Difference between revisions of "2023 AMC 10A Problems/Problem 4"
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==Solution 1== | ==Solution 1== | ||
| − | Lets 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=\text{\boxed{(D)12}}</math> | + | Lets 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=\text{\boxed{(D)12}}</math>. |
| + | |||
| + | ~zhenghua | ||
| + | |||
| + | == See Also == | ||
| + | {{AMC10 box|year=2022|ab=A|num-b=3|num-a=5}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:10, 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?
![\[\textbf{(A)}~9\qquad\textbf{(B)}~10\qquad\textbf{(C)}~11\qquad\textbf{(D)}~12\qquad\textbf{(E)}~13\]](http://latex.artofproblemsolving.com/c/a/a/caa0c870b300a94184a7cea172ecb5dba7a38f2c.png)
Solution 1
Lets 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
See Also
| 2022 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 | ||
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