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Difference between revisions of "2024 AMC 8 Problems/Problem 11"
 
(9 intermediate revisions by 5 users not shown)
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==Solution 2==
 
==Solution 2==
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<asy>
 
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size(10cm);
 
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draw((5,7)--(11,7)--(3,11)--cycle);
 
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draw((3,11)--(3,7)--(5,7),red);
 
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draw((3,7.5)--(3.5,7.5)--(3.5,7));
 
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label("$A(5,7)$", (5,7),S);
 
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label("$B(11,7)$", (11,7),S);
 
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label("$C(3,y)$", (3,11),W);
 
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label("$D(3,7)$", (3,7),SW);
 
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</asy>
 
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Label point <math>D(3,7)</math> as the point at which <math>CD\perp DA</math>. We now have <math>[\triangle ABC] = [\triangle BCD] - [\triangle ACD]</math>, 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 <math>\triangle ACD</math> are <math>y-7</math> and <math>5-3=2</math>. The two side lengths of <math>\triangle BCD</math> are <math>y-7</math> and <math>11-3 = 8.</math> Now,
 
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<cmath>[\triangle ABC] = 12  = \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2  = 3(y-7)</cmath>
 
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Dividing by <math>3</math> gives <math>y -7 = 4,</math> so <math>y = \boxed{\textbf{(D)\ 11}}.</math>
 
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-Benedict T (countmath1)
 
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==Solution 3==
 
 
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.
 
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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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>.
  
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~ cxsmi
+
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
  
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==Solution 4==
+
==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</math>. Expanding the determinant and rewriting the absolute valuese, 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>.
+
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>.
  
−
~ cxsmi (again!)
+
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] (again!)
  
−
==Video Solution (easy to digest) by Power Solve==
+
==Video Solution (A Clever Explanation You’ll Get Instantly)==
−
https://www.youtube.com/watch?v=2UIVXOB4f0o
+
https://youtu.be/5ZIFnqymdDQ?si=6FzUoSOA5moM-gDP&t=1191
 +
 
 +
~hsnacademy
  
 
==Video Solution by Math-X (First understand the problem!!!)==
 
==Video Solution by Math-X (First understand the problem!!!)==
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~Math-X
 
~Math-X
 +
 +
==Video Solution  (easy to digest) by Power Solve==
 +
https://www.youtube.com/watch?v=2UIVXOB4f0o
 +
 +
 
==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
 +
 +
==Video Solution by Daily Dose of Math (Certified, Simple, and Logical)==
 +
 +
https://youtu.be/8GHuS5HEoWc
 +
 +
~Thesmartgreekmathdude
 +
 +
==Video Solution by Dr. David==
 +
 +
https://youtu.be/0O4Y3RHzcR4
  
 
==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 18:08, 19 September 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 (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=6FzUoSOA5moM-gDP&t=1191

~hsnacademy

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

Video Solution by Daily Dose of Math (Certified, Simple, and Logical)

https://youtu.be/8GHuS5HEoWc

~Thesmartgreekmathdude

Video Solution by Dr. David

https://youtu.be/0O4Y3RHzcR4

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. AMC logo.png

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