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Let $I$ be the incentre of $\Delta ABC$ and let $P_1$, $P_2$, and $P_3$ be the circumradii of $\Delta BIC$, $\Delta CIA$, and $\Delta AIB$, respectively (i.e., $P_1 = O_1B$, $P_2 = O_2C$, $P_3=O_3A$, where $O_1$, $O_2$, and $O_3$ are circumcentres of $\Delta BIC$, $\Delta CIA$, and $\Delta AIB$, respectively). Let the inradius of $\Delta ABC$ be $r$ and let the circumradius of $ \Delta ABC$ be $R$. Prove that $P_1P_2P_3=2rR^2$.

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  • $\begingroup$ Please define all objects, even if they are introduced in the picture. (What are $O1$, $O2$, $O3$?) What are the own efforts? You may use a (central) dot in latex as \cdot instead of the bolder star. Indices may be accessed as P_1 for $P_1$, etc. $\endgroup$
    – dan_fulea
    Sep 4, 2018 at 21:18

2 Answers 2

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In fact, $O_1,O_2,O_3$ are the second intersection points of $AI,BI,CI$ respectively and the circumcircle $(ABC)$. Thus, you may readily obtain that $$p_1=2R\sin \frac{A}{2},~~~p_2=2R\sin \frac{B}{2},~~~p_3=2R\sin \frac{A}{2}.$$

Moreover, I guess you must know the formula on the incircle radius that $$r=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}.$$

Therefore, $$p_1p_2p_3=2\cdot4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\cdot R^2=2rR^2.$$

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Let $i$ denote the inversion about the incircle of the triangle $ABC$. Write $D$, $E$, and $F$ for $i(A)$, $i(B)$, and $i(C)$, respectively. Then, the lines $EF$, $FD$, and $DE$ are the images of the circumcircles of the triangles $BIC$, $CIA$, and $AIB$, respectively. The point $I$ is the orthocenter of the triangle $DEF$. Let $X$, $Y$, and $Z$ denote the feet of the altitudes from $D$, $E$, and $F$, respectively, of the triangle $DEF$.

Point Definitions

We have $$P_1=O_1I=\frac{r^2}{2\,IX}\,,$$ and similarly, $$P_2=O_2I=\frac{r^2}{2\,IY}\text{ and }P_3=O_3I=\frac{r^2}{2\,IZ}\,.$$ That is, to prove that $P_1\cdot P_2\cdot P_3=2\,r\,R^2$, we need to prove that $$IX\cdot IY\cdot IZ=\frac{r^5}{16\,R^2}\,.$$

Write $\alpha:=\dfrac12\,\angle BAC$, $\beta:=\dfrac12\,\angle ABC$, and $\gamma:=\dfrac12\,\angle BCA$. Note that $IA=\dfrac{r}{\sin(\alpha)}$. That is, $$ID=\frac{r^2}{IA}=r\,\sin(\alpha)\,.$$ Similarly, $IE=r\,\sin(\beta)$ and $IF=r\,\sin(\gamma)$. We have $\angle IEX=\gamma$, $\angle IFX=\alpha$, and $\angle IDX=\beta$, whence $$IX=IE\,\sin(\gamma)=r\,\sin(\beta)\,\sin(\gamma)\,,$$ $$IY=IF\,\sin(\alpha)=r\,\sin(\gamma)\,\sin(\alpha)\,,$$ and $$IZ=ID\,\sin(\beta)=r\,\sin(\alpha)\,\sin(\beta)\,.$$ Consequently, $$IX\cdot IY\cdot IZ=r^3\,\sin^2(\alpha)\,\sin^2(\beta)\,\sin^2(\gamma)\,.$$ It is well known that $r=4\,R\,\sin(\alpha)\,\sin(\beta)\,\sin(\gamma)$. Ergo, $IX\cdot IY\cdot IZ=\dfrac{r^5}{16\,R^2}$, as required.

P.S.: My solution is overly complicated, but I will leave it like this. In fact, it can be proven directly (without inversion) via trigonometry that $$P_1=O_1I=\frac{r}{2\,\sin(\beta)\,\sin(\gamma)}\,,$$ $$P_2=O_2I=\frac{r}{2\,\sin(\gamma)\,\sin(\alpha)}\,,$$ and $$P_3=O_3I=\frac{r}{2\,\sin(\alpha)\,\sin(\beta)}\,.$$

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