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Difference between revisions of "2023 AMC 10A Problems/Problem 4"
(→‎Video Solution by Math-X (First understand the problem!!!))
 
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~zhenghua
 
~zhenghua
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Sidenote: If there weren't a restriction on integer side lengths, the answer would be the decimal just less than 13, so the sum of the other 3 sides could be just more than 13. That would make the longest side 12.99999..., stopping at who knows how many 9's.
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~sidenote by mihikamishra
  
 
==Solution 2==
 
==Solution 2==
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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>.
  
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The smallest length that satisfies this is <math>a=\boxed {\textbf{(D) 12}}</math>
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The largest length that satisfies this is <math>a=\boxed {\textbf{(D) 12}}</math>
  
 
~not_slay
 
~not_slay
  
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== Solution 3 (Fast) ==
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~slight edits by e___
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The quadrilateral can by cyclic only when it is an isosceles triangle. Without Loss of Generality, lets 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>
 
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~[https://artofproblemsolving.com/wiki/index.php/User:South South]
 
  
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== Solution 4 ==
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== Solution 3 ==
 
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 <math>2</math> sides of the trapezoid equal to <math>4</math>. Next we can split the trapezoid into <math>5</math> triangles, where each base length of the triangle equals <math>4</math>. So the top side equals <math>8</math>, and the bottom side length equals <math>4+4+4</math> <math>=</math> <math>\boxed {\textbf{(D) 12}}</math>
 
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 <math>2</math> sides of the trapezoid equal to <math>4</math>. Next we can split the trapezoid into <math>5</math> triangles, where each base length of the triangle equals <math>4</math>. So the top side equals <math>8</math>, and the bottom side length equals <math>4+4+4</math> <math>=</math> <math>\boxed {\textbf{(D) 12}}</math>
 
~ kabbybear
 
~ kabbybear
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</asy>
 
</asy>
  
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==Video Solution by Little Fermat==
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https://youtu.be/h2Pf2hvF1wE?si=3Dc3LOUcyY0QLrPs&t=507
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~little-fermat
 
==Video Solution by Math-X (First understand the problem!!!)==
 
==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/cMgngeSmFCY?si=YBa-pxkomHrJErSz&t=597
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https://youtu.be/GP-DYudh5qU?si=ZOEgPD6mg6z9b02s&t=677
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 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
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https://www.youtube.com/watch?v=UDhZqXI2A6Y
  
 
==Video Solution==
 
==Video Solution==
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==Video Solution (easy to digest) by Power Solve==
 
==Video Solution (easy to digest) by Power Solve==
 
https://youtu.be/Od1Spf3TDBs
 
https://youtu.be/Od1Spf3TDBs
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==Video Solution (Fast and Easy) by Dr.Google (YT: Pablo's Math)==
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https://youtu.be/S3LquYpHrsY
  
 
== 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}}

Latest revision as of 06:18, 5 November 2024

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

~zhenghua

Sidenote: If there weren't a restriction on integer side lengths, the answer would be the decimal just less than 13, so the sum of the other 3 sides could be just more than 13. That would make the longest side 12.99999..., stopping at who knows how many 9's.

~sidenote by mihikamishra

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

~not_slay

~slight edits by e___

Solution 3

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 $2$ sides of the trapezoid equal to $4$. Next we can split the trapezoid into $5$ triangles, where each base length of the triangle equals $4$. So the top side equals $8$, and the bottom side length equals $4+4+4$ $=$ $\boxed {\textbf{(D) 12}}$ ~ kabbybear

[asy] size(120); draw((0,0)--(2,6),red); draw((2,6)--(5,6),red); draw((5,6)--(7,0),red); draw((0,0)--(7,0),red); draw((2,6)--(2.67,0),red); draw((2.67,0)--(3.5,6),red); draw((3.5,6)--(4.67,0),red); draw((4.67,0)--(5,6),red); [/asy]

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=3Dc3LOUcyY0QLrPs&t=507 ~little-fermat

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

https://youtu.be/GP-DYudh5qU?si=ZOEgPD6mg6z9b02s&t=677

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=UDhZqXI2A6Y

Video Solution

https://youtu.be/HCUAbodk_NA

~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

https://youtu.be/Od1Spf3TDBs

Video Solution (Fast and Easy) by Dr.Google (YT: Pablo's Math)

https://youtu.be/S3LquYpHrsY

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

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