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

Latest revision as of 10:17, 27 October 2025

Problem

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 $c=8$ . However, $(1, 3, 8)$ 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 ~infinitywasp (minor errors)

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, \begin{align*} CH^2&=a^2-x^2 \\ CH^2&= b^2 - (c+x)^2.\\ \end{align*} 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)

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

$a:b:c = 1: 2\cos(A) : 4\cos^2(A)-1$ $\cos(A)$ 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

Video Solution 1 by TheSpreadTheMathLove

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

Solution 4

Draw the circumcircle of $\triangle{ABC}$ and let the angle bisector of $\angle{B}$ meet the circle at $D$

By fact 5 we have $CD=AD, \angle{CBD}=\angle{ABD}=\angle{CAB}=\angle{DAC}, \angle{CBA}=\angle{DAB}$ , thus $AC=BD, CD=CB$

By Ptolemy, we have $c^2=(a+b)CD, CD=\frac{c^2}{a+b}=a, a^2+ab=c^2$ . Try some numbers and the answer is $(4,5,6)\implies \boxed{15}$

~Bluesoul

Solution 5

Let $\angle A=\theta$ , then $\angle B=2\theta$ .

Find $D$ on $AB$ such that $\angle ACD=\theta$ . Thus, $\angle CDB=\angle A+\angle ACD=2\theta$ . So $AD=CD=BD$ .

Find $E$ on $BD$ such that $CE\perp BD$ . Apparently, this gives $E$ as the mid-point of $BD$ .

Let the length of $BC$ be $x$ . Then $AB$ can be expressed as $AD+BD=AD+2BE=x+2x\cos 2\theta$ .

Since $CE=x\sin 2\theta$ , the length of $AC$ can be expressed as $2x\cos\theta$ (using double angle formula).

Now we need to determine the range of $\theta$ .

The above conditions are only valid if $\angle B$ is an acute angle.(the strict proof will be shown in the end*) So $\theta<45^\circ$ , this yields $\cos\theta\in\left(\frac{\sqrt{2}}{2},1\right)$ .

Let $\cos\theta=\frac{p}{q}$ , where $(p,q)=1$ .

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 $q=2$ , $\frac{1}{2}$ is not in the range.

Test $q=3$ , $\frac{1}{3}$ and $\frac{2}{3}$ are not in the range (since $0.67<0.71$ ).

Test $q=4$ , $\frac{3}{4}$ is in the range. In this case, the smallest $x$ that make side length integer is $4$ , since the side length is $x$ , $\frac{5}{4}x$ and $\frac{3}{2}x$ So the perimeter is $4+5+6=15$ .

So $\boxed{\textbf{(C) }15}$ is the correct answer.

When $q$ becomes bigger, a larger $x$ is required to integrate the length, thus cannot give the minimum perimeter.

If $\angle B>90^\circ$ , $\angle A+\angle B>135^\circ$ ,then $\angle C<45^\circ$ . This will result in point D on the extension of AB, meaning that $\angle CDB+\angle CBD<180^\circ$ . Hence, $2\angle B<180^\circ$ , $\angle B<90^\circ$ , which contradicts our condition. If $\angle B=90^\circ$ ,The triangle is isosceles right triangle. So, the ratio of sides is $1:1:\sqrt{2}$ , which, obviously means the length cannot be all integers.

~solution by Tonyttian [1] ~reformatted as LaTeX by Michaelihc [2]

Solution 6

Extend the angle bisector of $\angle B$ to point $P$ on $\overline{AC}.$ We have $\triangle BPC \sim ABC,$ so $\frac{m}{b} = \frac{b}{c},$ yielding $m = \frac{b^2}{c}.$ By the angle bisector theorem, $\frac{b}{a} = \frac{m}{n},$ so $\frac{ma}{b} = \frac{ab}{c} = n$ after substitution. We also have $m + n = c,$ so $\frac{b^2}{c} + \frac{ab}{c} = c \implies b^2 + ab = b(b + a) = c^2,$ for $a, b, c \in \mathbb{Z}_{> 0}.$ We can't have $b=1$ for any integral triangle. If $b = 2$ then $2a + 4 = c^2$ , which gives $c = 4$ and $a = 6,$ which fails the triangle inequality. If $b=3$ then $a = 9$ and $c = 6$ , which fails again. If $b = 4$ , then $(a, c) = (5, 6)$ , which works and yields a perimeter of $15$ . If $b=5$ , then $(a, c) = (15, 10),$ and if $b = 6$ , then $(a, c) = (18, 12),$ which fails again.

