Difference between revisions of "2024 AMC 12B Problems/Problem 22"
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− | Note that this is a 4-5 | + | ==Problem 22== |
+ | Let <math>\triangle{ABC}</math> be a triangle with integer side lengths and the property that <math>\angle{B} = 2\angle{A}</math>. What is the least possible perimeter of such a triangle? | ||
+ | |||
+ | <math> | ||
+ | \textbf{(A) }13 \qquad | ||
+ | \textbf{(B) }14 \qquad | ||
+ | \textbf{(C) }15 \qquad | ||
+ | \textbf{(D) }16 \qquad | ||
+ | \textbf{(E) }17 \qquad | ||
+ | </math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | Let <math>AB=c</math>, <math>BC=a</math>, <math>AC=b</math>. According to the law of sines, | ||
+ | <cmath>\frac{b}{a}=\frac{\sin \angle B}{\sin \angle A}</cmath> | ||
+ | <cmath>=2\cos \angle A</cmath> | ||
+ | |||
+ | According to the law of cosines, | ||
+ | <cmath>\cos \angle A=\frac{b^2+c^2-a^2}{2bc}</cmath> | ||
+ | |||
+ | Hence, | ||
+ | <cmath>\frac{b}{a}=\frac{b^2+c^2-a^2}{bc}</cmath> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Case <math>1</math>: <math>b=1</math> | ||
+ | |||
+ | Clearly, this case yields no valid solutions. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | Case <math>5</math>: <math>b=5</math> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC12 box|year=2024|ab=B|num-b=21|num-a=23}} | ||
+ | {{MAA Notice}} |
Revision as of 02:07, 16 November 2024
Contents
[hide]Problem 22
Let be a triangle with integer side lengths and the property that . What is the least possible perimeter of such a triangle?
Solution 1
Let , , . According to the law of sines,
According to the law of cosines,
Hence,
This simplifies to . We want to find the positive integer solution to this equation such that forms a triangle, and is minimized. We proceed by casework on the value of . Remember that .
Case :
Clearly, this case yields no valid solutions.
Case :
For this case, we must have and . However, does not form a triangle. Hence this case yields no valid solutions.
Case :
For this case, we must have and . However, does not form a triangle. Hence this case yields no valid solutions.
Case :
For this case, and , or and . As one can check, this case also yields no valid solutions
Case :
For this case, we must have and . There are no valid solutions
Case :
For this case, and , or and , or and . The only valid solution for this case is , which yields a perimeter of .
When , it is easy to see that . Hence , which means . Therefore, the answer is
~tsun26
Solution 2 (Similar to Solution 1)
Let , , . Extend to point on such that . This means is isosceles, so . Since is the exterior angle of , we have Thus, is isosceles, so Then, draw the altitude of , from to , and let this point be . Let . Then, by Pythagorean Theorem,
Case : .
This means , so the least possible values are , , but this does not work as it does not satisfy the triangle inequality. Similarly, , also does not satisfy it. Anything larger goes beyond the answer choices, so we stop checking this case.
Case : This means , so the least possible values for and are ,, but this does not satisfy the triangle inequality, and anything larger does not satisfy the answer choices.
Case : This means , and the least possible value for is , which occurs when . Unfortunately, this also does not satisfy the triangle inequality, and similarly, any means the perimeter will get too big.
Case : This means , so we have , so the least possible perimeter so far is .
Case : We have , so least possible value for is , which already does not work as , and the minimum perimeter is already.
Case : We have , so , which already does not work.
Then, notice that when , we also must have and , so , so the least possible perimeter is
~evanhliu2009
Solution 3 (Trigonometry)
cos(A) must be rational, let's evaluate some small values
case #1: cos(A) = invalid since =
case #2: cos(A) = invalid since <
case #3: cos(A) = give with side (9:12:7) , perimeter = 28
case #4: cos(A) = invalid since <
case #5: cos(A) = give with side (4:6:5), perimeter = 15
case #6: cos(A) = give with side (25:30:11)
case #7: cos(A) = give with side (25:40:39)
case #8: cos(A) = same as
case #9: cos(A) = give with side (9:15:16)
case #10: when a 7, b =2cos(A)*a > 2* *7 = 7 , a+b+c > 15
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.