Art of Problem Solving
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

2024 AMC 12B Problems/Problem 22: Difference between revisions

← Older editNewer edit →
Tsun26 (talk | contribs)
editor
136 edits
No edit summary
Tsun26 (talk | contribs)
editor
136 edits
Line 37: Line 37:


<math>\textbf{Case 5: b=5}</math>
<math>\textbf{Case 5: b=5}</math>
For this case, we must have <math>a=1</math> <math>c=24</math>. There are no valid solutions
For this case, we must have <math>a=1</math> and <math>c=24</math>. There are no valid solutions


<math>\textbf{Case 6: b=6}</math>
<math>\textbf{Case 6: b=6}</math>

Revision as of 03:56, 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, \[\frac{b}{a}=\frac{\sin \angle B}{\sin \angle A}\] \[=2\cos \angle A\]

According to the law of cosines, \[\cos \angle A=\frac{b^2+c^2-a^2}{2bc}\]

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 $b$. Remember that $a<a+c$.

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

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

$\textbf{Case 3: b=3}$ For this case, we must have $a=1$ and $c=9$. However, $(1, 3, 9)$ does not form a triangle. Hence this case yields no valid solutions.

$\textbf{Case 4: b=4}$ For this case, $a=1$ and $c=15$, or $a=2$ and $c=6$. As one can check, this case also yields no valid solutions

$\textbf{Case 5: b=5}$ For this case, we must have $a=1$ and $c=24$. There are no valid solutions

$\textbf{Case 6: b=6}$ For this case, $a=2$ and $c=16$, or $a=4$ and $c=5$, or $a=3$ and $c=9$. The only valid solution for this case is $(4, 6, 5)$.

It is safe to assume that $(4, 5, 6)$ will be the solution with least perimeter. Hence, the answer is $\fbox{\textbf{(C) }15}$

~tsun26

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12B_Problems/Problem_22&oldid=233764"