Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Added solution 3 to AMC 12B 2024 problem 15.)
m (Changed solution 3 to clarify logarithm properties.)
 
Line 59: Line 59:
  
  
−
1. Triangle 1. Legs = <math>\log_2 3</math> and <math>1</math>. Area = <math>\log_2 \sqrt{3}</math>.;
+
1. Triangle 1. Legs = <math>\log_2 3</math> and <math>1</math>. Area = <math>\frac{1}{2}\log_2 3 = \log_2 \sqrt{3}</math>.;
  
  
−
2. Triangle 2. Legs = <math>\log_2\frac{7}{3}</math> and <math>1</math>. Area = <math>\log_2 \frac{\sqrt{7}}{\sqrt{3}}</math>;
+
2. Triangle 2. Legs = <math>\log_2\frac{7}{3}</math> and <math>1</math>. Area = <math>\frac{1}{2}\log_2 \frac{7}{3} = \log_2 \frac{\sqrt{7}}{\sqrt{3}}</math>;
  
  
Line 69: Line 69:
  
 
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>.
 
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>.
+
This is equivalent to <math>\log_2 7*\frac{1}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{7}}*\frac{3}{7}</math> which simplifies to <math>\boxed{\textbf{(B)}\log_2 \frac{3}{\sqrt{7}}}</math>.
  
 
~mathwizard123123
 
~mathwizard123123
 +
 
==Video Solution 1 by SpreadTheMathLove==
 
==Video Solution 1 by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=jyupN3dT2yY&t=0s
 
https://www.youtube.com/watch?v=jyupN3dT2yY&t=0s

Latest revision as of 12:15, 17 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)

None

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 = $\frac{1}{2}\log_2 3 = \log_2 \sqrt{3}$.;


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


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


The area of the triangle is therefore $\log_2 7 - \log_2 \sqrt{3} - \log_2 \frac{\sqrt{7}}{\sqrt{3}} - \log_2\frac{7}{3}$. This is equivalent to $\log_2 7*\frac{1}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{7}}*\frac{3}{7}$ which simplifies to $\boxed{\textbf{(B)}\log_2 \frac{3}{\sqrt{7}}}$.

~mathwizard123123

Video Solution 1 by SpreadTheMathLove

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

See also

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12B_Problems/Problem_15&oldid=254458"