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Question Number 38281 by Tinkutara last updated on 23/Jun/18

Answered by ajfour last updated on 23/Jun/18

$$\frac{{ap}_{A\to BC}}{2}=\frac{{bp}_{B\to CA}}{2}=\frac{{cp}_{C\to AB}}{2}=△$$$$\frac{{ap}_{P\to BC}}{2}+\frac{{bp}_{P\to CA}}{2}+\frac{{cp}_{P\to AB}}{2}=\mathrm{\Delta }$$$$p_{P\to BC}=\frac{p_{A\to BC}}{r}andsoon..$$$$\Rightarrow \frac{{ap}_{P\to BC}}{2}=\frac{{ap}_{A\to BC}}{2r}=\frac{\mathrm{\Delta }}{r}$$$$\frac{{ap}_{P\to BC}}{2}=\frac{{bp}_{P\to CA}}{2}=\frac{{cp}_{P\to AB}}{2}=\frac{\mathrm{\Delta }}{r}$$$$So$$$$\frac{3\mathrm{\Delta }}{r}=\mathrm{\Delta }\Rightarrow r=3.$$$$$$

Commented by Tinkutara last updated on 24/Jun/18

Thank you very much Sir! I got the answer. �������� Very nice solution.

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