Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 27"
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== Solution == | == Solution == | ||
− | Notice that <math>P</math> is the [[incenter]] of the [[triangle]]. The [[incircle]] has [[radius]] <math>2</math>. Thus, using <math>rs | + | Notice that <math>P</math> is the [[incenter]] of the [[triangle]]. The [[incircle]] has [[radius]] <math>2</math>. Thus, using the [[area]] formula <math>A=rs</math> we have <math>2 \cdot s=24 \Longrightarrow s=12</math> and the perimeter is <math>24</math>. |
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 26|Previous Problem]] | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 26|Previous Problem]] | ||
− | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 28 | + | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 28|Next Problem]] |
* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]] | * [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]] | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |