Let $I$ be the incenter of the scalene $\Delta ABC$, such, $AB<AC$, and let $I'$ be the reflection of point $I$ in line $BC$. The angle bisector $AI$ meets $BC$ at $D$ and circumcircle of $\Delta ABC$ at $E$. The line $EI'$ meets the circumcircle at $F$. Prove, that, $\text{(i) } \frac{AI}{IE}=\frac{ID}{DE}$ $\text{(ii) } IA=IF$
Problem
Source: JBMO TST 3 2017 P3
Tags: geometry, incenter, geometric transformation, reflection, angle bisector, circumcircle
15.11.2018 11:48
Dear Mathlinkers, for (ii) https://artofproblemsolving.com/community/c6t48f6h1491907_an_incenter_problem Sincerely Jean-Louis
15.11.2018 13:54
Dear Mathlinkers, (i) refers to the Haruki's lemma Sincerely Jean-Louis
15.11.2018 13:56
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Quickies%205.pdf p. 20-25 Sincerely Jean-Louis
15.11.2018 18:34
My solution $(i)$ We know that $\frac{AI}{ID}=\frac{AC}{DC}$ and $EB=EC=EI$, $\triangle DEC \sim \triangle CEA$ It implies that $\frac{AI}{IE}=\frac{ID}{DE}$. $(ii)$ Draw diameter $AJ$, $JF$ cuts $AE$ at $L$. $EF$ cuts $BC$ at $K$. By changing angle we get $\angle I'IL=\angle EFJ \Rightarrow FILI'$ is cyclic On the other hand, $EI^2=EC^2=EK.EF \Rightarrow \angle IKE=\angle FIE=\angle LI'E \Rightarrow KI \parallel LI'$ By Menelaus and Thales, we obtain $\frac{AI}{IE}=\frac{ID}{DE}=\frac{KI'}{KE}=\frac{IL}{IE} \Rightarrow IL=IA=IF$ Q.E.D
15.11.2018 20:56
nguyendangkhoa17112003 wrote: My solution $(i)$ We know that $\frac{AI}{ID}=\frac{AC}{DC}$ and $EB=EC=EI$, $\triangle DEC \sim \triangle CEA$ It implies that $\frac{AI}{IE}=\frac{ID}{DE}$. $(ii)$ Draw diameter $AJ$, $JF$ cuts $AE$ at $L$. $EF$ cuts $BC$ at $K$. By changing angle we get $\angle I'IL=\angle EFJ \Rightarrow FILI'$ is cyclic On the other hand, $EI^2=EC^2=EK.EF \Rightarrow \angle IKE=\angle FIE=\angle LI'E \Rightarrow KI \parallel LI'$ By Menelaus and Thales, we obtain $\frac{AI}{IE}=\frac{ID}{DE}=\frac{KI'}{KE}=\frac{IL}{IE} \Rightarrow IL=IA=IF$ Q.E.D Nice Mine was a two page bash
24.06.2020 09:38
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/78.1%20Relations%20metriques.pdf p. 44... Sincerely Jean-Louis
12.08.2024 01:27
solved also here