If $b \geq 7$ then $a+b>7,$ yielding a minimum perimeter of $\boxed{\textbf{(C)\ 15}},$ which was already achieved.

-Benedict T (countmath1)

Solution 7 (No Casework, Trig)

Let $AB=c$ , $BC=a$ , $AC=b$ .

Our goal is to find

$\min_{a,b,c \in \mathbb{N}} a+b+c$

By the law of sines,

$\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}$

Substituting $B=2A$ and $C = 180^\circ - A - B$ :

$\frac{a}{\sin(A)} = \frac{b}{\sin(2A)} = \frac{c}{\sin(180^\circ-3A)} = \frac{c}{\sin(3A)}$

We will attempt to cancel the $\sin(A)$ factor in the denominator. Using multiple-angle trigonometric identities,

$\sin(2A) = 2\sin(A)\cos(A)$ $\sin(3A) = \sin(A)(4\cos^2(A)-1)$

we obtain:

$\frac{a}{\sin(A)} = \frac{b}{2\sin(A)\cos(A)} = \frac{c}{\sin(A)(4\cos^2(A)-1))}$

which simplifies to:

$a = \frac{b}{2\cos(A)} = \frac{c}{(4\cos^2(A)-1))}$

We notice that the term $2\cos(A)$ appears multiple times. For convenience, let $t:=2\cos(A)$ . Then, rewriting $b$ and $c$ in terms of $a$ , we get:

$b=ta$ $c=(t^2-1)a$

We now rewrite the original goal in terms of $t$ and $a$ . Notice that since $a,b,c \in \mathbb{N}$ , then the perimeter $a+b+c \in \mathbb{N}$ . Hence,

$a+b+c = a + ta + (t^2-1)a$ $= at(t+1) \in \mathbb{N}$

This means that $t \in \mathbb{Q}$ , so let $t := \frac{m}{n}$ with $m,n \in \mathbb{N}$ . To minimize

$a+b+c = at(t+1) = a\left(\frac{m^2}{n^2}+\frac{m}{n}\right)$

we set $a = n^2$ to ensure the perimeter is an integer, and the objective becomes

$\min_{m,n \in \mathbb{N}} m(m+n)$

Notice that to have a valid triangle, $c=(t^2-1)a>0$ implies $t^2 - 1 > 0$ so $t > 1$ .

In addition, due to the triangle inequality, we must have $c < a + b$ , such that

$(t^2-1)a < a + ta$

Since $a>0$ , we can safely divide through, yielding $t > 2$ . Overall, we have $1<t<2$ , or equivalently, $n<m<2n$ . The smallest pair satisfying this condition is $n=2$ , $m=3$ , yielding a perimeter of $15$ . (Notice, to have a valid integer between $n$ and $2n$ , there must be a gap of at least $1$ in between, implying $2n-n-1=n-1\ge 1$ , so setting $n=2$ yields the minimum.) Increasing $m$ or $n$ will increase the perimeter, so this is the global minimum. Hence the answer is:

