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: <...")
 
(Solution 1 (Shoelace Theorem))
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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>
<math>C(\log _{2} 7, 3)</math>
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<math>C(\log _{2} 7, 3)</math>.
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From here we setup Shoelace Theorem and obtain:
 
From here we setup Shoelace Theorem and obtain:
<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).
 
Following log properties and simplifying gives (B).
  
~PeterDoesPhysics
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~MendenhallIsBald

Revision as of 01:06, 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