AMC 12 B, 2024

Answer (B): Circumscribe $\triangle ABC$ by rectangle $AGCD$, with $D$ on the y-axis, and project point $B$ onto $\overline{AG}$ and $\overline{CG}$, producing points $E$ and $F$, respectively, as shown in the figure below.

The area of rectangle $AGCD$ is $2\log_2 7$, so the area of $\triangle ACG$ is $\log_2 7$. The requested area is
$$\text{Area}(\triangle ABC) = \text{Area}(\triangle ACG) - \text{Area}(\triangle ABE) - \text{Area}(\triangle BCF) - \text{Area}(\triangle BEGF).$$
Note that
$$\text{Area}(\triangle ABE) = \frac{1}{2} \log_2 3 \cdot 1 = \log_2 \sqrt{3},$$
$$\text{Area}(\triangle BCF) = \frac{1}{2} (\log_2 7 - \log_2 3) \cdot 1 = \log_2 \sqrt{\frac{7}{3}}, \text{ and}$$
$$\text{Area}(\triangle BEGF) = (\log_2 7 - \log_2 3) \cdot 1 = \log_2 \frac{7}{3}.$$
Therefore
$$\begin{aligned} \text{Area}(\triangle ABC) &= \log_2 7 - \log_2 \sqrt{3} - \log_2 \sqrt{\frac{7}{3}} - \log_2 \frac{7}{3} \\ &= \log_2 \left( \frac{7}{\sqrt{3} \cdot \sqrt{\frac{7}{3} \cdot \frac{7}{3}}} \right) \\ &= \log_2 \frac{3}{\sqrt{7}}. \end{aligned}$$

The area of triangle $\triangle ABC$ is given by
$$\begin{aligned} \frac{1}{2} \det \begin{bmatrix} 1 & 0 & 1 \\ 1 & \log_2 3 & 2 \\ 1 & \log_2 7 & 3 \end{bmatrix} &= \frac{1}{2} \det \begin{bmatrix} 1 & 0 & 1 \\ -1 & \log_2 3 & 0 \\ -2 & \log_2 7 & 0 \end{bmatrix} \\ &= \frac{1}{2} \det \begin{bmatrix} -1 & \log_2 3 \\ -2 & \log_2 7 \end{bmatrix} \\ &= \frac{1}{2} (-\log_2 7 + \log_2 9) \\ &= \log_2 \frac{3}{\sqrt{7}}. \end{aligned}$$

The cross product of two vectors in the x-y plane is a vector in the z direction whose magnitude is twice the area of the triangle determined by the two vectors. The area of the triangle with vertices $A(0, 1, 0)$, $B(\log_2 3, 2, 0)$, and $C(\log_2 7, 3, 0)$ is the z-component of
$$\begin{align*} \frac{1}{2}(B - A) \times (C - A) &= \frac{1}{2}(\log_2 3, 1, 0) \times (\log_2 7, 2, 0) \\ &= \frac{1}{2}(1 \cdot 0 - 0 \cdot 2, \log_2 7 \cdot 0 - \log_2 3 \cdot 0, \log_2 3 \cdot 2 - 1 \cdot \log_2 7) \\ &= (0, 0, \log_2 3 - \log_2 \sqrt{7}) = (0, 0, \log_2 \frac{3}{\sqrt{7}}). \end{align*}$$
Thus the area of the triangle is $\log_2 \frac{3}{\sqrt{7}}$.

Define the points $H(0, 2)$, $I(0, 3)$, and $J(\log_2 3, 3)$.

Then $HBCI$ is a trapezoid, which can be decomposed into right triangle $\triangle CJB$ and rectangle $HBJI$. Because
$$\log_2 3 > \log_2 \sqrt{7} = \frac{1}{2} \log_2 7,$$
point $B$ is to the right of line $AC$. Therefore
$$\begin{align*} \text{Area}(\triangle ABC) &= \text{Area}(\triangle ABH) + \text{Area}(HBCI) - \text{Area}(\triangle ACI) \\ &= \frac{1}{2} \log_2 3 + \frac{1}{2} (\log_2 7 - \log_2 3) + \log_2 3 - \log_2 7 \\ &= \log_2 3 - \frac{1}{2} \log_2 7 \\ &= \log_2 \frac{3}{\sqrt{7}}. \end{align*}$$