Difference between revisions of "2024 AMC 12B Problems/Problem 15"

Revision as of 15:04, 16 July 2025

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, ShortPeopleFartalot

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

Solution 3 (Geometry)

In the graph above, the biggest triangle has legs $\log_2 7$ and $2$ . Its area is then $\log_2 7$ .

There are 3 smaller shapes as well:

1. Triangle 1. Legs = $\log_2 3$ and $1$ . Area = $\log_2 \sqrt{3}$ .;

2. Triangle 2. Legs = $\log_2\frac{7}{3}$ and $1$ . Area = $\log_2 \frac{\sqrt{7}}{\sqrt{3}}$ ;

3. Rectangle. Legs = $\log_2\frac{7}{3}$ and $1$ . Area = $\log_2\frac{7}{3}$ .;

The area of the triangle is therefore $\log_2 7 - \log_2 \sqrt{3} - \log_2 \frac{\sqrt{7}}{\sqrt{3}} - \log_2\frac{7}{3}$ . Following log properties and simplifying gives $\boxed{\textbf{(B)}\log_2 \frac{3}{\sqrt{7}}}$ .

~mathwizard123123

Video Solution 1 by SpreadTheMathLove

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

@@ Line 50: / Line 50: @@
 ~[https://artofproblemsolving.com/wiki/index.php/User:Athmyx Athmyx]
+==Solution 3 (Geometry)==
+[[File:AMC 12B 2024 Problem15.png|None]]
+In the graph above, the biggest triangle has legs <math>\log_2 7</math> and <math>2</math>. Its area is then <math>\log_2 7</math>.
+There are 3 smaller shapes as well:
+. Triangle 1. Legs = <math>\log_2 3</math> and <math>1</math>. Area = <math>\log_2 \sqrt{3}</math>.;
+. Triangle 2. Legs = <math>\log_2\frac{7}{3}</math> and <math>1</math>. Area = <math>\log_2 \frac{\sqrt{7}}{\sqrt{3}}</math>;
+. Rectangle. Legs = <math>\log_2\frac{7}{3}</math> and <math>1</math>. Area = <math>\log_2\frac{7}{3}</math>.;
+The area of the triangle is therefore <math>\log_2 7 - \log_2 \sqrt{3} - \log_2 \frac{\sqrt{7}}{\sqrt{3}} - \log_2\frac{7}{3}</math>.
+Following log properties and simplifying gives <math>\boxed{\textbf{(B)}\log_2 \frac{3}{\sqrt{7}}}</math>.
+~mathwizard123123
 ==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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