(A.Myakishev, 9--10) Given triangle $ ABC$. One of its excircles is tangent to the side $ BC$ at point $ A_1$ and to the extensions of two other sides. Another excircle is tangent to side $ AC$ at point $ B_1$. Segments $ AA_1$ and $ BB_1$ meet at point $ N$. Point $ P$ is chosen on the ray $ AA_1$ so that $ AP=NA_1$. Prove that $ P$ lies on the incircle.
Problem
Source: Sharygin contest 2008. The correspondence round. Problem 13
Tags: geometry, incenter, geometry proposed
plane geometry
03.09.2008 18:08
Denote AA1∩⊙I = Q , where ⊙I is the incircle of △ABC O1,O2,O3 are midpoint of BC,CA,AB respectively I’ is the incenter of △O1O2O3 AA1∩O2O3=M It is well-known N is called Nagel point so we have I’I=I’N I’M⊥O2O3 IQ/IaA1=r/Ra=AI/AIa =〉IQ∥IaA1 =〉 IQ⊥O2O3 =〉IQ∥I’M =〉 QM=MN =〉 QA=NA1 =〉 P=Q Done!
Attachments:
livetolovemath030894
04.04.2010 10:15
You can see here http://www.mathlinks.ro/viewtopic.php?t=337716
huynguyen
25.12.2015 10:04
Sorry to revive.But I have another solution Let the incircle be tangent to $BC,CA$ at $D E$. Draw the diameter $DD'$ of $(I)$. Let $O_a$ be $A$-excircle of triangle $ABC$,this circle is tangent to the extension of $AC$ at $A_2$. The homothety center $A$ sends $(I)\rightarrow (O_a)$ will send $D'\rightarrow A_1, E\rightarrow A_2$. This means $A,D',A_1$ are collinear and $\frac{D'A}{D'A_1}=\frac{AE}{EA_2}$ Note that $EA_2=EC+CA_2=EC+CA_1=CD+CA_1=CD+BD=BC$. Applyinh Menelaus theorem for triangle $AA_1C$ with line $BNB_1$: $\frac{NS}{NA}.\frac{B_1A}{B_1C}.\frac{BC}{BA_1}=1$ Note that $B_1A=EC=DC=BA_1$. Hence: $\frac{NA_1}{NA}=\frac{B_1C}{BC}=\frac{AE}{BC}=\frac{D'A}{D'A_1}$ But $NA_1+NA=AA_1=D'A+D'A_1$ so it implies that $NA_1=AD'$, or we conclude that $D'$ is exactly $P$. Done!
jayme
21.10.2021 09:29
Dear Mathlinkers, for a nice synthetic proof... http://jl.ayme.pagesperso-orange.fr/Docs/26.%200.%20Segments.pdf p. 6-7. Sincerely Jean-Louis
Moubinool
26.10.2021 13:22
jayme wrote: Dear Mathlinkers, for a nice synthetic proof... see here p. 6-7. Sincerely Jean-Louis