Difference between revisions of "2024 AMC 8 Problems/Problem 11"
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+ | ==Problem== | ||
+ | The coordinates of <math>\triangle ABC</math> are <math>A(5,7)</math>, <math>B(11,7)</math>, and <math>C(3,y)</math>, with <math>y>7</math>. The area of <math>\triangle ABC</math> is 12. What is the value of <math>y</math>? | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | draw((3,11)--(11,7)--(5,7)--(3,11)); | ||
+ | |||
+ | dot((5,7)); | ||
+ | label("$A(5,7)$",(5,7),S); | ||
+ | |||
+ | dot((11,7)); | ||
+ | label("$B(11,7)$",(11,7),S); | ||
+ | |||
+ | dot((3,11)); | ||
+ | label("$C(3,y)$",(3,11),NW); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | |||
+ | <math>\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12</math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | The triangle has base <math>6,</math> which means its height satisfies | ||
+ | <cmath>\dfrac{6h}{2}=3h=12.</cmath> | ||
+ | This means that <math>h=4, </math> so the answer is <math>7+4=\boxed{(D) 11}</math> | ||
+ | |||
+ | ==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. | ||
+ | |||
+ | <math>\frac{1}{2}|6y - 42| = 12</math> | ||
+ | |||
+ | <math>|6y-42| = 24</math> | ||
+ | |||
+ | <math>6y - 42 = 24</math> OR <math>6y - 42 = -24</math> | ||
+ | |||
+ | <math>6y = 66</math> OR <math>6y = 18</math> | ||
+ | |||
+ | <math>y = 11</math> OR <math>y = 3</math> | ||
+ | |||
+ | However, since, as stated in the problem, <math>y > 7</math>, our only valid solution is <math>\boxed{\textbf{(D)} 11}</math>. | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] (again!) | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/BaE00H2SHQM?si=qhPbhu8o5hamBrtb&t=2315 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==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 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2024|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 17:11, 18 August 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 (easy to digest) by Power Solve
- 7 Video Solution by NiuniuMaths (Easy to understand!)
- 8 Video Solution 3 by SpreadTheMathLove
- 9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 10 Video Solution by Interstigation
- 11 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 (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
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.