Difference between revisions of "2023 AMC 10A Problems/Problem 4"
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==Solution 1== | ==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 <math>\frac{26}{2}-1=13-1=\boxed {\textbf{(D) 12}}</math> | + | 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 <math>\frac{26}{2}-1=13-1=\boxed {\textbf{(D) 12}}</math> | ||
~zhenghua | ~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== | ==Solution 2== | ||
Line 16: | Line 22: | ||
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 | + | The largest length that satisfies this is <math>a=\boxed {\textbf{(D) 12}}</math> |
~not_slay | ~not_slay | ||
+ | |||
+ | ~slight edits by e___ | ||
== Solution 3 == | == 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> | |
+ | ~ 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 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 == | == 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 20:38, 22 October 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Video Solution by Math-X (First understand the problem!!!)
- 6 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 7 Video Solution
- 8 Video Solution by SpreadTheMathLove
- 9 Video Solution (easy to digest) by Power Solve
- 10 Video Solution (Fast and Easy) by Dr.Google (YT: Pablo's Math)
- 11 See Also
Problem
A quadrilateral has all integer sides lengths, a perimeter of , and one side of length . What is the greatest possible length of one side of this quadrilateral?
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
~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 and the other sides are .
By the Generalised Polygon Inequality, . We also have .
Combining these two, we get .
The largest length that satisfies this is
~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 sides of the trapezoid equal to . Next we can split the trapezoid into triangles, where each base length of the triangle equals . So the top side equals , and the bottom side length equals ~ kabbybear
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
~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
Video Solution (Fast and Easy) by Dr.Google (YT: Pablo's Math)
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.