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

@@ Line 1: / Line 1: @@
-==Problem 22==
+==Problem==
 Let <math>\triangle{ABC}</math> be a triangle with integer side lengths and the property that <math>\angle{B} = 2\angle{A}</math>. What is the least possible perimeter of such a triangle?
@@ Line 13: / Line 13: @@
 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}=2\cos \angle A</cmath>
-<cmath>=2\cos \angle A</cmath>
 According to the law of cosines,
@@ Line 34: / Line 33: @@
 Case <math>3</math>: <math>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.
+For this case, we must have <math>a=1</math> and <math>c=8</math>. However, <math>(1, 3, 8)</math> does not form a triangle. Hence this case yields no valid solutions.
 Case <math>4</math>: <math>b=4</math>
@@ Line 51: / Line 50: @@
 ~tsun26
+~infinitywasp (minor errors)
 ==Solution 2 (Similar to Solution 1)==
@@ Line 85: / Line 85: @@
 ==Solution 3 (Trigonometry)==
-<cmath> \frac{a}{sin(A)} = \frac{a}{sin(B)} = \frac{c}{sin(C)} </cmath>
+<cmath> \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} </cmath>
-<cmath> \frac{a}{sin(A)} = \frac{a}{sin(2A)} = \frac{c}{sin(3A)} </cmath>
+<cmath> \frac{a}{\sin(A)} = \frac{b}{\sin(2A)} = \frac{c}{\sin(\pi- 3A)} = \frac{c}{\sin(3A)} </cmath>
-<cmath> \frac{a}{sin(A)} = \frac{a}{2sin(A)cos(A)} = \frac{c}{3sin(A) - 4sin^3(A)} </cmath>
+<cmath> \frac{a}{\sin(A)} = \frac{b}{2\sin(A)\cos(A)} = \frac{c}{3\sin(A) - 4\sin^3(A)} </cmath>
-<cmath> b = 2cos(A)a </cmath>
+<cmath> b = 2\cos(A)a </cmath>
-<cmath> c = (3  - 4sin^2(A) ) a = (4cos^2(A)-1)a </cmath>
+<cmath> c = (3 - 4\sin^2(A) ) a = (4\cos^2(A)-1)a </cmath>
-<cmath> A <= 60 ,  \frac{1}{2} <= cos(A)  <=1 </cmath>
+<cmath> A+B = A+2A < 180^{\circ} </cmath>
+<cmath> A < 60^{\circ} ,  \frac{1}{2} <  cos(A)  < 1 </cmath>
-<cmath> a:b:c = 1: 2cos(A) : 4cos^2(A)-1 </cmath>
+<cmath> a:b:c = 1: 2\cos(A) : 4\cos^2(A)-1 </cmath>
-cos(A) must be rational, let's evaluate some small values
+<math>\cos(A)</math> must be rational, let's evaluate some small values
-case #1: cos(A) = <math>\frac{1}{2}</math> invalid  c= 0
+case #1: <math>\cos(A) = \frac{1}{2}</math> invalid since = <math>\frac{1}{2}</math>
-case #2: cos(A) = <math>\frac{1}{3}</math> invalid since c < 4   <math> \cdot \frac{1}{9} -1 < 0 </math>
+case #2: <math>\cos(A) = \frac{1}{3}</math> invalid since < <math>\frac{1}{2}</math>
-case #3: cos(A) = <math>\frac{2}{3}</math> give <math>{1: \frac{4}{3} :  \frac{7}{9}  }</math> with side (9:12:7) , perimeter = 28
+case #3: <math>\cos(A) = \frac{2}{3}</math> give <math>{1: \frac{4}{3} :  \frac{7}{9}  }</math> with side <math>9:12:7</math> , perimeter = 28
-case #4: cos(A) = <math>\frac{1}{4}</math> invalid  c<0
+case #4: <math>\cos(A) = \frac{1}{4}</math> invalid since < <math>\frac{1}{2}</math>
-case #5: cos(A) = <math>\frac{3}{4}</math> give <math>{1: \frac{3}{2} :  \frac{5}{4}  }</math> with side (4:6:5)
+case #5: <math>\cos(A) = \frac{3}{4}</math> give <math>{1: \frac{3}{2} :  \frac{5}{4}  }</math> with side <math>4:6:5</math>, perimeter = 15
+case #6: <math>\cos(A) = \frac{3}{5}</math> give <math>{1: \frac{6}{5} :  \frac{11}{25}  }</math> with side <math>25:30:11</math>
+case #7: <math>\cos(A) = \frac{4}{5}</math> give <math>{1: \frac{8}{5} :  \frac{39}{25}  }</math> with side <math>25:40:39</math>
-for denominators 5 and above, the fraction denominators getting larger  and perimeter will be larger than 15
+case #8: <math>\cos(A) = \frac{4}{6}</math> same as <math>\frac{2}{3}</math>
+case #9: <math>\cos(A) = \frac{5}{6}</math> give <math>{1: \frac{5}{3} :  \frac{16}{9}  }</math> with side (9:15:16)
+case #10: when <math>a \geq 7, b =2cos(A)*a > 2*  \frac{1}{2}*7 = 7 , a+b+c > 15</math>
-<math>\boxed{\textbf{(C) }15}.</math>
+<math>\boxed{\textbf{(C) }15}</math>
 ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
-Alternative Solution ：
+==Video Solution 1 by TheSpreadTheMathLove==
+https://www.youtube.com/watch?v=N7cNhAx9ifE&t=0s
+==Solution 4==
+Draw the circumcircle of <math>\triangle{ABC}</math> and let the angle bisector of <math>\angle{B}</math> meet the circle at <math>D</math>
+By fact 5 we have <math>CD=AD, \angle{CBD}=\angle{ABD}=\angle{CAB}=\angle{DAC}, \angle{CBA}=\angle{DAB}</math>, thus <math>AC=BD, CD=CB</math>
+By Ptolemy, we have <math>c^2=(a+b)CD, CD=\frac{c^2}{a+b}=a, a^2+ab=c^2</math>. Try some numbers and the answer is <math>(4,5,6)\implies \boxed{15}</math>
+~Bluesoul
+==Solution 5==
+Let <math>\angle A=\theta</math>, then <math>\angle B=2\theta</math>.
+Find <math>D</math> on <math>AB</math> such that <math>\angle ACD=\theta</math>. Thus, <math>\angle CDB=\angle A+\angle ACD=2\theta</math>. So <math>AD=CD=BD</math>.
+Find <math>E</math> on <math>BD</math> such that <math>CE\perp BD</math>. Apparently, this gives <math>E</math> as the mid-point of <math>BD</math>.
-Let ∠A=θ， then ∠B=2θ
+Let the length of <math>BC</math> be <math>x</math>. Then <math>AB</math> can be expressed as <math>AD+BD=AD+2BE=x+2x\cos 2\theta</math>.
-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.
+Since <math>CE=x\sin 2\theta</math>, the length of <math>AC</math> can be expressed as <math>2x\cos\theta</math> (using double angle formula).
-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)
+Now we need to determine the range of <math>\theta</math>.
-So θ＜45°， this yield cosθ∈（\frac{\sqrt{2} }{2} ，I）
+The above conditions are only valid if <math>\angle B</math> is an acute angle.(the strict proof will be shown in the end*) So <math>\theta<45^\circ</math>, this yields <math>\cos\theta\in\left(\frac{\sqrt{2}}{2},1\right)</math>.
-Let cosθ=\frac{p}{q} ，where （p，q）=1
-We also know that \cos 2θ=2\cos^{2} θ-1
+Let <math>\cos\theta=\frac{p}{q}</math>, where <math>(p,q)=1</math>.
-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
+To minimize the perimeter, the denominator needs to be as small as possible. In this way, a small <math>x</math> can be used to integrate the side length.
-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.
+Test <math>q=2</math>, <math>\frac{1}{2}</math> is not 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
+Test <math>q=3</math>, <math>\frac{1}{3}</math> and <math>\frac{2}{3}</math> are not in the range (since <math>0.67<0.71</math>).
-When q becomes bigger， a larger x is required to integrate the length， thus can not give the minimum perimeter.
+Test <math>q=4</math>, <math>\frac{3}{4}</math> is in the range. In this case, the smallest <math>x</math> that make side length integer is <math>4</math>, since the side length is <math>x</math>, <math>\frac{5}{4}x</math> and <math>\frac{3}{2}x</math> So the perimeter is <math>4+5+6=15</math>.
+So <math>\boxed{\textbf{(C) }15}</math> is the correct answer.
+When <math>q</math> becomes bigger, a larger <math>x</math> is required to integrate the length, thus cannot give the minimum perimeter.
+If <math>\angle B>90^\circ</math>, <math>\angle A+\angle B>135^\circ</math>,then<math>\angle C<45^\circ</math>. This will result in point D on the extension of AB, meaning that <math>\angle CDB+\angle CBD<180^\circ</math>. Hence, <math>2\angle B<180^\circ</math>,<math>\angle B<90^\circ</math>, which contradicts our condition. If <math>\angle B=90^\circ</math>,The triangle is isosceles right triangle. So, the ratio of sides is <math>1:1:\sqrt{2}</math>, which, obviously means the length cannot be all integers.
+~solution by Tonyttian [https://artofproblemsolving.com/wiki/index.php/User:Tonyttian]
+~reformatted as LaTeX by Michaelihc [https://artofproblemsolving.com/wiki/index.php/User:Michaelihc]
+==Solution 6==
+Extend the angle bisector of <math>\angle B</math> to point <math>P</math> on <math>\overline{AC}.</math> We have <math>\triangle BPC \sim ABC,</math> so <math>\frac{m}{b} = \frac{b}{c},</math> yielding <math>m = \frac{b^2}{c}.</math> By the angle bisector theorem, <math>\frac{b}{a} = \frac{m}{n},</math> so <math>\frac{ma}{b} = \frac{ab}{c} = n</math> after substitution. We also have <math>m + n = c,</math> so
