2024 AMC 12B Problem 15

Attempt Problem 15 of the 2024 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12B solutions, or check the answer key.

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15.

A triangle in the coordinate plane has vertices $A(\log_2 1, \log_2 2),$ $B(\log_2 3, \log_2 4),$ and $C(\log_2 7, \log_2 8).$ What is the area of $\triangle ABC?$

$\log_2 \dfrac{\sqrt3}{7}$

$\log_2 \dfrac{3}{\sqrt7}$

$\log_2 \dfrac{7}{\sqrt3}$

$\log_2 \dfrac{11}{\sqrt7}$

$\log_2 \dfrac{11}{\sqrt3}$

Answer: B

Concepts:shoelace formula logarithm triangle area

Difficulty rating: 1800

Solution:

The vertices are $A = (0, 1),$ $B = (\log_2 3, 2),$ $C = (\log_2 7, 3).$ By the shoelace formula, $\begin{gathered} [\triangle ABC]\\ {}=\tfrac12\,\bigl|2\log_2 3-\log_2 7\bigr|. \end{gathered}$

This equals $\tfrac12\log_2 \dfrac{9}{7} = \log_2 \sqrt{\tfrac{9}{7}} = \log_2 \dfrac{3}{\sqrt7}.$

Thus, the correct answer is B.

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