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

Revision as of 11:20, 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 $AB$ ?

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

Solution 1

Using the rectangle with 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

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:

$\begin{align*} y&=\sqrt{200\sqrt{2}} \\ &=\sqrt{100\sqrt{8}} \\ &=10\sqrt[4]{8} \end{align*}$

~mathboy282

Video Solution 1 by Power Solve

https://youtu.be/8abGnAJZ3AM

@@ 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>
-==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.
@@ Line 15: / Line 15: @@
 ~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 contains a fourth root. The only such option is <math>\boxed{\textbf{(D) }10\sqrt[4]{8}}</math>.
 ==Solution 2==

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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