2024 AMC 12B Problems/Problem 22

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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 $a<a+c$.

Case $1$: $b=1$

Clearly, this case yields no valid solutions.

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.

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.

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

Case $5$: $b=5$

For this case, we must have $a=1$ and $c=24$. There are no valid solutions

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 $(4, 6, 5)$, which yields a perimeter of $15$.

When $b\ge 7$, it is easy to see that $a+c>7$. Hence $a+b+c>14$, which means $a+b+c\ge15$. Therefore, the answer is $\fbox{\textbf{(C) }15}$

~tsun26

Solution 2 (Similar to Solution 1)

Let $\overline{BC}=a$, $\overline{AC}=b$, $\overline{AB}=c$. Extend $C$ to point $D$ on $\overline{AB}$ such that $\angle ACD = \angle CAD$. This means $\triangle CDA$ is isosceles, so $CD=DA$. Since $\angle CDB$ is the exterior angle of $\triangle CDA$, we have \[\angle CDB=m+m=2m=\angle CBD.\] Thus, $\triangle CBD$ is isosceles, so $CB=CD=DA=a.$ Then, draw the altitude of $\triangle CBD$, from $C$ to $\overline{BD}$, and let this point be $H$. Let $BH=HD=x$. Then, by Pythagorean Theorem, CH2=a2x2CH2=b2(c+x)2. Thus, \[a^2-x^2=b^2-(c-x)^2.\] Solving for $x$, we have $x=\frac{a^2-b^2+c^2}{2c}.$ Since $2x=c-a$, we have \[c-a=\frac{a^2-b^2+c^2}{c},\] and simplifying, we get $b^2=a^2+ac.$ Now we can consider cases on what $a$ is. (Note: Although there looks to be quite a few cases, they are just trivial and usually only take up to a few seconds max).

Case $1$: $a=1$.

This means $b^2=c+1$, so the least possible values are $b=2$, $c=3$, but this does not work as it does not satisfy the triangle inequality. Similarly, $b=3$, $c=8$ also does not satisfy it. Anything larger goes beyond the answer choices, so we stop checking this case.

Case $2$: $a=2$ This means $b^2=2c+4$, so the least possible values for $b$ and $c$ are $b=4$,$c=6$, but this does not satisfy the triangle inequality, and anything larger does not satisfy the answer choices.

Case $3$: $a=3$ This means $b^2=3c+9$, and the least possible value for $b$ is $b=6$, which occurs when $c=9$. Unfortunately, this also does not satisfy the triangle inequality, and similarly, any $b > 6$ means the perimeter will get too big.

Case $4$: $a=4$ This means $b^2=4c+16$, so we have $b=6,c=5,a=4$, so the least possible perimeter so far is $4+5+6=15$.

Case $5$: $a=5$ We have $b^2=5c+25$, so least possible value for $b$ is $b=10$, which already does not work as $a=5$, and the minimum perimeter is $15$ already.

Case $6$: $a=6$ We have $b^2=6c+36$, so $b=10$, which already does not work.

Then, notice that when $a\geq 7$, we also must have $b\geq8$ and $c\geq1$, so $a+b+c \geq 16$, so the least possible perimeter is $\boxed{\textbf{(C) }15}.$

~evanhliu2009

Solution 3 (Trigonometry)

\[\frac{a}{sin(A)} = \frac{a}{sin(B)} = \frac{c}{sin(C)}\] \[\frac{a}{sin(A)} = \frac{a}{sin(2A)} = \frac{c}{sin(\pi- 3A)} = \frac{c}{ sin(3A)}\] \[\frac{a}{sin(A)} = \frac{a}{2sin(A)cos(A)} = \frac{c}{3sin(A) - 4sin^3(A)}\] \[b = 2cos(A)a\] \[c = (3  - 4sin^2(A) ) a = (4cos^2(A)-1)a\] \[A+B = A+2A < 180^\circ\] \[A < 60^\circ ,  \frac{1}{2} <  cos(A)  < 1\]

\[a:b:c = 1: 2cos(A) : 4cos^2(A)-1\] cos(A) must be rational, let's evaluate some small values

case #1: cos(A) = $\frac{1}{2}$ invalid since = $\frac{1}{2}$

case #2: cos(A) = $\frac{1}{3}$ invalid since < $\frac{1}{2}$

case #3: cos(A) = $\frac{2}{3}$ give ${1: \frac{4}{3} :  \frac{7}{9}  }$ with side (9:12:7) , perimeter = 28

case #4: cos(A) = $\frac{1}{4}$ invalid since < $\frac{1}{2}$

case #5: cos(A) = $\frac{3}{4}$ give ${1: \frac{3}{2} :  \frac{5}{4}  }$ with side (4:6:5), perimeter = 15

case #6: cos(A) = $\frac{3}{5}$ give ${1: \frac{6}{5} :  \frac{11}{25}  }$ with side (25:30:11)

case #7: cos(A) = $\frac{4}{5}$ give ${1: \frac{8}{5} :  \frac{39}{25}  }$ with side (25:40:39)

case #8: cos(A) = $\frac{4}{6}$ same as $\frac{2}{3}$

case #9: cos(A) = $\frac{5}{6}$ give ${1: \frac{5}{3} :  \frac{16}{9}  }$ with side (9:15:16)

case #10: when a $\geq$ 7, b =2cos(A)*a > 2* $\frac{1}{2}$*7 = 7 , a+b+c > 15

$\boxed{\textbf{(C) }15}$

~luckuso

Alternative Solution :

Let ∠A=θ, then ∠B=2θ Find D on AB such that ∠ACD=θ Thus, ∠CDB=∠A+∠ACD=2θ So AD=CD=BD Find E on BD such that CE⊥BD Apparently, this gives E the mid-point of BD Let the length of BC be x, Then AB can be expressed as AD+BD=AD+2BE=x+2xcos2θ Since CE=xsin2θ The length of AC can be expressed as \frac{CE}{sinθ} =2xcosθ(using double angle formula) Since the length of the sides of this triangle is all integers, x, x+2xcos2θ, 2xcosθ is all integers.

Now we need to determine the range of θ. We know 3θ<180°, so θ<60° Also, the above conditions are only valid if ∠B is an acute angle.(the strict proof will be shown in the end) So θ<45°, this yield cosθ∈(\frac{\sqrt{2} }{2} ,I)

Let cosθ=\frac{p}{q} ,where (p,q)=1 We also know that \cos 2θ=2\cos^{2} θ-1 To minimize the perimeter, the denominator needs to be as small as possible. In this way, a small x can be used to integrate the side length. Test q=2, one half is not in the range Test q=3,one third and two thirds are not in the range (since 0.67<0.71) Test q=4,three fourths is in the range. In this case, the smallest x that make side length integer is 4,since the side length is x,\frac{5x}{4} and \frac{3x}{2} So the perimeter= 4+5+6=15 When q becomes bigger, a larger x is required to integrate the length, thus can not give the minimum perimeter.

Video Solution 1 by TheSpreadTheMathLove

https://www.youtube.com/watch?v=N7cNhAx9ifE&t=0s

See also

2024 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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