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The radii $r_1,r_2,r_3$ of ex-scribed circles of the triangle $ABC$ are in harmonic progression. If the area of the triangle is $24$ sq.cm and its perimeter is $24$ cm, then what is the length of the smallest side?

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I think that you mean excircles instead of ex-scribed circles. (this seems like the direct Romanian to English translation :) ) See mathworld.wolfram.com/Excenter.html for more terminology. Also, try and motivate your question a bit, even if it's pure homework. Write down what you have tried until now. –  Beni Bogosel Mar 12 '12 at 11:46
    
I don't know where to begin, and its really important me to know how its done. –  Tomarinator Mar 12 '12 at 11:54
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up vote 2 down vote accepted

If you denote $r_a,r_b,r_c$ the radii of the excircles corresponding to the sides $a,b,c$ then you have the formulas

$$ r_a=\frac{S}{p-a},r_b=\frac{S}{p-b}, r_c=\frac{S}{p-c}$$ where $S$ is the area of the triangle and $p$ is the semiperimeter.

The fact that $r_a,r_b,r_c$ are in harmonic progression means that

$$ \frac{1}{r_a}+\frac{1}{r_c}=\frac{2}{r_b}$$

This will give you easily the fact that $a,b,c$ are in fact in arithmetic progression, i.e. $2b=a+c$. You know the perimeter, so you can find $b$. Write $a=b-r,c=b+r$. Substitute in Heron's formula for the area, and find $r$. Then you can find the smallest side of the triangle.

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