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| − | ==Problem==
| + | #redirect[[2023 AMC 12A Problems/Problem 4]] |
| − | A quadrilateral has all integer sides lengths, a perimeter of <math>26</math>, and one side of length <math>4</math>. What is the greatest possible length of one side of this quadrilateral?
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| − | <math>\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13</math>
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| − | ==Solution 1==
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| − | Let's use the triangle inequality. We know that for a triangle, the 2 shorter sides must always be longer than the longest side. Similarly for a convex quadrilateral, the shortest 3 sides must always be longer than the longest side. Thus, the answer is <math>\frac{26}{2}-1=13-1=\boxed {\textbf{(D) 12}}</math>
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| − | ~zhenghua
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| − | ==Solution 2==
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| − | Say the chosen side is <math>a</math> and the other sides are <math>b,c,d</math>.
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| − | By the Generalised Polygon Inequality, <math>a<b+c+d</math>. We also have <math>a+b+c+d=26\Rightarrow b+c+d=26-a</math>.
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| − | Combining these two, we get <math>a<26-a\Rightarrow a<13</math>.
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| − | The smallest length that satisfies this is <math>a=\boxed {\textbf{(D) 12}}</math>
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| − | ~not_slay
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| − | == Solution 3 (Fast) ==
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| − | By Brahmagupta's Formula, the area of the rectangle is defined by <math>\sqrt{(s-a)(s-b)(s-c)(s-d)}</math> where <math>s</math> is the semi-perimeter. If the perimeter of the rectangle is <math>26</math>, then the semi-perimeter will be <math>13</math>. The area of the rectangle must be positive so the difference between the semi-perimeter and a side length must be greater than <math>0</math> as otherwise, the area will be <math>0</math> or negative. Therefore, the longest a side can possibly be in this rectangle is <math>\boxed {\textbf{(D) 12}}</math>
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| − | ~[https://artofproblemsolving.com/wiki/index.php/User:South South]
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| − | == See Also ==
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| − | {{AMC10 box|year=2023|ab=A|num-b=3|num-a=5}}
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| − | {{MAA Notice}}
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