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

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~evanhliu2009
~evanhliu2009
==Solution 3 (Trigonometry)==
<cmath> \frac{a}{sin(A)} = \frac{a}{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{a}{2sin(A)cos(A)} = \frac{c}{3sin(A) - 4sin^3(A)} </cmath>
<cmath> b = 2cos(A)a </cmath>
<cmath> c = (3  - 4sin^2(A) ) a = (4cos^2(A)-1)a </cmath>
<cmath> A <= 60 ,  \frac{1}{2} <= cos(A)  <=1 </cmath>
<cmath> cos(A) =  \frac{3}{4} </cmath>
<cmath> a:b:c = 1: 2cos(A) : 4cos^2(A)-1 </cmath>
cos(A) must be rational, let's evaluate some small values
case #1: cos(A) = <math>\frac{1}{2}</math> invalid  c= 0
case #2: cos(A) = <math>\frac{1}{3}</math> invalid   
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 #4: cos(A) = <math>\frac{1}{4}</math> invalid  c<0
case #5: cos(A) = <math>\frac{3}{4}</math> give <math>{1: \frac{3}{2} :  \frac{5}{4}  }</math> with side (4:6:5)
for denominators 5 and above, the fraction denominators getting larger  and perimeter will be larger than 15
<math>\boxed{\textbf{(C) }15}.</math>
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
==See also==
{{AMC12 box|year=2024|ab=B|num-b=21|num-a=23}}
{{MAA Notice}}

Revision as of 09:24, 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

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)

\[\frac{a}{sin(A)} = \frac{a}{sin(B)} = \frac{c}{sin(C)}\] \[\frac{a}{sin(A)} = \frac{a}{sin(2A)} = \frac{c}{sin(3A)}\] \[\frac{a}{sin(A)} = \frac{a}{2sin(A)cos(A)} = \frac{c}{3sin(A) - 4sin^3(A)}\] \[b = 2cos(A)a\] \[c = (3 - 4sin^2(A) ) a = (4cos^2(A)-1)a\] \[A <= 60 , \frac{1}{2} <= cos(A) <=1\] \[cos(A) = \frac{3}{4}\] \[a:b:c = 1: 2cos(A) : 4cos^2(A)-1\] cos(A) must be rational, let's evaluate some small values

case #1: cos(A) = $\frac{1}{2}$ invalid c= 0

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

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 c<0

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

for denominators 5 and above, the fraction denominators getting larger and perimeter will be larger than 15

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

~luckuso

See also

2024 AMC 12B (Problems • Answer Key • Resources)
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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