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

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==Solution 1==
 
==Solution 1==
  
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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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Using the rectangle with area <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.
  
 
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>.
 
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>.

Revision as of 22:36, 9 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 1

Using the rectangle with area $1$, 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

Remark

We know that the area is an integer, so after finding $y=x\sqrt{2}$, AB must contains a fourth root. The only such option is $\boxed{\textbf{(D) }10\sqrt[4]{8}}$.

Solution 2

Let the rectangle's height be $x,$ the length $AB=y.$ The entire rectangle has an area of $200.$ We will be using this fact for ratios.

Note that the short side of the rectangle with area 32 will have a height of $\sqrt{\frac{32}{200}}\cdot x = \frac{2}{5}x.$ We use $x$ because it is apparent that the height of the rectangle with area $32$ is the shorter side, corresponding with $x.$

Similarly, the long side of the rectangle with area 36 has a height of $\sqrt{\frac{36}{200}}\cdot y = \frac{3\sqrt{2}}{10}y.$

Noting that the total height of the big rectangle has height $x,$ we have the equation $\frac25 x + \frac{3\sqrt{2}}{10}y = x \Rightarrow x=\frac{y}{\sqrt{2}}.$

Since the area $xy=\frac{y^2}{\sqrt{2}}$ is equal to 200, we have:

\begin{align*} y&=\sqrt{200\sqrt{2}} \\ &=\sqrt{100\sqrt{8}} \\ &=\boxed{\textbf{(D) }10\sqrt[4]{8}}. \end{align*}

~mathboy282

Solution 3 - ruler (last effort)

Since we are given that the diagram is drawn to scale, we can use a ruler to estimate the approximate length of $AB$.

We can start by measuring the lengths of the rectangle with area $1$, which may vary per viewing medium. For the sake of the solution, we will use side lengths ~$\frac{7}{10}$ cm and ~$\frac{5}{10}$ cm.

To get the scale ratio from centimeters to the units in the problem, we need to find a ratio $x$ such that $\frac{7x}{10}\cdot\frac{5x}{10}=1$

Solving this equation, we get $x=\frac{10}{\sqrt{35}}$

We then measure out the length of $AB$(will vary), to get ~$10.2$ cm. We multiply this length by our early ratio to get $\frac{10 \cdot 10.2}{\sqrt{35}} = \frac{102}{\sqrt{35}} \approx \frac{102}{6} \approx 17$

The answer choice closest to this would be $10\sqrt[4]{8}\approx16.8$, so therefore the closest answer is $\boxed{\textbf{(D) }10\sqrt[4]{8}}$.

~shreyan.chethan

Remark

The specific numbers used in the solution above vary per test medium, but the method should still work

Video Solution by Pi Academy

https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE

Video Solution 1 by Power Solve == https://youtu.be/8abGnAJZ3AM

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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