Difference between revisions of "2023 AMC 10A Problems/Problem 4"
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+ | == Solution 3 == | ||
+ | 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>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 21:37, 9 November 2023
Contents
[hide]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
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 |
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