Difference between revisions of "2024 AMC 12B Problems/Problem 22"
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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>. | 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>. | + | 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 | ~tsun26 |
Revision as of 10:29, 14 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 c= 0
case #2: cos(A) = invalid
case #3: cos(A) = give with side (9:12:7) , perimeter = 28
case #4: cos(A) = invalid c<0
case #5: cos(A) = give with side (4:6:5)
for denominators 5 and above, the fraction denominators getting larger and perimeter will be larger than 15
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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