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 $\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}$ .

~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: $\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$ The coordinates are: $A(0, 1)$ , $B(\log_2 3, 2)$ , $C(\log_2 7, 3)$

Taking a numerical value into account: $\text{Area} = \frac{1}{2} \left| 0 \cdot (2 - 3) + \log_2 3 \cdot (3 - 1) + \log_2 7 \cdot (1 - 2) \right|$ Simplify: $= \frac{1}{2} \left| 0 + \log_2 3 \cdot 2 + \log_2 7 \cdot (-1) \right|$ $= \frac{1}{2} \left| \log_2 (3^2) - \log_2 7 \right|$ $= \frac{1}{2} \left| \log_2 \frac{9}{7} \right|$ Thus, the area is: $\text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right|$ = $\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}$

~Athmyx

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jyupN3dT2yY&t=0s

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 12 Problems and Solutions

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2024 AMC 12B Problems/Problem 15

Contents

Problem

Solution 1 (Shoelace Theorem)

Solution 2 (Determinant)

Video Solution 1 by SpreadTheMathLove

See also