Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 27"

m
 
(One intermediate revision by one other user not shown)
Line 6: Line 6:
  
 
== Solution ==
 
== Solution ==
Notice that <math>P</math> is the [[incenter]] of the [[triangle]]. The [[incircle]] has [[radius]] <math>2</math>. Thus, using <math>rs=A</math>, we have <math>2 \cdot s=24 \Longrightarrow s=12</math> and the perimeter is <math>24</math>.
+
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>.
  
 
----
 
----
  
 
* [[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/Next Problem]]
+
* [[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]]

Latest revision as of 16:27, 18 August 2006

Problem

Suppose $\triangle ABC$ is a triangle with area 24 and that there is a point $P$ inside $\triangle ABC$ which is distance 2 from each of the sides of $\triangle ABC$. What is the perimeter of $\triangle ABC$?

$\mathrm{(A) \ } 12 \qquad \mathrm{(B) \ }24 \qquad \mathrm{(C) \ }36 \qquad \mathrm{(D) \ }12\sqrt{2} \qquad \mathrm{(E) \ }12\sqrt{3}$

Solution

Notice that $P$ is the incenter of the triangle. The incircle has radius $2$. Thus, using the area formula $A=rs$ we have $2 \cdot s=24 \Longrightarrow s=12$ and the perimeter is $24$.