3
$\begingroup$

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 $ ABB_1 $ and $ ACC_1 $ meet on a bisector of $ \angle BAC. $

What I thought: Circle $(ABB_1)$ has $\omega_1=b(s-a)$ and circle $(ACC_1)$ 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.

$\endgroup$

1 Answer 1

3
$\begingroup$

Let $D$ be on $\overline{BC}$ so that $\overline{AD}$ bisects $\angle BAC$. Let $\varphi$ 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 $\varphi(B)=C$, $\varphi(C)=B$, $\varphi(\overline{AD})=\overline{AD}$. Let also $B_2=\varphi(B_1)$, $C_2=\varphi(C_1)$. Notice that $\varphi((ABB_1))=\overline{CB_2}$, $\varphi((ACC_1))=\overline{BC_2}$. We now wish to prove that $\overline{AD}$, $\overline{BC_2}$, $\overline{CB_2}$ concur. We do this by Ceva on $\bigtriangleup ABC$.

Problem diagram

Let $a=\overline{BC}$, $b=\overline{CA}$, $c=\overline{AB}$, $s=\frac{a+b+c}{2}$. We have $$\frac{|\overline{AC_2}|}{|\overline{C_2B}|}=\frac{|\overline{AC_2}|}{|\overline{AC_2}|-|\overline{AB}|}=\frac{\frac{r^2}{s-c}}{\frac{r^2}{s-c}-\frac{r^2}{b}}=\frac{b}{b+c-s}.$$ Likewise, $$\frac{|\overline{CB_2}|}{|\overline{B_2A}|}=\frac{|\overline{AB_2}|-|\overline{AC}|}{|\overline{B_2A}|}=\frac{\frac{r^2}{s-b}-\frac{r^2}{c}}{\frac{r^2}{s-b}}=\frac{b+c-s}{c}.$$And finally, $$\frac{|\overline{BD}|}{|\overline{DC}|}=\frac{c}{b},$$by the Bisector Theorem. Cross-multiplying yields what we wanted to prove.

$\endgroup$
1
  • $\begingroup$ Very good! Thanks! $\endgroup$ Commented Oct 29, 2019 at 18:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .