Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 10A Problems/Problem 4"

(undo)
(Tag: Removed redirect)
Line 24: Line 24:
  
 
~[https://artofproblemsolving.com/wiki/index.php/User:South South]
 
~[https://artofproblemsolving.com/wiki/index.php/User:South South]
 +
 +
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/cMgngeSmFCY?si=YBa-pxkomHrJErSz&t=597
 +
  
 
== 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 14:58, 10 November 2023

Problem

A quadrilateral has all integer sides lengths, a perimeter of $26$, and one side of length $4$. What is the greatest possible length of one side of this quadrilateral?

$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$

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 $\frac{26}{2}-1=13-1=\boxed {\textbf{(D) 12}}$

~zhenghua

Solution 2

Say the chosen side is $a$ and the other sides are $b,c,d$.

By the Generalised Polygon Inequality, $a<b+c+d$. We also have $a+b+c+d=26\Rightarrow b+c+d=26-a$.

Combining these two, we get $a<26-a\Rightarrow a<13$.

The smallest length that satisfies this is $a=\boxed {\textbf{(D) 12}}$

~not_slay

Solution 3 (Fast)

By Brahmagupta's Formula, the area of the rectangle is defined by $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ where $s$ is the semi-perimeter. If the perimeter of the rectangle is $26$, then the semi-perimeter will be $13$. The area of the rectangle must be positive so the difference between the semi-perimeter and a side length must be greater than $0$ as otherwise, the area will be $0$ or negative. Therefore, the longest a side can possibly be in this rectangle is $\boxed {\textbf{(D) 12}}$

~South


Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/cMgngeSmFCY?si=YBa-pxkomHrJErSz&t=597


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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10A_Problems/Problem_4&oldid=201910"