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

(Solution 1 (Shoelace Theorem))
(Solution 2)
Line 26: Line 26:
  
  
==Solution 2==
+
==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:
 
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:
  
\[
+
<cmath>
 
\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|
 
\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|
\]
+
</cmath>
  
 
The coordinates are:
 
The coordinates are:
  
- \( A(0, 1) \)
+
- <cmath>A(0, 1)</cmath>
- \( B(\log_2 3, 2) \)
+
- <cmath>B(\log_2 3, 2)</cmath>
- \( C(\log_2 7, 3) \)
+
- <cmath>C(\log_2 7, 3)</cmath>
  
take a numerical value into account
+
Taking a numerical value into account:
  
\[
+
<cmath>
 
\text{Area} = \frac{1}{2} \left| 0 \cdot (2 - 3) + \log_2 3 \cdot (3 - 1) + \log_2 7 \cdot (1 - 2) \right|
 
\text{Area} = \frac{1}{2} \left| 0 \cdot (2 - 3) + \log_2 3 \cdot (3 - 1) + \log_2 7 \cdot (1 - 2) \right|
\]
+
</cmath>
  
 
Simplify:
 
Simplify:
  
\[
+
<cmath>
 
= \frac{1}{2} \left| 0 + \log_2 3 \cdot 2 + \log_2 7 \cdot (-1) \right|
 
= \frac{1}{2} \left| 0 + \log_2 3 \cdot 2 + \log_2 7 \cdot (-1) \right|
\]
+
</cmath>
  
\[
+
<cmath>
 
= \frac{1}{2} \left| \log_2 (3^2) - \log_2 7 \right|
 
= \frac{1}{2} \left| \log_2 (3^2) - \log_2 7 \right|
\]
+
</cmath>
  
\[
+
<cmath>
 
= \frac{1}{2} \left| \log_2 \frac{9}{7} \right|
 
= \frac{1}{2} \left| \log_2 \frac{9}{7} \right|
\]
+
</cmath>
  
 
Thus, the exact area is:
 
Thus, the exact area is:
  
\[
+
<cmath>
 
\text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right|
 
\text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right|
\]
+
</cmath>
  
\[
+
<cmath>
 
\boxed{\text{B. \log_2 \frac{3}{\sqrt{7}}}}
 
\boxed{\text{B. \log_2 \frac{3}{\sqrt{7}}}}
\]
+
</cmath>
 +
 
 +
---
 +
 
 +
This format with `<math></math>` will make Obsidian recognize all the mathematical expressions as standalone blocks, rendering them properly in mathematical form.

Revision as of 05:23, 14 November 2024

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:

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

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 exact area is:

\[\text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right|\]

\[\boxed{\text{B. \log_2 \frac{3}{\sqrt{7}}}}\] (Error compiling LaTeX. Unknown error_msg)

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This format with `$$ (Error compiling LaTeX. Unknown error_msg)` will make Obsidian recognize all the mathematical expressions as standalone blocks, rendering them properly in mathematical form.