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

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(5 intermediate revisions by 4 users not shown)
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<cmath> \frac{a}{sin(A)} = \frac{a}{sin(B)} = \frac{c}{sin(C)} </cmath>
 
<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}{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{a}{2sin(A)cos(A)} = \frac{c}{3sin(A) - 4sin^3(A)} </cmath>
 
<cmath> b = 2cos(A)a </cmath>
 
<cmath> b = 2cos(A)a </cmath>
 
<cmath> c = (3  - 4sin^2(A) ) a = (4cos^2(A)-1)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> 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: 2cos(A) : 4cos^2(A)-1 </cmath>
 
cos(A) must be rational, let's evaluate some small values
 
cos(A) must be rational, let's evaluate some small values
  
−
case #1: cos(A) = <math>\frac{1}{2}</math> invalid c= 0
+
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 c < 4  <math> \cdot \frac{1}{9} -1 < 0 </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 #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 #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)  
+
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)  
  
−
for denominators 5 and above, the fraction denominators getting larger and perimeter will be larger than 15
+
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)
  
−
<math>\boxed{\textbf{(C) }15}.</math>
+
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]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
Line 141: Line 148:
 
So the perimeter= 4+5+6=15
 
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.
 
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>
 +
 +
~Bluesoul
 +
 +
==See also==
 +
{{AMC12 box|year=2024|ab=B|num-b=21|num-a=23}}
 +
{{MAA Notice}}

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

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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