num-b=15|num-a=17}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:00, 8 November 2024

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 $AB$?

Screenshot 2024-11-08 2.08.49 PM.png

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

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 $(x\sqrt{2})(x) = 1$, so $x=\frac{1}{\sqrt[4]{2}}$, and $y=\sqrt[4]{2}$. Therefore, $AB = 4y\sqrt{2}+12x = 4(\sqrt[4]{2})(\sqrt{2})+\frac{12}{\sqrt[4]{2}} = \boxed{\textbf{(D) }10\sqrt[4]{8}}$

See also

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
All AMC 10 Problems and Solutions

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