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2024 AMC 12B Problem 15

Problem 15 of 25IntermediateAlgebraGeometry

A triangle in the coordinate plane has vertices A(log21,log22),A(\log_2 1, \log_2 2), B(log23,log24),B(\log_2 3, \log_2 4), and C(log27,log28).C(\log_2 7, \log_2 8). What is the area of ABC?\triangle ABC?

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Solution

The vertices are A=(0,1),A = (0, 1), B=(log23,2),B = (\log_2 3, 2), C=(log27,3).C = (\log_2 7, 3). By the shoelace formula, [ABC]=122log23log27. \begin{gathered} [\triangle ABC]\\ {}=\tfrac12\,\bigl|2\log_2 3-\log_2 7\bigr|. \end{gathered} This equals 12log297=log297=log237.\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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Concepts: shoelace formula · logarithm · triangle area

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.