Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 3 - ruler (last effort))
 
(21 intermediate revisions by 11 users not shown)
Line 6: Line 6:
 
<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>
  
−
==Solution==
+
==Solution 1==
  
−
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.
+
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>.
Line 15: Line 15:
  
 
~i_am_suk_at_math_2
 
~i_am_suk_at_math_2
 +
 +
===Remark===
 +
 +
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>.
 +
 +
==Solution 2==
 +
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.
 +
 +
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>
 +
 +
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>
 +
 +
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>
 +
 +
Since the area <math>xy=\frac{y^2}{\sqrt{2}}</math> is equal to 200, we have:
 +
 +
<cmath>\begin{align*}
 +
y&=\sqrt{200\sqrt{2}} \\
 +
&=\sqrt{100\sqrt{8}} \\
 +
&=\boxed{\textbf{(D) }10\sqrt[4]{8}}.
 +
\end{align*}</cmath>
 +
 +
~mathboy282
 +
 +
==Solution 3 - ruler (last effort)==
 +
 +
Given that the diagram is drawn to scale, we can use a ruler to estimate the length of <math>AB</math>.
 +
 +
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.
 +
 +
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>
 +
 +
Solving this equation, we get <math>x=\frac{10}{\sqrt{35}}</math>
 +
 +
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>
 +
 +
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>.
 +
 +
~shreyan.chethan
 +
 +
Note:
 +
Never ever use this in an actual test
 +
 +
==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==
 
==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}}

Latest revision as of 15:41, 24 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

Note: Never ever use this in an actual test

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_16&oldid=235502"