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 22"

(→‎Solution 1)
Line 14: Line 14:
 
Let <math>AB=c</math>, <math>BC=a</math>, <math>AC=b</math>. According to the law of sines,
 
Let <math>AB=c</math>, <math>BC=a</math>, <math>AC=b</math>. According to the law of sines,
 
<cmath>\frac{b}{a}=\frac{\sin \angle B}{\sin \angle A}</cmath>
 
<cmath>\frac{b}{a}=\frac{\sin \angle B}{\sin \angle A}</cmath>
−
<cmath>=2\cos</cmath>
+
<cmath>=2\cos \angleA</cmath>
 +
 
 +
According to the law of cosines,
 +
<cmath>\cos \angleA=\frac{b^2+c^2-a^2}{2bc}</cmath>
 +
 
 +
Hence,
 +
<cmath>\frac{b}{a}=\frac{b^2+c^2-a^2}{bc}</cmath>
 +
 
 +
This simplifies to <math>b^2=a(a+c)</math>. We want to find the positive integer solution <math>(a, b, c)</math> to this equation such that <math>a, b, c</math> forms a triangle, and <math>a+b+c</math> is minimized. We proceed by casework on the value of <math>b</math>.
 +
 
 +
<math>\textbf{Case 1: b=1}</math>
 +
Clearly, this case yields no valid solutions.
 +
 
 +
<math>\textbf{Case 2: b=2}</math>
 +
Since <math>a<a+c</math>, we must have <math>a=1</math> and <math>c=3</math>. However, <math>(1, 2, 3)</math> does not form a triangle. Hence this case yields no valid solutions.

Revision as of 03:39, 14 November 2024

Problem 22

Let $\triangle{ABC}$ be a triangle with integer side lengths and the property that $\angle{B} = 2\angle{A}$. What is the least possible perimeter of such a triangle?

$\textbf{(A) }13 \qquad \textbf{(B) }14 \qquad \textbf{(C) }15 \qquad \textbf{(D) }16 \qquad \textbf{(E) }17 \qquad$

Solution 1

Let $AB=c$, $BC=a$, $AC=b$. According to the law of sines, \[\frac{b}{a}=\frac{\sin \angle B}{\sin \angle A}\] 

\[=2\cos \angleA\] (Error compiling LaTeX. Unknown error_msg)

According to the law of cosines, 

\[\cos \angleA=\frac{b^2+c^2-a^2}{2bc}\] (Error compiling LaTeX. Unknown error_msg)

Hence, \[\frac{b}{a}=\frac{b^2+c^2-a^2}{bc}\]

This simplifies to $b^2=a(a+c)$. We want to find the positive integer solution $(a, b, c)$ to this equation such that $a, b, c$ forms a triangle, and $a+b+c$ is minimized. We proceed by casework on the value of $b$.

$\textbf{Case 1: b=1}$ Clearly, this case yields no valid solutions.

$\textbf{Case 2: b=2}$ Since $a<a+c$, we must have $a=1$ and $c=3$. However, $(1, 2, 3)$ does not form a triangle. Hence this case yields no valid solutions.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12B_Problems/Problem_22&oldid=233760"