2023 AMC 10A Problems/Problem 4

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

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.

Page

Toolbox

Search

2023 AMC 10A Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

See Also