Diego Barr

2022-10-12

Let ${A}_{1},{B}_{1},{C}_{1}$ be the tangency points of the intersection of the excircles of a triangle $ABC$ with the sides $BC,CA,AB,$ respectively. Prove that the circumcircles of $AB{B}_{1}$ and $AC{C}_{1}$ meet on a bisector of $\mathrm{\angle}BAC.$

What I thought: Circle $(AB{B}_{1})$ has ${\omega}_{1}=b(s-a)$ and circle $(AC{C}_{1})$ has ${v}_{2}=c(s-a)$. Their radical axis is given by $-yc(s-a)+zb(s-a)$ which is equation of A angle bisector and we are done.

What I thought: Circle $(AB{B}_{1})$ has ${\omega}_{1}=b(s-a)$ and circle $(AC{C}_{1})$ has ${v}_{2}=c(s-a)$. Their radical axis is given by $-yc(s-a)+zb(s-a)$ which is equation of A angle bisector and we are done.

Alannah Yang

Beginner2022-10-13Added 22 answers

Let $D$ be on $\overline{BC}$ so that $\overline{AD}$ bisects $\mathrm{\angle}BAC$. Let $\phi $ be the transformation of the plane formed by composing an inversion with center $A$ and radius $r=\sqrt{AB\cdot AC}$, with a reflection through $\overline{AD}$. Notice that $\phi (B)=C$, $\phi (C)=B$, $\phi (\overline{AD})=\overline{AD}$. Let also ${B}_{2}=\phi ({B}_{1})$, ${C}_{2}=\phi ({C}_{1})$. Notice that $\phi ((AB{B}_{1}))=\overline{C{B}_{2}}$, $\phi ((AC{C}_{1}))=\overline{B{C}_{2}}$. We now wish to prove that $\overline{AD}$, $\overline{B{C}_{2}}$, $\overline{C{B}_{2}}$ concur. We do this by Ceva on $\u25b3ABC$.

Let $a=\overline{BC}$, $b=\overline{CA}$, $c=\overline{AB}$, $s=\frac{a+b+c}{2}$. We have

$\frac{|\overline{A{C}_{2}}|}{|\overline{{C}_{2}B}|}=\frac{|\overline{A{C}_{2}}|}{|\overline{A{C}_{2}}|-|\overline{AB}|}=\frac{\frac{{r}^{2}}{s-c}}{\frac{{r}^{2}}{s-c}-\frac{{r}^{2}}{b}}=\frac{b}{b+c-s}.$

Likewise,

$\frac{|\overline{C{B}_{2}}|}{|\overline{{B}_{2}A}|}=\frac{|\overline{A{B}_{2}}|-|\overline{AC}|}{|\overline{{B}_{2}A}|}=\frac{\frac{{r}^{2}}{s-b}-\frac{{r}^{2}}{c}}{\frac{{r}^{2}}{s-b}}=\frac{b+c-s}{c}.$

And finally,

$\frac{|\overline{BD}|}{|\overline{DC}|}=\frac{c}{b},$

by the Bisector Theorem. Cross-multiplying yields what we wanted to prove.

Let $a=\overline{BC}$, $b=\overline{CA}$, $c=\overline{AB}$, $s=\frac{a+b+c}{2}$. We have

$\frac{|\overline{A{C}_{2}}|}{|\overline{{C}_{2}B}|}=\frac{|\overline{A{C}_{2}}|}{|\overline{A{C}_{2}}|-|\overline{AB}|}=\frac{\frac{{r}^{2}}{s-c}}{\frac{{r}^{2}}{s-c}-\frac{{r}^{2}}{b}}=\frac{b}{b+c-s}.$

Likewise,

$\frac{|\overline{C{B}_{2}}|}{|\overline{{B}_{2}A}|}=\frac{|\overline{A{B}_{2}}|-|\overline{AC}|}{|\overline{{B}_{2}A}|}=\frac{\frac{{r}^{2}}{s-b}-\frac{{r}^{2}}{c}}{\frac{{r}^{2}}{s-b}}=\frac{b+c-s}{c}.$

And finally,

$\frac{|\overline{BD}|}{|\overline{DC}|}=\frac{c}{b},$

by the Bisector Theorem. Cross-multiplying yields what we wanted to prove.

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