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 TopicsToolkit 48 — Logarithms: Definition and PropertiesToolkit 55 — Shoelace TheoremHints (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) log273
