Difference between revisions of "2023 AMC 10A Problems/Problem 4"
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| + | ==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>. | ||
| + | Combining these two, we get <math>a<26-a\Rightarrow a<13</math>. | ||
| + | The smallest length that satisfies this is <math>a=\boxed {\textbf{(D) 12}}</math> | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2023|ab=A|num-b=3|num-a=5}} | {{AMC10 box|year=2023|ab=A|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:01, 9 November 2023
Contents
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 
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 | ||
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