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

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

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2023 AMC 10A Problems/Problem 4

Contents

Problem

Solution 1

Solution 2

Solution 3 (Fast)

See Also