AMC 12B 2024 (Problem 15)A triangle in the coordinate plane has vertices A(log21,log22)A(\log_2 1,\log_2 2)A(log21,log22), B(log23,log24)B(\log_2 3,\log_2 4)B(log23,log24), and C(log27,log28)C(\log_2 7,\log_2 8)C(log27,log28). What is the area of △ABC\triangle ABC△ABC?(A) log237\text{(A)}\;\log_2\frac{\sqrt{3}}{7}(A)log273(B) log237\text{(B)}\;\log_2\frac{3}{\sqrt{7}}(B)log273(C) log273\text{(C)}\;\log_2\frac{7}{\sqrt{3}}(C)log237(D) log2117\text{(D)}\;\log_2\frac{11}{\sqrt{7}}(D)log2711(E) log2113\text{(E)}\;\log_2\frac{11}{\sqrt{3}}(E)log2311Related TopicsCoreToolkit 55 — Shoelace TheoremMajorToolkit 48 — Logarithms: Definition and PropertiesCheck AnswerYour answer:ABCDECheckHints (3)Hint 1log21=0,log22=1,log24=2,log28=3\log_2 1=0,\quad \log_2 2=1,\quad \log_2 4=2,\quad \log_2 8=3log21=0,log22=1,log24=2,log28=3Hint 2[ABC]=12∣0⋅2+3log23+log27−log23−2log27−3⋅0∣[ABC]=\frac12\left|0\cdot2+3\log_2 3+\log_2 7-\log_2 3-2\log_2 7-3\cdot0\right|[ABC]=21∣0⋅2+3log23+log27−log23−2log27−3⋅0∣=12∣2log23−log27∣=\frac12\left|2\log_2 3-\log_2 7\right|=21∣2log23−log27∣Hint 3[ABC]=12∣2log23−log27∣=12∣log29−log27∣=12∣log297∣=12log297=log237[ABC]=\frac12\left|2\log_2 3-\log_2 7\right|=\frac12\left|\log_2 9-\log_2 7\right|=\frac12\left|\log_2\frac97\right|=\frac12\log_2\frac97=\log_2\frac{3}{\sqrt7}[ABC]=21∣2log23−log27∣=21∣log29−log27∣=21log279=21log279=log273Final Answer(B) log237(B)\ \log_2\frac{3}{\sqrt7}(B) log273Related Problems (10)AMC 12B 2025 (Problem 7)AMC 12B 2024 (Problem 8)AMC 12A 2024 (Problem 8)AMC 12A 2022 (Problem 14)AMC 12A 2023 (Problem 19)AMC 12A 2025 (Problem 7)AMC 12A 2025 (Problem 17)AIME I 2026 (Problem 6)View all related problems →
