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Difference between revisions of "2024 AMC 8 Problems/Problem 11"

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==Problem==
 
==Problem==
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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>?
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<asy>
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draw((3,11)--(11,7)--(5,7)--(3,11));
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dot((5,7));
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label("$A(5,7)$",(5,7),S);
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dot((11,7));
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label("$B(11,7)$",(11,7),S);
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dot((3,11));
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label("$C(3,y)$",(3,11),NW);
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</asy>
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<math>\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12</math>
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==Solution 1==
 
==Solution 1==
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D
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The triangle has base <math>6,</math> which means its height satisfies
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<cmath>\dfrac{6h}{2}=3h=12.</cmath>
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This means that <math>h=4, </math> so the answer is <math>7+4=\boxed{(D) 11}</math>
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==Solution 2==
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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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<math>\frac{1}{2}|6y - 42| = 12</math>
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<math>|6y-42| = 24</math>
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<math>6y - 42 = 24</math> OR <math>6y - 42 = -24</math>
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<math>6y = 66</math> OR <math>6y = 18</math>
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<math>y = 11</math> OR <math>y = 3</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>.
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~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
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==Solution 3==
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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!!!)==
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https://youtu.be/BaE00H2SHQM?si=qhPbhu8o5hamBrtb&t=2315
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~Math-X
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==Video Solution  (easy to digest) by Power Solve==
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https://www.youtube.com/watch?v=2UIVXOB4f0o
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 +
 
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==Video Solution by NiuniuMaths (Easy to understand!)==
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https://www.youtube.com/watch?v=V-xN8Njd_Lc
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 +
~NiuniuMaths
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==Video Solution 3 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=RRTxlduaDs8
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== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
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https://www.youtube.com/watch?v=-64aBL-lEVg
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==Video Solution by Interstigation==
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https://youtu.be/ktzijuZtDas&t=1063
 +
 
 +
==See Also==
 +
{{AMC8 box|year=2024|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 17:11, 18 August 2024

Problem

The coordinates of $\triangle ABC$ are $A(5,7)$, $B(11,7)$, and $C(3,y)$, with $y>7$. The area of $\triangle ABC$ is 12. What is the value of $y$?

[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]


$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$

Solution 1

The triangle has base $6,$ which means its height satisfies \[\dfrac{6h}{2}=3h=12.\] This means that $h=4,$ so the answer is $7+4=\boxed{(D) 11}$

Solution 2

By the Shoelace Theorem, $\triangle ABC$ has area \[\frac{1}{2}|(y \cdot 11 + 7 \cdot 5 + 7 \cdot 3) - (3 \cdot 7 + 11 \cdot 7 + 5 \cdot y)| = \frac{1}{2}|(11y + 56) - (98 + 5y)| = \frac{1}{2}|6y - 42|\]. From the problem, this is equal to $12$. We now solve for y.

$\frac{1}{2}|6y - 42| = 12$

$|6y-42| = 24$

$6y - 42 = 24$ OR $6y - 42 = -24$

$6y = 66$ OR $6y = 18$

$y = 11$ OR $y = 3$

However, since, as stated in the problem, $y > 7$, our only valid solution is $\boxed{\textbf{(D)} 11}$.

~ cxsmi

Solution 3

As in the figure, the triangle is determined by the vectors $\begin{bmatrix}-2 \\ y-7\end{bmatrix}$ and $\begin{bmatrix}6\\0\end{bmatrix}$. 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 $\frac{1}{2}|\begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix}| = 12 \implies \begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix} = \pm 24$. Expanding the determinants, we find that $-6(y-7) = 24$ or $-6(y-7) = -24$. Solving each equation individually, we find that $y = 3$ or $y = 11$. However, the problem states that $y > 7$, so the only valid solution is $\boxed{\textbf{(D)} 11}$.

~ 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

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