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

Latest revision as of 17:16, 19 December 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 $a<a+c$ .

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, \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)

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

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

@@ Line 165: / Line 165: @@
 ～Tonyttian [https://artofproblemsolving.com/wiki/index.php/User:Tonyttian]
+==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)
 ==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
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

Page

Toolbox

Search