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

(→‎Solution 1)
(→‎Solution 2)
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==Solution 2==
 
==Solution 2==
 
Say the chosen side is <math>a</math> and the other sides are <math>b,c,d</math>.
 
Say the chosen side is <math>a</math> and the other sides are <math>b,c,d</math>.
 +
 
By the Generalised Polygon Inequality, <math>a<b+c+d</math>. We also have <math>a+b+c+d=26\Rightarrow b+c+d=26-a</math>.
 
By the Generalised Polygon Inequality, <math>a<b+c+d</math>. We also have <math>a+b+c+d=26\Rightarrow b+c+d=26-a</math>.
 +
 
Combining these two, we get <math>a<26-a\Rightarrow a<13</math>.
 
Combining these two, we get <math>a<26-a\Rightarrow a<13</math>.
 +
 
The smallest length that satisfies this is <math>a=\boxed {\textbf{(D) 12}}</math>
 
The smallest length that satisfies this is <math>a=\boxed {\textbf{(D) 12}}</math>
 +
 +
~not_slay
  
 
== 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 21:01, 9 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

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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