2024 AMC 12B Problem 15

Problem 15 of 25IntermediateAlgebra Geometry

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?

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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.

Tagged: shoelace formula · logarithm · triangle area