Difference between revisions of "2005 AMC 8 Problems/Problem 15"
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==Solution== | ==Solution== | ||
− | Let <math>b</math> be the base of the isosceles triangles, and let <math>a</math> be the lengths of the other legs. From this, <math>2a+b=23</math> and <math>b=23-2a</math>. From triangle inequality, <math>2a>b</math>, then plug in the value from the previous equation to get <math>2a>23-2a</math> or <math>a>5.75</math>. The maximum value of <math>a</math> occurs when <math>b=1</math>, in which from the first equation <math>a=11</math>. Thus, <math>a</math> can have integer side lengths from <math>6</math> to <math>11</math>, and there are <math>\boxed{\textbf{( | + | Let <math>b</math> be the base of the isosceles triangles, and let <math>a</math> be the lengths of the other legs. From this, <math>2a+b=23</math> and <math>b=23-2a</math>. From triangle inequality, <math>2a>b</math>, then plug in the value from the previous equation to get <math>2a>23-2a</math> or <math>a>5.75</math>. The maximum value of <math>a</math> occurs when <math>b=1</math>, in which from the first equation <math>a=11</math>. Thus, <math>a</math> can have integer side lengths from <math>6</math> to <math>11</math>, and there are <math>\boxed{\textbf{(C)}\ 6}</math> triangles. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=14|num-a=16}} | {{AMC8 box|year=2005|num-b=14|num-a=16}} |
Revision as of 14:45, 20 June 2013
Problem
How many different isosceles triangles have integer side lengths and perimeter 23?
Solution
Let be the base of the isosceles triangles, and let be the lengths of the other legs. From this, and . From triangle inequality, , then plug in the value from the previous equation to get or . The maximum value of occurs when , in which from the first equation . Thus, can have integer side lengths from to , and there are triangles.
See Also
2005 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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All AJHSME/AMC 8 Problems and Solutions |