Difference between revisions of "2023 AMC 12A Problems/Problem 4"
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| − | A quadrilateral has all | + | ==Problem== |
| + | 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> | ||
| + | |||
| + | ==Solution 1== | ||
| + | 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> | ||
| + | |||
| + | ~zhenghua | ||
| + | |||
| + | ==Solution 2== | ||
| + | Say the chosen side is <math>a</math> and the other sides are <math>b,c,d</math>. | ||
| + | |||
| + | 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> | ||
| + | |||
| + | ~not_slay | ||
| + | |||
| + | == Solution 3 (Fast) == | ||
| + | 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> | ||
| + | |||
| + | ~[https://artofproblemsolving.com/wiki/index.php/User:South South] | ||
| + | |||
| + | == See Also == | ||
| + | {{AMC10 box|year=2023|ab=A|num-b=3|num-a=5}} | ||
| + | {{AMC12 box|year=2023|ab=A|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Revision as of 22:06, 9 November 2023
Problem
A quadrilateral has all integer sides lengths, a perimeter of 
, and one side of length 
. What is the greatest possible length of one side of this quadrilateral?
Solution 1
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 
~zhenghua
Solution 2
Say the chosen side is 
and the other sides are 
.
By the Generalised Polygon Inequality, 
. We also have 
.
Combining these two, we get 
.
The smallest length that satisfies this is 
~not_slay
Solution 3 (Fast)
By Brahmagupta's Formula, the area of the rectangle is defined by 
where 
is the semi-perimeter. If the perimeter of the rectangle is , then the semi-perimeter will be 
. The area of the rectangle must be positive so the difference between the semi-perimeter and a side length must be greater than 
as otherwise, the area will be or negative. Therefore, the longest a side can possibly be in this rectangle is 
See Also
| 2023 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
| 2023 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
