2024 AMC 12B Problems/Problem 22
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[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 | |
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