Difference between revisions of "2024 AMC 8 Problems/Problem 11"
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==Solution 2== | ==Solution 2== | ||
− | + | By the Shoelace Theorem, <math>\triangle ABC</math> has area <cmath>\frac{1}{2}|(y \cdot 11 + 7 \cdot 5 + 7 \cdot 3) - (3 \cdot 7 + 11 \cdot 7 + 5 \cdot y)| = \frac{1}{2}|(11y + 56) - (98 + 5y)| = \frac{1}{2}|6y - 42|.</cmath> From the problem, this is equal to <math>12</math>. We now solve for y. | |
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− | By the | ||
<math>\frac{1}{2}|6y - 42| = 12</math> | <math>\frac{1}{2}|6y - 42| = 12</math> | ||
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However, since, as stated in the problem, <math>y > 7</math>, our only valid solution is <math>\boxed{\textbf{(D)} 11}</math>. | However, since, as stated in the problem, <math>y > 7</math>, our only valid solution is <math>\boxed{\textbf{(D)} 11}</math>. | ||
− | ~ cxsmi | + | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] |
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+ | ==Solution 3== | ||
+ | As in the figure, the triangle is determined by the vectors <math>\begin{bmatrix}-2 \\ y-7\end{bmatrix}</math> and <math>\begin{bmatrix}6\\0\end{bmatrix}</math>. Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined by those vectors; the triangle has half the area of that parallelogram. Then we must have that <math>\frac{1}{2}|\begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix}| = 12 \implies \begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix} = \pm 24</math>. Expanding the determinants, we find that <math>-6(y-7) = 24</math> or <math>-6(y-7) = -24</math>. Solving each equation individually, we find that <math>y = 3</math> or <math>y = 11</math>. However, the problem states that <math>y > 7</math>, so the only valid solution is <math>\boxed{\textbf{(D)} 11}</math>. | ||
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+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] (again!) | ||
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+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/BaE00H2SHQM?si=qhPbhu8o5hamBrtb&t=2315 | ||
+ | |||
+ | ~Math-X | ||
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+ | ==Video Solution (A Clever Explanation You’ll Get Instantly)== | ||
+ | https://youtu.be/5ZIFnqymdDQ?si=6FzUoSOA5moM-gDP&t=1191 | ||
+ | |||
+ | ~hsnacademy | ||
+ | |||
==Video Solution (easy to digest) by Power Solve== | ==Video Solution (easy to digest) by Power Solve== | ||
https://www.youtube.com/watch?v=2UIVXOB4f0o | https://www.youtube.com/watch?v=2UIVXOB4f0o | ||
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==Video Solution by NiuniuMaths (Easy to understand!)== | ==Video Solution by NiuniuMaths (Easy to understand!)== | ||
https://www.youtube.com/watch?v=V-xN8Njd_Lc | https://www.youtube.com/watch?v=V-xN8Njd_Lc | ||
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==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/ktzijuZtDas&t=1063 | https://youtu.be/ktzijuZtDas&t=1063 | ||
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+ | ==Video Solution by Daily Dose of Math (Certified, Simple, and Logical)== | ||
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+ | https://youtu.be/8GHuS5HEoWc | ||
+ | |||
+ | ~Thesmartgreekmathdude | ||
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+ | ==Video Solution by Dr. David== | ||
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+ | https://youtu.be/0O4Y3RHzcR4 | ||
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+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/_r1Zh4HGA7g | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2024|num-b=10|num-a=12}} | {{AMC8 box|year=2024|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:54, 20 November 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 (A Clever Explanation You’ll Get Instantly)
- 7 Video Solution (easy to digest) by Power Solve
- 8 Video Solution by NiuniuMaths (Easy to understand!)
- 9 Video Solution 3 by SpreadTheMathLove
- 10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 11 Video Solution by Interstigation
- 12 Video Solution by Daily Dose of Math (Certified, Simple, and Logical)
- 13 Video Solution by Dr. David
- 14 Video Solution by WhyMath
- 15 See Also
Problem
The coordinates of are , , and , with . The area of is 12. What is the value of ?
Solution 1
The triangle has base which means its height satisfies This means that so the answer is
Solution 2
By the Shoelace Theorem, has area From the problem, this is equal to . We now solve for y.
OR
OR
OR
However, since, as stated in the problem, , our only valid solution is .
~ cxsmi
Solution 3
As in the figure, the triangle is determined by the vectors and . Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined by those vectors; the triangle has half the area of that parallelogram. Then we must have that . Expanding the determinants, we find that or . Solving each equation individually, we find that or . However, the problem states that , so the only valid solution is .
~ cxsmi (again!)
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=qhPbhu8o5hamBrtb&t=2315
~Math-X
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=6FzUoSOA5moM-gDP&t=1191
~hsnacademy
Video Solution (easy to digest) by Power Solve
https://www.youtube.com/watch?v=2UIVXOB4f0o
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=RRTxlduaDs8
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=-64aBL-lEVg
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=1063
Video Solution by Daily Dose of Math (Certified, Simple, and Logical)
~Thesmartgreekmathdude
Video Solution by Dr. David
Video Solution by WhyMath
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.