2024 AMC 12B Problems/Problem 15

Problem

A triangle in the coordinate plane has vertices $A(\log_21,\log_22)$ , $B(\log_23,\log_24)$ , and $C(\log_27,\log_28)$ . What is the area of $\triangle ABC$ ?

$\textbf{(A) }\log_2\frac{\sqrt3}7\qquad \textbf{(B) }\log_2\frac3{\sqrt7}\qquad \textbf{(C) }\log_2\frac7{\sqrt3}\qquad \textbf{(D) }\log_2\frac{11}{\sqrt7}\qquad \textbf{(E) }\log_2\frac{11}{\sqrt3}\qquad$

Solution 1 (Shoelace Theorem)

We rewrite: $A(0,1)$ $B(\log _{2} 3, 2)$ $C(\log _{2} 7, 3)$ .

From here we setup Shoelace Theorem and obtain: $\frac{1}{2}(2(\log _{2} 3) - log _{2} 7)$ .

Following log properties and simplifying gives (B).

~MendenhallIsBald

Solution 2 (Determinant)

To calculate the area of a triangle formed by three points $ A(x_1, y_1) $, $ B(x_2, y_2) $, and $ C(x_3, y_3) $ on a Cartesian coordinate plane, you can use the following formula:

$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$