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 2)
Line 26: Line 26:
  
 
Since the area <math>xy=\frac{y^2}{\sqrt{2}}</math> is equal to 200, we have:
 
Since the area <math>xy=\frac{y^2}{\sqrt{2}}</math> is equal to 200, we have:
−
\begin{align*}
+
 
−
y&=\sqrt{200\sqrt{2} \\
+
<cmath>\begin{align*}
 +
y&=\sqrt{200\sqrt{2}} \\
 
&=\sqrt{100\sqrt{8}} \\
 
&=\sqrt{100\sqrt{8}} \\
−
&=10\sqrt[4]{8}.
+
&=10\sqrt[4]{8}
−
\end{align*}
+
\end{align*}</cmath>
  
 
~mathboy282
 
~mathboy282

Revision as of 20:21, 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 $(x\sqrt{2})(x) = 1$, 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

Solution 2

Let the height of the rectangle by $x,$ 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 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 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}} \\ &=10\sqrt[4]{8} \end{align*}

~mathboy282

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=232276"