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

(→‎Solution 2 (Determinant))
(→‎Solution 2 (Determinant))
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==Solution 2 (Determinant)==   
 
==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>
 
<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>
 
</cmath>
−
 
 
The coordinates are:<math>A(0, 1)</math>, <math>B(\log_2 3, 2)</math>, <math>C(\log_2 7, 3)</math>
 
The coordinates are:<math>A(0, 1)</math>, <math>B(\log_2 3, 2)</math>, <math>C(\log_2 7, 3)</math>
  
 
Taking a numerical value into account:
 
Taking a numerical value into account:
−
 
 
<cmath>
 
<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>
 
</cmath>
−
 
 
Simplify:
 
Simplify:
−
 
 
<cmath>
 
<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|
Line 52: Line 47:
 
= \frac{1}{2} \left| \log_2 \frac{9}{7} \right|
 
= \frac{1}{2} \left| \log_2 \frac{9}{7} \right|
 
</cmath>
 
</cmath>
−
 
 
Thus, the area is:<math>\text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right|</math> = <math>\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}</math>
 
Thus, the area is:<math>\text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right|</math> = <math>\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}</math>
  
 
~Athmyx
 
~Athmyx

Revision as of 05:28, 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 area is:$\text{Area} = \frac{1}{2} \left| \log_2 \frac{9}{7} \right|$ = $\boxed{\textbf{(B) }\log_2 \frac{3}{\sqrt{7}}}$

~Athmyx

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