circumradii wanted in terms of inradii, <A=60^o (Mathematical Excalibur 399)

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circumradii wanted in terms of inradii, <A=60^o (Mathematical Excalibur 399)

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#1 h Apr 8, 2020, 11:37 PM

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Let $ABC$ be a triangle for which $\angle BAC=60^o$ . Let P be the point of intersection of the bisector of $\angle ABC$ and the side $AC$ . Let $Q$ be the point of intersection of the bisector of $\angle ACB$ and the side $AB$ . Let $r_1$ and $r_2$ be the radii of the incircles of triangles $ABC$ and $APQ$ respectively. Determine the radius of the circumcircle of triangle $APQ$ in terms of $r_1$ and $r_2$ with proof.

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#2 h Apr 10, 2020, 7:00 PM

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Given $\angle BAC=60^{\circ}$ ; if $\angle O_{1}BC = \beta$ , then $\angle O_{1}CB=60^{\circ}-\beta$
and $\angle BO_{1}C = \angle PO_{1}Q=120^{\circ}$ , so $APQO_{1}$ is concyclic (red circle).

$\sin 30^{\circ}=\frac{r_{2}}{AO_{2}}=\frac{r_{1}}{AO_{1}}$ , so $AO_{1}=2r_{1}$ and $AO_{2}=2r_{2}$ .
$\triangle APO_{1}\ :\ \frac{O_{1}P}{\sin 30^{\circ}}=2R$ , the radius $R$ of the red circle, so $O_{1}P=R \quad\quad(1)$ .

$\triangle APQ$ with incenter $O_{2}$ .
1) $\angle O_{1}AP=\angle O_{1}AQ=\angle O_{1}PQ \Rightarrow O_{1}P=O_{1}Q \quad\quad(2)$ .
2) $\angle QPO_{2}=\angle O_{2}PA$ ,
rewriting as a difference $\angle O_{1}PO_{2}-\angle O_{1}PQ = \angle O_{1}O_{2}P-\angle O_{1}AP$ , because $\angle O_{1}AP+\angle O_{2}PA=\angle O_{1}O_{2}P$ .
Because (2), we have $\angle O_{1}PO_{2} = \angle O_{1}O_{2}P \Rightarrow O_{1}P=O_{1}O_{2}$ .
(1) $R=O_{1}P=O_{1}O_{2}=AO_{1}-AO_{2}=2r_{1}-2r_{2}$ .

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