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I am trying to derive a relation between radius of those outer circles and radius of the incircle. Those outer circles are tangent to the incircle and respective sides.

I have tried and failed miserably. Any help will be appreciated.

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2 Answers 2

up vote 3 down vote accepted

This is more an extended hint than an answer, but was getting too long for a comment.

Look at the bottom left-hand corner. The angle bisector goes through the centres of the incircle and the smaller circle. If you look at the triangles which are cut off by the bisector, one of the sides of the triangle, and the radii of the two circles you will have similar right-angled triangles.

If you use $d$ for the distance from the vertex to the centre of the small circle, you should be able to express this in terms of the radii of the circles. When you have eliminated $d$ you should get a relationship which allows you to express the smaller radius in terms of the inradius and half the angle at the vertex.

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I get $$sin(A/2)=\frac{r-r_1}{r+r_1}$$ and similarly 2 other expressions. What would be the best way to eliminate angles? –  Awesome Apr 22 at 13:02

You can use Descartes 4 Circles Theorem. Descartes' theorem says: If four circles are tangent to each other at six distinct points and the circles have curvatures $k_i$ (for $i = 1,\cdots, 4$), then $k_i$ satisfies the following relation: $$ (k_1+k_2+k_3+k_4)^2=2(k_1^2+k_2^2+k_3^2+k_4^2), $$ where $k_i=\pm\dfrac{1}{r_i}$, $r_i$ is the radius of circle. The equation can also be written as: $$ k_4=k_1+k_2+k_3\pm2\sqrt{k_1k_2+k_2k_3+k_1k_3}, $$ or $$ \frac{1}{r_4}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}\pm2\sqrt{\frac{1}{r_1r_2}+\frac{1}{r_2r_3}+\frac{1}{r_1r_3}}. $$ The generalization to $n$ dimensions or variables is referred to as the Soddy–Gosset theorem. $$ \left(\sum_{i=1}^{n+2}k_i\right)^2=n\sum_{i=1}^{n+2}k_i^2. $$ For detail explanation and complete proof of Descartes' theorem (also to answer your question), you may refer to these sites: 1, 2, or download this journal. $$\\$$


$$\large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

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