The incircle of a non-isosceles triangle $ABC$ touches $AB$ at point $C'$. The circle with diameter $BC'$ meets the incircle and the bisector of angle $B$ again at points $A_1$ and $A_2$ respectively. The circle with diameter $AC'$ meets the incircle and the bisector of angle $A$ again at points $B_1$ and $B_2$ respectively. Prove that lines $AB, A_1B_1, A_2B_2$ concur. (E. H. Garsia)
Problem
Source: 2014 Sharygin Geometry Olympiad Final Round 10.6
Tags: geometry, concurrency, concurrent
jayme
04.08.2018 12:04
Dear Mathlinkers, http://www.artofproblemsolving.com/community/c6t48f6h1123040_concyclic_points_easy and then think to the three chords theorem (Monge). Sincerely Jean-Louis
MeineMeinung
03.12.2019 13:37
jayme wrote: Dear Mathlinkers, http://www.artofproblemsolving.com/community/c6t48f6h1123040_concyclic_points_easy and then think to the three chords theorem (Monge). Sincerely Jean-Louis Dear Louis, could you elaborate more on how to apply Monge to this problem?
Let $DJ$ be the diameter of the incircle. Since $\angle JA_1D + \angle DA_1D = 90^{\circ} + 90^{\circ} = 180^{\circ}$ then $A,A_1,J$ are collinear and similarly, $B,B_1,J$ are collinear.
Now let $Z = A_2B_2 \cap AB$ then by Menelaus, we have $\frac{BZ}{ZA} \times \frac{AB_2}{B_2I} \times \frac{IA_2}{A_2B} = 1 \implies \frac{BZ}{ZA} \times \frac{AD^2}{DI^2} \times \frac{DI^2}{DB^2} = 1 \implies \frac{BZ}{ZA} = \frac{DB^2}{AD^2}$.
Clearly, from right triangle $ADJ$ dan $BDJ$, we have $\frac{AB_1}{B_1J} = \frac{AD^2}{DJ^2}$ and $\frac{BA_1}{A_1J} = \frac{BD^2}{DJ^2}$. Thus
$\frac{BZ}{ZA} \times \frac{AB_1}{B_1J} \times \frac{JA_1}{BA_1} = \frac{DB^2}{AD^2} \times \frac{AD^2}{DJ^2} \times \frac{DJ^2}{BD^2} = 1$. So, $A_1,B_1,Z$ are collinear and we are done.
amar_04
04.12.2019 00:44
Better Invert... parmenides51 wrote: The incircle of a non-isosceles triangle $ABC$ touches $AB$ at point $C'$. The circle with diameter $BC'$ meets the incircle and the bisector of angle $B$ again at points $A_1$ and $A_2$ respectively. The circle with diameter $AC'$ meets the incircle and the bisector of angle $A$ again at points $B_1$ and $B_2$ respectively. Prove that lines $AB, A_1B_1, A_2B_2$ concur. (E. H. Garsia) The triangle $ABC$ is causing trouble here, so let's omit it by solving restating the above problem as this. Same Problem wrote: Let two circles $\omega_1$ and $\omega_2$ be externally tangent at $C'$. Let $AC',BC'$ be the diameters of $\omega_1$ and $\omega_2$ respectively. Let another circle $\omega_3$ be tangent to the line $AB$ at $C'$ with center $O$. Let $\omega_1\cap\omega_3=B_1$ and $\omega_2\cap\omega_3=A_1$. Also let $AO\cap\omega_1=B_2$ and $BO\cap\omega_2=A_2$. Then $AB,A_1B_1,A_2B_2$ will be concurrent. Claim 1:- $A,B_2,A_2,B$ are concyclic . First of all note that $\omega_1,\omega_3$ and $\omega_2,\omega_3$ are orthogonal. So by an Inversion (let this map be $\Psi$) around $\omega_3$ we get $$\begin{cases} \Psi:B_2\leftrightarrow A \\ \Psi:A_2\leftrightarrow B\end{cases}\implies A,B_2,A_2,B \text{ are concyclic}$$Note that one can avoid Inversion by just using Similarity of triangles, but Inversion is more natural here. Claim 2:- $A,B_1,A_1,B$ are concyclic. For this Invert around $C'$ with an arbitary radius, let this map be $\Psi$ again. So, $\Psi$ transforms $\omega_1$ and $\omega_2$ to parallel lines also it sends $\omega_3$ to a line perpendicular to those parallel lines, let the Inverse of $X$ be $X'$ from now onwards. So, by this map $$\begin{cases} \Psi:A\leftrightarrow \omega_1'\cap AB=A' \\ \Psi:B\leftrightarrow \omega_2'\cap AB=B'\end{cases}$$Note here that $A'B_1'A_1'B'$ is a rectangle which is a cyclic quadrilateral, hence, Inverting back we get that $AB_1A_1B$ is a cyclic quadrilateral. Note that by Simple angle chasing you also get the claim. Returning to the main proof :- Let $A_1B_1\cap AB=T$ and $A_2B_2\cap AB=T'$. So, $$TC'^2=TB_1.TA_1=TA.TB=T'A.T'B\implies \boxed{T\equiv T'}$$Hence we get that $AB,A_1B_1,A_2B_2$ are concurrent. $\blacksquare$.