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Question Number 154415 by mr W last updated on 18/Sep/21

Commented by mr W last updated on 18/Sep/21

$$\left[Q154210\right]$$

Answered by mr W last updated on 18/Sep/21

Commented by mr W last updated on 18/Sep/21

$$sayE,F,Garethecentersofidentical$$$$circleswithradiusr.$$$$Zisthecircumcenterof\mathrm{\Delta }EFG,since$$$$ZE=ZF=ZG=r,andthecircum-$$$$radiusof\mathrm{\Delta }EFGis{r}_{out}=r.$$$$saythecenterofincircleof\mathrm{\Delta }EFG$$$$isYandtheinradiusof\mathrm{\Delta }EFGis{r}_{in}.$$$$thedistancesfromYtothesidesof$$$$\mathrm{\Delta }EFGare{r}_{in}.$$$$thedistancesfromYtothesidesof$$$$\mathrm{\Delta }ABCare{r}_{in}+r.thatmeansYis$$$$alsotheincenterof\mathrm{\Delta }ABC,and$$$$R_{in}=r_{in}+r.$$$$itisobviourthat\mathrm{\Delta }ABC\sim \mathrm{\Delta }EFG,$$$$therefore$$$$\frac{circumradiusof\mathrm{\Delta }EFG}{circumradiusof\mathrm{\Delta }ABC}=\frac{inradiusof\mathrm{\Delta }EFG}{inradiusof\mathrm{\Delta }ABC}$$$$\Rightarrow \frac{r_{out}}{R_{out}}=\frac{r_{in}}{R_{in}}$$$$\Rightarrow \frac{r}{R_{out}}=\frac{r_{in}}{r_{in}+r}$$$$\Rightarrow \frac{1}{R_{out}}=\frac{r_{in}}{(r_{in}+r)r}$$$$\frac{1}{R_{in}}+\frac{1}{R_{out}}=\frac{1}{r_{in}+r}+\frac{r_{in}}{(r_{in}+r)r}=\frac{1}{r}$$$$\Rightarrow \frac{1}{r}=\frac{1}{R_{in}}+\frac{1}{R_{out}}$$

Commented by Tawa11 last updated on 18/Sep/21

$$\mathrm{Weldone}\mathrm{sir}$$

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