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

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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 (Shoelace Theorem)==
 
==Solution (Shoelace Theorem)==
 
We rewrite:
 
We rewrite:

Revision as of 01:01, 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 (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).

~PeterDoesPhysics

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