Difference between revisions of "2024 AMC 12B Problems/Problem 22"

(Solution 5)
(Add analytical solution)
 
Line 174: Line 174:
  
 
-Benedict T (countmath1)
 
-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==
 
==See also==
 
{{AMC12 box|year=2024|ab=B|num-b=21|num-a=23}}
 
{{AMC12 box|year=2024|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:21, 13 January 2025

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

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 ∠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 2xcosθ(using double angle formula)

Now we need to determine the range of θ.

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 yields cosθ∈($\frac{\sqrt{2}}{2}$ ,1)

Let cosθ=$\frac{p}{q}$ ,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, 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,$\frac{5}{4}$x and $\frac{3}{2}$x So the perimeter= 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 can not give the minimum perimeter.

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

~Tonyttian [1]

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:

\[\boxed{\textbf{(C)\ 15}}\]

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png