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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 \angleA</cmath>
+
<cmath>=2\cos \angle A</cmath>
  
 
According to the law of cosines,
 
According to the law of cosines,
−
<cmath>\cos \angleA=\frac{b^2+c^2-a^2}{2bc}</cmath>
+
<cmath>\cos \angle A=\frac{b^2+c^2-a^2}{2bc}</cmath>
  
 
Hence,
 
Hence,
 
<cmath>\frac{b}{a}=\frac{b^2+c^2-a^2}{bc}</cmath>
 
<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>.
+
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>. Remember that <math>a<a+c</math>.
  
 
<math>\textbf{Case 1: b=1}</math>
 
<math>\textbf{Case 1: b=1}</math>
Line 28: Line 28:
  
 
<math>\textbf{Case 2: b=2}</math>
 
<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.
+
For this case, 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.
 +
 +
<math>\textbf{Case 3: b=3}</math>
 +
For this case, we must have <math>a=1</math> and <math>c=9</math>. However, <math>(1, 3, 9)</math> does not form a triangle. Hence this case yields no valid solutions.
 +
 
 +
<math>\textbf{Case 4: b=4}</math>
 +
For this case, <math>a=1</math> and <math>c=15</math>, or <math>a=2</math> and <math>c=6</math>. As one can check, this case also yields no valid solutions
 +
 
 +
<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
 +
 
 +
<math>\textbf{Case 6: b=6}</math>
 +
For this case, <math>a=2</math> and <math>c=16</math>, or <math>a=4</math> and <math>c=5</math>, or <math>a=3</math> and <math>c=9</math>. The only valid solution for this case is <math>(4, 6, 5).
 +
 
 +
It is safe to assume that </math>(4, 5, 6)$ will be the solution with least perimeter. Hence, the answer is

Revision as of 03:54, 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 $a=1$ $c=24$. 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$ (Error compiling LaTeX. Unknown error_msg)(4, 5, 6)$ will be the solution with least perimeter. Hence, the answer is

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