Difference between revisions of "2024 AMC 8 Problems/Problem 11"
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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] |
==Solution 4== | ==Solution 4== |
Revision as of 12:38, 4 April 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution (easy to digest) by Power Solve
- 7 Video Solution by Math-X (First understand the problem!!!)
- 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 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
Label point as the point at which . We now have , where the brackets denote areas. On the right hand side, both of these triangles are right, so we can just compute the two sides of each triangle. The two side lengths of are and . The two side lengths of are and Now,
Dividing by gives so
-Benedict T (countmath1)
Solution 3
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 4
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 (easy to digest) by Power Solve
https://www.youtube.com/watch?v=2UIVXOB4f0o
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=qhPbhu8o5hamBrtb&t=2315
~Math-X
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
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.