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

(Never Gonna Give You Up)
 
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Never Gonna Give You Up
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==Problem==
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Song by
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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 <math>AB</math>?
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Rick Astley
 
  
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We're no strangers to love
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[[File:Screenshot 2024-11-08 2.08.49 PM.png]]
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You know the rules and so do I (do I)
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A full commitment's what I'm thinking of
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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>
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You wouldn't get this from any other guy
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I just wanna tell you how I'm feeling
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==Solution 1==
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Gotta make you understand
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Never gonna give you up
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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.
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Never gonna let you down
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Never gonna run around and desert you
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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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Never gonna make you cry
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Never gonna say goodbye
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Since <math>xy=1</math>, we have <math>(x\sqrt{2})(x) = 1</math>, which we can evaluate <math>x</math> as <math>x=\frac{1}{\sqrt[4]{2}}</math>. From this, we can plug back in to <math>xy=1</math> to find <math>y=\sqrt[4]{2}</math>. Substituting into <math>AB</math>, we have <math>AB = 4y\sqrt{2}+12x = 4(\sqrt[4]{2})(\sqrt{2})+\frac{12}{\sqrt[4]{2}}</math> which can be evaluated to <math>\boxed{\textbf{(D) }10\sqrt[4]{8}}</math>.
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Never gonna tell a lie and hurt you
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We've known each other for so long
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~i_am_suk_at_math_2
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Your heart's been aching, but you're too shy to say it (say it)
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Inside, we both know what's been going on (going on)
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===Remark===
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We know the game and we're gonna play it
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And if you ask me how I'm feeling
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We know that the area is an integer, so after finding <math>y=x\sqrt{2}</math>, AB must contain a fourth root. The only such option is <math>\boxed{\textbf{(D) }10\sqrt[4]{8}}</math>.
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Don't tell me you're too blind to see
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Never gonna give you up
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==Solution 2==
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Never gonna let you down
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Let the rectangle's height be <math>x,</math> the length <math>AB=y.</math> The entire rectangle has an area of <math>200.</math> We will be using this fact for ratios.
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Never gonna run around and desert you
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Never gonna make you cry
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Note that the short side of the rectangle with area 32 will have a height of <math>\sqrt{\frac{32}{200}}\cdot x = \frac{2}{5}x.</math> We use <math>x</math> because it is apparent that the height of the rectangle with area <math>32</math> is the shorter side, corresponding with <math>x.</math>
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Never gonna say goodbye
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Never gonna tell a lie and hurt you
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Similarly, the long side of the rectangle with area 36 has a height of <math>\sqrt{\frac{36}{200}}\cdot y = \frac{3\sqrt{2}}{10}y.</math>
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Never gonna give you up
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Never gonna let you down
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Noting that the total height of the big rectangle has height <math>x,</math> we have the equation <math>\frac25 x + \frac{3\sqrt{2}}{10}y = x \Rightarrow x=\frac{y}{\sqrt{2}}.</math>
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Never gonna run around and desert you
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Never gonna make you cry
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Since the area <math>xy=\frac{y^2}{\sqrt{2}}</math> is equal to 200, we have:
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Never gonna say goodbye
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Never gonna tell a lie and hurt you
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<cmath>\begin{align*}
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(Ooh, give you up)
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y&=\sqrt{200\sqrt{2}} \\
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(Ooh, give you up)
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&=\sqrt{100\sqrt{8}} \\
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(Ooh) Never gonna give, never gonna give (give you up)
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&=\boxed{\textbf{(D) }10\sqrt[4]{8}}.
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(Ooh) Never gonna give, never gonna give (give you up)
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\end{align*}</cmath>
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We've known each other for so long
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Your heart's been aching, but you're too shy to say it (to say it)
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~mathboy282
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Inside, we both know what's been going on (going on)
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We know the game and we're gonna play it
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==Solution 3 - ruler (last effort)==
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I just wanna tell you how I'm feeling
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Gotta make you understand
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Given that the diagram is drawn to scale, we can use a ruler to estimate the length of <math>AB</math>.
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Never gonna give you up
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Never gonna let you down
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We start by measuring the lengths of the rectangle with area <math>1</math>, which may vary per viewing medium. For the sake of the solution, we use side lengths ~<math>\frac{7}{10}</math> cm and ~<math>\frac{5}{10}</math> cm.
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Never gonna run around and desert you
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Never gonna make you cry
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To get the scale ratio from centimeters to the units in the problem, we need to find a ratio <math>x</math> such that <math>\frac{7x}{10}\cdot\frac{5x}{10}=1</math>
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Never gonna say goodbye
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Never gonna tell a lie and hurt you
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Solving this equation, we get <math>x=\frac{10}{\sqrt{35}}</math>
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Never gonna give you up
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Never gonna let you down
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We then measure the length of <math>AB</math>(will vary), to get ~<math>10.2</math> cm. We multiply this length by our early ratio to get <math>\frac{10 \cdot 10.2}{\sqrt{35}} = \frac{102}{\sqrt{35}} \approx \frac{102}{6} \approx 17</math>
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Never gonna run around and desert you
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Never gonna make you cry
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The answer choice closest to this would be <math>10\sqrt[4]{8}\approx16.8</math>, so therefore the closest answer is <math>\boxed{\textbf{(D) }10\sqrt[4]{8}}</math>.
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Never gonna say goodbye
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Never gonna tell a lie and hurt you
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~shreyan.chethan
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Never gonna give you up
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Never gonna let you down
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==Remark==
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Never gonna run around and desert you
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The specific numbers used in the solution above vary per test medium, but the method should still work.
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Never gonna make you cry
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Never gonna say goodbye
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== Video Solution 1 by Pi Academy ==
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Never gonna tell a lie and hurt you
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https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
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== Video Solution 2 by Power Solve ==
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https://youtu.be/8abGnAJZ3AM
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== Video Solution 3 by Innovative Minds (Similar to Solution 1 above)==
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https://youtu.be/EepOGN0_Rgw
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=6SQ74nt3ynw
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==See also==
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{{AMC10 box|year=2024|ab=A|num-b=15|num-a=17}}
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{{MAA Notice}}

Revision as of 21:48, 13 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 contain 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)

Given that the diagram is drawn to scale, we can use a ruler to estimate the length of $AB$.

We start by measuring the lengths of the rectangle with area $1$, which may vary per viewing medium. For the sake of the solution, we 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 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 1 by Pi Academy

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

Video Solution 2 by Power Solve

https://youtu.be/8abGnAJZ3AM

Video Solution 3 by Innovative Minds (Similar to Solution 1 above)

https://youtu.be/EepOGN0_Rgw

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

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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