+<cmath>\frac{b^2}{c} + \frac{ab}{c} = c \implies b^2 + ab = b(b + a) = c^2,</cmath>
+for <math>a, b, c \in \mathbb{Z}_{> 0}.</math> We can't have <math>b=1</math> for any integral triangle. If <math>b = 2</math> then <math>2a + 4 = c^2</math>, which gives <math>c = 4</math> and <math>a = 6,</math> which fails the triangle inequality. If <math>b=3</math> then <math>a = 9</math> and <math>c = 6</math>, which fails again. If <math>b = 4</math>, then <math>(a, c) = (5, 6)</math>, which works and yields a perimeter of <math>15</math>. If <math>b=5</math>, then <math>(a, c) = (15, 10),</math> and if <math>b = 6</math>, then <math>(a, c) = (18, 12),</math> which fails again.
+If <math>b \geq 7</math> then <math>a+b>7,</math> yielding a minimum perimeter of <math>\boxed{\textbf{(C)\ 15}},</math> which was already achieved.
+-Benedict T (countmath1)
+==Solution 7 (No Casework, Trig)==
+Let <math>AB=c</math>, <math>BC=a</math>, <math>AC=b</math>.
+Our goal is to find
+<cmath>\min_{a,b,c \in \mathbb{N}} a+b+c</cmath>
+By the law of sines,
+<cmath> \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} </cmath>
+Substituting <math>B=2A</math> and <math>C = 180^\circ - A - B</math>:
+<cmath> \frac{a}{\sin(A)} = \frac{b}{\sin(2A)} = \frac{c}{\sin(180^\circ-3A)} = \frac{c}{\sin(3A)} </cmath>
+We will attempt to cancel the <math>\sin(A)</math> factor in the denominator. Using multiple-angle trigonometric identities,
+<cmath>\sin(2A) = 2\sin(A)\cos(A)</cmath>
+<cmath>\sin(3A) = \sin(A)(4\cos^2(A)-1)</cmath>
+we obtain:
+<cmath> \frac{a}{\sin(A)} = \frac{b}{2\sin(A)\cos(A)} = \frac{c}{\sin(A)(4\cos^2(A)-1))} </cmath>
+which simplifies to:
+<cmath> a = \frac{b}{2\cos(A)} = \frac{c}{(4\cos^2(A)-1))} </cmath>
+We notice that the term <math>2\cos(A)</math> appears multiple times. For convenience, let <math>t:=2\cos(A)</math>. Then, rewriting <math>b</math> and <math>c</math> in terms of <math>a</math>, we get:
+<cmath>b=ta</cmath>
+<cmath>c=(t^2-1)a</cmath>
+We now rewrite the original goal in terms of <math>t</math> and <math>a</math>. Notice that since <math>a,b,c \in \mathbb{N}</math>, then the perimeter <math>a+b+c \in \mathbb{N}</math>. Hence,
+<cmath> a+b+c = a + ta + (t^2-1)a </cmath>
+<cmath>= at(t+1) \in \mathbb{N}</cmath>
+This means that <math>t \in \mathbb{Q}</math>, so let <math>t := \frac{m}{n}</math> with <math>m,n \in \mathbb{N}</math>. To minimize
+<cmath>a+b+c = at(t+1) = a\left(\frac{m^2}{n^2}+\frac{m}{n}\right)</cmath>
+we set <math>a = n^2</math> to ensure the perimeter is an integer, and the objective becomes
+<cmath>\min_{m,n \in \mathbb{N}} m(m+n) </cmath>
+Notice that to have a valid triangle, <math>c=(t^2-1)a>0</math> implies <math>t^2 - 1 > 0</math> so <math>t > 1</math>.
+In addition, due to the triangle inequality, we must have <math>c < a + b</math>, such that
+<cmath> (t^2-1)a < a + ta </cmath>
+Since <math>a>0</math>, we can safely divide through, yielding <math> t > 2</math>. Overall, we have <math>1<t<2</math>, or equivalently, <math>n<m<2n</math>. The smallest pair satisfying this condition is <math>n=2</math>, <math>m=3</math>, yielding a perimeter of <math>15</math>. (Notice, to have a valid integer between <math>n</math> and <math>2n</math>, there must be a gap of at least <math>1</math> in between, implying <math>2n-n-1=n-1\ge 1</math>, so setting <math>n=2</math> yields the minimum.) Increasing <math>m</math> or <math>n</math> will increase the perimeter, so this is the global minimum. Hence the answer is:
+<cmath>\boxed{\textbf{(C)\ 15}}</cmath>
+==See also==
+{{AMC12 box|year=2024|ab=B|num-b=21|num-a=23}}
+{{MAA Notice}}

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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