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

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(→‎Solution 2 (Determinant))
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</cmath>
 
</cmath>
  
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The coordinates are:
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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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- <cmath>A(0, 1)</cmath>
 
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- <cmath>B(\log_2 3, 2)</cmath>
 
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- <cmath>C(\log_2 7, 3)</cmath>
 
  
 
Taking a numerical value into account:
 
Taking a numerical value into account:
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= \frac{1}{2} \left| 0 + \log_2 3 \cdot 2 + \log_2 7 \cdot (-1) \right|
 
= \frac{1}{2} \left| 0 + \log_2 3 \cdot 2 + \log_2 7 \cdot (-1) \right|
 
</cmath>
 
</cmath>
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<cmath>
 
<cmath>
 
= \frac{1}{2} \left| \log_2 (3^2) - \log_2 7 \right|
 
= \frac{1}{2} \left| \log_2 (3^2) - \log_2 7 \right|
 
</cmath>
 
</cmath>
−
 
 
<cmath>
 
<cmath>
 
= \frac{1}{2} \left| \log_2 \frac{9}{7} \right|
 
= \frac{1}{2} \left| \log_2 \frac{9}{7} \right|

Revision as of 05:27, 14 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 (B).


~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

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