Difference between revisions of "2023 AMC 10A Problems/Problem 4"
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− | + | Without Loss of Generality, let's assume that this quadrilateral is a trapezoid. We can assume this as if we inscribe a trapezoid in a triangle, the base can be the diameter of the circle which is the longest chord in the circle, therefore maximizing the side length. By Brahmagupta's Formula, the area of the quadrilateral 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 quadrilateral is <math>26</math>, then the semi-perimeter will be <math>13</math>. The area of the quadrilateral 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 be in this quadrilateral is <math>\boxed {\textbf{(D) 12}}</math> | |
~[https://artofproblemsolving.com/wiki/index.php/User:South South] | ~[https://artofproblemsolving.com/wiki/index.php/User:South South] |
Revision as of 21:36, 17 April 2024
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 sum of the 2 shorter sides must always be longer than the longest side. This is because if the longest side were to be as long as the sum of the other sides, or longer, we would only have a line.
Similarly, for a convex quadrilateral, the sum of 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 largest length that satisfies this is
~not_slay
~slight edits by e___
Solution 3 (Fast)
Without Loss of Generality, let's assume that this quadrilateral is a trapezoid. We can assume this as if we inscribe a trapezoid in a triangle, the base can be the diameter of the circle which is the longest chord in the circle, therefore maximizing the side length. By Brahmagupta's Formula, the area of the quadrilateral is defined by where is the semi-perimeter. If the perimeter of the quadrilateral is , then the semi-perimeter will be . The area of the quadrilateral 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 be in this quadrilateral is
Solution 4
This is an AMC 10 problem 4, so there is no need for any complex formulas. The largest singular side length from a quadrilateral comes from a trapezoid. So we can set the sides of the trapezoid equal to . Next we can split the trapezoid into triangles, where each base length of the triangle equals . So the top side equals , and the bottom side length equals ~ kabbybear
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=YBa-pxkomHrJErSz&t=597
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=UDhZqXI2A6Y
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=Iu0AJ2rof7k
Video Solution (easy to digest) by Power Solve
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.