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I have a triangle $ABC$ with inradius $r$. Points $D$, $E$, $F$ are chosen on side $BC$, $CA$, $AB$, respectively, such that $\triangle AFE$, $\triangle BDF$, and $\triangle CED$ have same inradius $r_1$. Compute the inradius of $\triangle DEF$ in terms of $r$ and $r_1$.

I have an approach in mind by taking the various side lengths as x,y,z and so on but I know for sure it will turn out to be lengthy. Any better method please?

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$$\begin{eqnarray*}r_1\left(p_{AFE}+p_{BDF}+p_{CED}\right)&=&2\left([AFE]+[BDF]+[CED]\right)\\&=&2\left([ABC]-[DEF]\right)\\&=&r\,p_{ABC}-r_{DEF}\,p_{DEF}\\&=&r_1\left(p_{ABC}+p_{DEF}\right)\end{eqnarray*}$$ leads to $$ r_{DEF}\, p_{DEF} = (r-r_1) p_{ABC} - r_1 p_{DEF}. $$ Now the interesting fact is that the perimeter of $DEF$ is fixed by $r,r_1,p_{ABC}$.

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By considering the distances of $D,E,F$ from the tangency points marked with a red cross we have that $p_DEF$ equals the perimeter of $I_A I_B I_C$, with $I_A,I_B,I_C$ being the incenters of $AEF,BDF,CDE$. It follows that

$$ p_{DEF} = \frac{r-r_1}{r} p_{ABC} $$ and

$$ r_{DEF} = r- r_1.$$

Another interesting fact is that $D,E,F$ and $I_A,I_B,I_C$ belong to the same ellipse.

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  • $\begingroup$ How did you compute the perimeter of triangle $I_AI_BI_C$? $\endgroup$ – saisanjeev Oct 3 '18 at 14:22
  • $\begingroup$ @saisanjeev: $I_A I_B I_C$ is a scaled version of $ABC$. The dilation factor is clearly $\frac{r-r_1}{r}$. $\endgroup$ – Jack D'Aurizio Oct 3 '18 at 15:42

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