Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $24$ and its perimeter is $24$ cm, then what is the length of the smallest side?

share|cite|improve this question
I think that you mean excircles instead of ex-scribed circles. (this seems like the direct Romanian to English translation :) ) See 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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.