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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 27"
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== Solution ==
 
== Solution ==
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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>.
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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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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 $24$.

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