Let $ABC$ be a triangle, and $A_1, B_1, C_1$ be points on the sides $BC, CA, AB$, respectively, such that $AA_1, BB_1, CC_1$ are the internal angle bisectors of $\triangle ABC$. The circumcircle $k' = (A_1B_1C_1)$ touches the side $BC$ at $A_1$. Let $B_2$ and $C_2$, respectively, be the second intersection points of $k'$ with lines $AC$ and $AB$. Prove that $|AB| = |AC|$ or $|AC_1| = |AB_2|$.
Problem
Source: 2020 Macedonian National Olympiad
Tags: geometry, Ugly, Angle Bisectors, angle bisector, circumcircle
Jalil_Huseynov
12.06.2021 10:49
We can find values of $AB_{1} ,AC_{1}, BA_{1}, BC_{1},CA_{1}, CB_{1}$ with the lenght of sides of $\triangle ABC$. And then using circumcircle of $\triangle A_{1}B_{1}C_{1}$ we can evalute $AC_{1}$ and $AB_{2}$. But it needs very bash and I understand why it's bumped still.
Mickey14
12.06.2021 12:16
This is basically, as expected, just Power of a Point and the angle bisector theorem. Indeed, we have \[\begin{cases}BC_1\cdot BC_2=BA_1^2\\ CB_1\cdot CB_2=CA_1^2\end{cases}\iff \begin{cases}AC_1\cdot \frac{a}{b}\cdot (c-AC_2)=BA_1^2\\ AB_1\cdot \frac{a}c\cdot (b-AB_2)=CA_1^2\end{cases}\]Therefore \[\frac{AC_1}{AB_1}\cdot\frac{c}{b}\cdot \frac{c-AC_2}{b-AB_2}=\frac{BA_1^2}{CA_1^2}=\frac{c^2}{b^2}\iff \frac{AC_1}{AB_1}\cdot \frac{c-AC_2}{b-AB_2}=\frac{c}{b}\]\[\iff bc\cdot AC_1-b\cdot AC_1\cdot AC_2=bc\cdot AB_1-c\cdot AB_1\cdot AB_2 \]Use now $b=AB_1+AB_1\cdot \frac ac, c=AC_1+AC_1\cdot \frac ab$ to obtain \[\left(AB_1+AB_1\cdot \frac ac\right)\cdot c\cdot AC_1-b\cdot AC_1\cdot AC_2=\left(AC_1+AC_1\cdot \frac ab\right)\cdot b\cdot AB_1-c\cdot AB_1\cdot AB_2\]\[\iff c\cdot AB_1\cdot AC_1-b\cdot AC_1\cdot AC_2=b\cdot AB_1\cdot AC_1-c\cdot AB_1\cdot AB_2\]Power of a point also yields $AB_1\cdot AB_2=AC_1\cdot AC_2$. Hence, the equation becomes \[c\cdot AC_1-b\cdot AB_2=b\cdot AC_1-c\cdot AB_2\iff (b-c)(AC_1-AB_2)=0\]