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Difference between revisions of "2024 AMC 12B Problems/Problem 15"

(Created page with "==Solution (Shoelace Theorem)== We rewrite: <math>A(0,1)</math> <math>B(\log _{2} 3, 2)</math> <math>C(\log _{2} 7, 3)</math> From here we setup Shoelace Theorem and obtain: <...")
 
 
(14 intermediate revisions by 2 users not shown)
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==Solution (Shoelace Theorem)==
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==Problem==
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A triangle in the coordinate plane has vertices <math>A(\log_21,\log_22)</math>, <math>B(\log_23,\log_24)</math>, and <math>C(\log_27,\log_28)</math>. What is the area of <math>\triangle ABC</math>?
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<math>
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\textbf{(A) }\log_2\frac{\sqrt3}7\qquad
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\textbf{(B) }\log_2\frac3{\sqrt7}\qquad
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\textbf{(C) }\log_2\frac7{\sqrt3}\qquad
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\textbf{(D) }\log_2\frac{11}{\sqrt7}\qquad
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\textbf{(E) }\log_2\frac{11}{\sqrt3}\qquad
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</math>
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==Solution 1 (Shoelace Theorem)==
 
We rewrite:
 
We rewrite:
 
<math>A(0,1)</math>
 
<math>A(0,1)</math>
 
<math>B(\log _{2} 3, 2)</math>
 
<math>B(\log _{2} 3, 2)</math>
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<math>C(\log _{2} 7, 3)</math>
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<math>C(\log _{2} 7, 3)</math>.
 +
 
 
From here we setup Shoelace Theorem and obtain:
 
From here we setup Shoelace Theorem and obtain:
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<math>\frac{1}{2}(2(\log _{2} 3) - log _{2} 7)</math>
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<math>\frac{1}{2}(2(\log _{2} 3) - log _{2} 7)</math>.
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Following log properties and simplifying gives (B).
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 +
Following log properties and simplifying gives <math>\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}</math>.
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 +
 
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~MendenhallIsBald
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==Solution 2 (Determinant)== 
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To calculate the area of a triangle formed by three points \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) on a Cartesian coordinate plane, you can use the following formula:
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<cmath>
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\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
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</cmath>
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The coordinates are:<math>A(0, 1)</math>, <math>B(\log_2 3, 2)</math>, <math>C(\log_2 7, 3)</math>
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Taking a numerical value into account:
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<cmath>
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\text{Area} = \frac{1}{2} \left| 0 \cdot (2 - 3) + \log_2 3 \cdot (3 - 1) + \log_2 7 \cdot (1 - 2) \right|
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</cmath>
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Simplify:
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<cmath>
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= \frac{1}{2} \left| 0 + \log_2 3 \cdot 2 + \log_2 7 \cdot (-1) \right|
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</cmath>
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<cmath>
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= \frac{1}{2} \left| \log_2 (3^2) - \log_2 7 \right|
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</cmath>
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<cmath>
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= \frac{1}{2} \left| \log_2 \frac{9}{7} \right|
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</cmath>
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Thus, the area is:<math>\text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right|</math> = <math>\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx]
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==Video Solution 1 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=jyupN3dT2yY&t=0s
  
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~PeterDoesPhysics
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==See also==
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{{AMC12 box|year=2024|ab=B|num-b=14|num-a=16}}
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{{MAA Notice}}

Latest revision as of 19:11, 15 November 2024

Problem

A triangle in the coordinate plane has vertices $A(\log_21,\log_22)$, $B(\log_23,\log_24)$, and $C(\log_27,\log_28)$. What is the area of $\triangle ABC$?

$\textbf{(A) }\log_2\frac{\sqrt3}7\qquad \textbf{(B) }\log_2\frac3{\sqrt7}\qquad \textbf{(C) }\log_2\frac7{\sqrt3}\qquad \textbf{(D) }\log_2\frac{11}{\sqrt7}\qquad \textbf{(E) }\log_2\frac{11}{\sqrt3}\qquad$


Solution 1 (Shoelace Theorem)

We rewrite: $A(0,1)$ $B(\log _{2} 3, 2)$ $C(\log _{2} 7, 3)$.

From here we setup Shoelace Theorem and obtain: $\frac{1}{2}(2(\log _{2} 3) - log _{2} 7)$.

Following log properties and simplifying gives $\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}$.


~MendenhallIsBald

Solution 2 (Determinant)

To calculate the area of a triangle formed by three points \( A(x_1, y_1) \), \( B(x_2, y_2) \), and \( C(x_3, y_3) \) on a Cartesian coordinate plane, you can use the following formula: \[\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|\] The coordinates are:$A(0, 1)$, $B(\log_2 3, 2)$, $C(\log_2 7, 3)$

Taking a numerical value into account: \[\text{Area} = \frac{1}{2} \left| 0 \cdot (2 - 3) + \log_2 3 \cdot (3 - 1) + \log_2 7 \cdot (1 - 2) \right|\] Simplify: \[= \frac{1}{2} \left| 0 + \log_2 3 \cdot 2 + \log_2 7 \cdot (-1) \right|\] \[= \frac{1}{2} \left| \log_2 (3^2) - \log_2 7 \right|\] \[= \frac{1}{2} \left| \log_2 \frac{9}{7} \right|\] Thus, the area is:$\text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right|$ = $\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}$

~Athmyx

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jyupN3dT2yY&t=0s

See also

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 AMC 12 Problems and Solutions

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