Difference between revisions of "2024 AMC 10A Problems/Problem 16"
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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>. | 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>, so <math>x=\frac{1}{\sqrt[4]{2}}</math>, and <math>y=\sqrt[4]{2}</math>. Therefore, <math>AB = 4y\sqrt{2}+12x = 4(\sqrt[4]{2})(\sqrt{2})+\frac{12}{\sqrt[4]{2}} | + | Since <math>xy=1</math>, we have <math>(x\sqrt{2})(x) = 1</math>, so <math>x=\frac{1}{\sqrt[4]{2}}</math>, and <math>y=\sqrt[4]{2}</math>. Therefore, <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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| + | ~i_am_suk_at_math_2 | ||
==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 18:01, 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 
?
![$\textbf{(A) }4+4\sqrt5\qquad\textbf{(B) }10\sqrt2\qquad\textbf{(C) }5+5\sqrt5\qquad\textbf{(D) }10\sqrt[4]{8}\qquad\textbf{(E) }20$](http://latex.artofproblemsolving.com/2/f/a/2fada6ff95f8cf1420e501b8f1b4aa0030154bba.png)
Solution
Using the rectangle with side length 
, 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 
, so ![$x=\frac{1}{\sqrt[4]{2}}$](http://latex.artofproblemsolving.com/f/7/8/f782abc4ce6d469063f7f3e3fec359411daca3e4.png)
, and ![$y=\sqrt[4]{2}$](http://latex.artofproblemsolving.com/c/c/c/cccf6318bd19a375ddd42202ab59ce80c2d2c000.png)
. Therefore, ![$AB = 4y\sqrt{2}+12x = 4(\sqrt[4]{2})(\sqrt{2})+\frac{12}{\sqrt[4]{2}}$](http://latex.artofproblemsolving.com/f/1/2/f12d6396cd9b4fb592b8e3f888d02b5d7cd9a0a1.png)
which can be evaluated to ![$\boxed{\textbf{(D) }10\sqrt[4]{8}}$](http://latex.artofproblemsolving.com/b/1/e/b1e4021ab0e1e65adb3cd0c843ca0c66498cdefc.png)
.
~i_am_suk_at_math_2
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. 
