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

Latest revision as of 16:05, 22 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 $a<a+c$ .

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

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 Solution 4

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

So 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

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

@@ Line 24: / Line 24: @@
 This simplifies to <math>b^2=a(a+c)</math>. We want to find the positive integer solution <math>(a, b, c)</math> to this equation such that <math>a, b, c</math> forms a triangle, and <math>a+b+c</math> is minimized. We proceed by casework on the value of <math>b</math>. Remember that <math>a<a+c</math>.
-<math>\textbf{Case 1: b=1}</math>
+Case <math>1</math>: <math>b=1</math>
 Clearly, this case yields no valid solutions.
-<math>\textbf{Case 2: b=2}</math>
+Case <math>2</math>: <math>b=2</math>
 For this case, we must have <math>a=1</math> and <math>c=3</math>. However, <math>(1, 2, 3)</math> does not form a triangle. Hence this case yields no valid solutions.
-<math>\textbf{Case 3: b=3}</math>
+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.
-<math>\textbf{Case 4: b=4}</math>
+Case <math>4</math>: <math>b=4</math>
 For this case, <math>a=1</math> and <math>c=15</math>, or <math>a=2</math> and <math>c=6</math>. As one can check, this case also yields no valid solutions
-<math>\textbf{Case 5: b=5}</math>
+Case <math>5</math>: <math>b=5</math>
-For this case, we must have <math>a=1</math> <math>c=24</math>. There are no valid solutions
+For this case, we must have <math>a=1</math> and <math>c=24</math>. There are no valid solutions
+Case <math>6</math>: <math>b=6</math>
+For this case, <math>a=2</math> and <math>c=16</math>, or <math>a=4</math> and <math>c=5</math>, or <math>a=3</math> and <math>c=9</math>. The only valid solution for this case is <math>(4, 6, 5)</math>, which yields a perimeter of <math>15</math>.
+When <math>b\ge 7</math>, it is easy to see that <math>a+c>7</math>. Hence <math>a+b+c>14</math>, which means <math>a+b+c\ge15</math>. Therefore, the answer is <math>\fbox{\textbf{(C) }15}</math>
+~tsun26
+==Solution 2 (Similar to Solution 1)==
+Let <math>\overline{BC}=a</math>, <math>\overline{AC}=b</math>, <math>\overline{AB}=c</math>. Extend <math>C</math> to point <math>D</math> on <math>\overline{AB}</math> such that <math>\angle ACD = \angle CAD</math>. This means <math>\triangle CDA</math> is isosceles, so <math>CD=DA</math>. Since <math>\angle CDB</math> is the exterior angle of <math>\triangle CDA</math>, we have <cmath>\angle CDB=m+m=2m=\angle CBD.</cmath> Thus, <math>\triangle CBD</math> is isosceles, so <math>CB=CD=DA=a.</math> Then, draw the altitude of <math>\triangle CBD</math>, from <math>C</math> to <math>\overline{BD}</math>, and let this point be <math>H</math>. Let <math>BH=HD=x</math>. Then, by Pythagorean Theorem,
+\begin{align*}
+CH^2&=a^2-x^2 \\
+CH^2&= b^2 - (c+x)^2.\\
+\end{align*}
+Thus, <cmath>a^2-x^2=b^2-(c-x)^2.</cmath> Solving for <math>x</math>, we have <math>x=\frac{a^2-b^2+c^2}{2c}.</math> Since <math>2x=c-a</math>, we have <cmath>c-a=\frac{a^2-b^2+c^2}{c},</cmath> and simplifying, we get <math>b^2=a^2+ac.</math> Now we can consider cases on what <math>a</math> 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 <math>1</math>: <math>a=1</math>.
+This means <math>b^2=c+1</math>, so the least possible values are <math>b=2</math>, <math>c=3</math>, but this does not work as it does not satisfy the triangle inequality. Similarly, <math>b=3</math>, <math>c=8</math> also does not satisfy it. Anything larger goes beyond the answer choices, so we stop checking this case.
+Case <math>2</math>: <math>a=2</math>
+This means <math>b^2=2c+4</math>, so the least possible values for <math>b</math> and <math>c</math> are <math>b=4</math>,<math>c=6</math>, but this does not satisfy the triangle inequality, and anything larger does not satisfy the answer choices.
+Case <math>3</math>: <math>a=3</math>
+This means <math>b^2=3c+9</math>, and the least possible value for <math>b</math> is <math>b=6</math>, which occurs when <math>c=9</math>. Unfortunately, this also does not satisfy the triangle inequality, and similarly, any <math>b > 6</math> means the perimeter will get too big.
+Case <math>4</math>: <math>a=4</math>
+This means <math>b^2=4c+16</math>, so we have <math>b=6,c=5,a=4</math>, so the least possible perimeter so far is <math>4+5+6=15</math>.
+Case <math>5</math>: <math>a=5</math>
+We have <math>b^2=5c+25</math>, so least possible value for <math>b</math> is <math>b=10</math>, which already does not work as <math>a=5</math>, and the minimum perimeter is <math>15</math> already.
+Case <math>6</math>: <math>a=6</math>
+We have <math>b^2=6c+36</math>, so <math>b=10</math>, which already does not work.
+Then, notice that when <math>a\geq 7</math>, we also must have <math>b\geq8</math> and <math>c\geq1</math>, so <math>a+b+c \geq 16</math>, so the least possible perimeter is <math>\boxed{\textbf{(C) }15}.</math>
+~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(\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> b = 2cos(A)a </cmath>
+<cmath> c = (3  - 4sin^2(A) ) a = (4cos^2(A)-1)a </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>
+cos(A) must be rational, let's evaluate some small values
+case #1: cos(A) = <math>\frac{1}{2}</math> invalid since = <math>\frac{1}{2}</math>
+case #2: cos(A) = <math>\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 #4: cos(A) = <math>\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), perimeter = 15
+case #6: cos(A) = <math>\frac{3}{5}</math> give <math>{1: \frac{6}{5} :  \frac{11}{25}  }</math> with side (25:30:11)
+case #7: cos(A) = <math>\frac{4}{5}</math> give <math>{1: \frac{8}{5} :  \frac{39}{25}  }</math> with side (25:40:39)
+case #8: cos(A) = <math>\frac{4}{6}</math> same as <math>\frac{2}{3}</math>
+case #9: cos(A) = <math>\frac{5}{6}</math> give <math>{1: \frac{5}{3} :  \frac{16}{9}  }</math> with side (9:15:16)
+case #10: when a <math>\geq</math> 7, b =2cos(A)*a > 2*  <math>\frac{1}{2}</math>*7 = 7 , a+b+c > 15
+<math>\boxed{\textbf{(C) }15}</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist 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
+Solution 4
+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
+ So 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
+==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>
-<math>\textbf{Case 6: b=6}</math>
+~Bluesoul
-For this case, <math>a=2</math> and <math>c=16</math>, or <math>a=4</math> and <math>c=5</math>, or <math>a=3</math> and <math>c=9</math>. The only valid solution for this case is <math>(4, 6, 5).
-It is safe to assume that </math>(4, 5, 6)$ will be the solution with least perimeter. Hence, the answer is
+==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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