Difference between revisions of "2024 AMC 12B Problems/Problem 22"
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+ | ==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 |
Revision as of 08:10, 14 November 2024
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 .
Clearly, this case yields no valid solutions.
For this case, we must have and . However, does not form a triangle. Hence this case yields no valid solutions.
For this case, we must have and . However, does not form a triangle. Hence this case yields no valid solutions.
For this case, and , or and . As one can check, this case also yields no valid solutions
For this case, we must have and . There are no valid solutions
For this case, and , or and , or and . The only valid solution for this case is .
It is safe to assume that will be the solution with least perimeter. Hence, 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