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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>?
| + | In how many ways can the integers <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, and <math>6</math> be arranged in a line so that the following statement is true? If <math>2</math> is not adjacent to <math>3</math>, then <math>3</math> is not adjacent to <math>4</math>. |
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| − | [[File:Screenshot 2024-11-08 2.08.49 PM.png]]
| + | <math>\textbf{(A)}~480 \qquad\textbf{(B)}~504 \qquad\textbf{(C)}~528 \qquad\textbf{(D)}~572 \qquad\textbf{(E)}~600</math> |
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| − | <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== |
| | + | We take the contrapositive, where the statement becomes "If <math>3</math> is adjacent to <math>4</math>, then <math>2</math> is adjacent to <math>3</math>." |
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| − | ==Solution 1== | + | If <math>3</math> is adjacent to <math>4</math>, then <math>2</math> is also adjacent to <math>3</math> so we can group the three numbers together in two ways: <math>234</math> or <math>432</math>. Then, the arrangements consists of this block of numbers and then three other numbers, which can be ordered in <math>4!</math> ways. Hence, in this case, the total count is <math>2 \cdot 4! = 48</math>. If <math>3</math> is not adjacent to <math>4</math>, there are no restrictions, so we can place <math>3</math> and <math>4</math> in <math>2 \cdot \tbinom{5}{2} = 20</math> ways and then place the rest in <math>4! = 24</math> ways, for a total of <math>480</math> configurations. |
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| − | 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.
| + | Overall, our total count is <math>480 + 48 = \boxed{\textbf{(C)}~528}</math>. |
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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>.
| + | ~joshualiu315 |
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| − | 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>.
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| − | ~i_am_suk_at_math_2 | |
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| − | ===Remark===
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| − | 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>.
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| − | ==Solution 2==
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| − | 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.
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| − | 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>
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| − | 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>
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| − | 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>
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| − | Since the area <math>xy=\frac{y^2}{\sqrt{2}}</math> is equal to 200, we have:
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| − | <cmath>\begin{align*}
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| − | y&=\sqrt{200\sqrt{2}} \\
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| − | &=\sqrt{100\sqrt{8}} \\
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| − | &=\boxed{\textbf{(D) }10\sqrt[4]{8}}.
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| − | \end{align*}</cmath>
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| − | ~mathboy282
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| − | ==Solution 3 - ruler (last effort)==
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| − | Given that the diagram is drawn to scale, we can use a ruler to estimate the length of <math>AB</math>.
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| − | 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.
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| − | 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>
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| − | Solving this equation, we get <math>x=\frac{10}{\sqrt{35}}</math>
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| − | 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>
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| − | 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>.
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| − | ~shreyan.chethan
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| − | Note:
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| − | Never ever use this in an actual test
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| − | ==Remark==
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| − | The specific numbers used in the solution above vary per test medium, but the method should still work.
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| − | == Video Solution 1 by Pi Academy ==
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| − | https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
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| − | == Video Solution 2 by Power Solve ==
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| − | https://youtu.be/8abGnAJZ3AM
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| − | == Video Solution 3 by Innovative Minds (Similar to Solution 1 above)==
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| − | https://youtu.be/EepOGN0_Rgw
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| − | ==Video Solution by SpreadTheMathLove==
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| − | https://www.youtube.com/watch?v=6SQ74nt3ynw
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| | ==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}} |
Problem
In how many ways can the integers 
, 
, 
, 
, 
, and 
be arranged in a line so that the following statement is true? If is not adjacent to , then is not adjacent to .
Solution
We take the contrapositive, where the statement becomes "If is adjacent to , then is adjacent to ."
If is adjacent to , then is also adjacent to so we can group the three numbers together in two ways: 
or 
. Then, the arrangements consists of this block of numbers and then three other numbers, which can be ordered in 
ways. Hence, in this case, the total count is 
. If is not adjacent to , there are no restrictions, so we can place and in 
ways and then place the rest in 
ways, for a total of 
configurations.
Overall, our total count is 
.
~joshualiu315
See also
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
, 
, 
, 
, 
, and 
be arranged in a line so that the following statement is true? If 
or 
. Then, the arrangements consists of this block of numbers and then three other numbers, which can be ordered in 
ways. Hence, in this case, the total count is 
. If 
ways and then place the rest in 
ways, for a total of 
configurations.
.
