Difference between revisions of "2024 AMC 10A Problems/Problem 16"
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==Problem== | ==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 <math>AB</math>? | 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>? | ||
+ | |||
+ | [[File:Screenshot 2024-11-08 2.08.49 PM.png]] | ||
<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 1== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~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 | ||
+ | |||
+ | ==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}} |
Revision as of 21:48, 13 November 2024
Contents
[hide]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 ?
Solution 1
Using the rectangle with area , let its short side be and the long side be . 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 is times that of the rectangle with area ), as they are all similar to each other.
The side opposite on the large rectangle is hence written as . However, can be written as . Since the two lengths are equal, we can write , or . Therefore, we can write .
Since , we have , which we can evaluate as . From this, we can plug back in to to find . Substituting into , we have which can be evaluated to .
~i_am_suk_at_math_2
Remark
We know that the area is an integer, so after finding , AB must contain a fourth root. The only such option is .
Solution 2
Let the rectangle's height be the length The entire rectangle has an area of We will be using this fact for ratios.
Note that the short side of the rectangle with area 32 will have a height of We use because it is apparent that the height of the rectangle with area is the shorter side, corresponding with
Similarly, the long side of the rectangle with area 36 has a height of
Noting that the total height of the big rectangle has height we have the equation
Since the area is equal to 200, we have:
~mathboy282
Solution 3 - ruler (last effort)
Given that the diagram is drawn to scale, we can use a ruler to estimate the length of .
We start by measuring the lengths of the rectangle with area , which may vary per viewing medium. For the sake of the solution, we use side lengths ~ cm and ~ cm.
To get the scale ratio from centimeters to the units in the problem, we need to find a ratio such that
Solving this equation, we get
We then measure the length of (will vary), to get ~ cm. We multiply this length by our early ratio to get
The answer choice closest to this would be , so therefore the closest answer is .
~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
Video Solution 3 by Innovative Minds (Similar to Solution 1 above)
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.