The incircle of triangle $ABC$ touches its sides $AB, BC, CA$ in points $C_1, A_1, B_1$ respectively. The point $B_2$ is symmetric to $B_1$ with respect to line $A_1C_1$, lines $BB_2$ and $AC$ meet in point $B_3$. points $A_3$ and $C_3$ may be defined analogously. Prove that points $A_3, B_3$ and $C_3$ lie on a line, which passes through the circumcentre of a triangle $ABC$. proposed by L. Emelyanov
Problem
Source: 239MO 2004, grade 10-11, problem 5
Tags: geometry, circumcircle, incenter, analytic geometry, Euler, geometry solved
darij grinberg
12.12.2004 13:38
I always love the triangle geometry problems of the 239MO. For this one, the line through the points $A_3$, $B_3$, $C_3$ is the famous OI line of triangle ABC (this is the line joining the circumcenter with the incenter of triangle ABC). See Lev Emelyanov, On the Intercepts of the OI-Line, Forum Geometricorum, 4 (2004) p. 81-84 for a trigonometric solution, Lev Emelyanov and Tatiana Emelyanova, A note on the Schiffler point, Forum Geometricorum, 3 (2003) p. 113-116 for a solution using barycentric coordinates, and Hyacinthos message #7256 for a synthetic proof. Must the problems of the 239MO always be new problems? PS. The links to Forum Geometricorum above sometimes don't work - the FAU servers are unpredictable... Darij
grobber
12.12.2004 20:24
I know this is now redundant, but here's the sketch of a proof that $A_3$ lies on $OI$ (without some computations, which are pretty mild). Let's turn the problem around alltogether, and look at it from the point of view of the intouch triangle. It's well-known that the orthocenter of the intouch triangle lies on $OI$, so we have the following problem: Given a triangle $ABC$, let $UVT$ be its tangential triangle. If $A'$ is the symmetric of $A$ wrt $BC$, show that $VT$ and $UA'$ intersect on the Euler line of $ABC$. [$U,V,T$ correspond to $A,B,C$ respectively] Let $M=UO\cap VT$, and $D=AH\cap BC$. After we observe that $A'H\|UO$, all we need to show is that $\frac{A'H}{HA}=\frac{UO}{OM}$. This is easily reduced to $\frac{UB}{VT}=\frac{DH}{HA}$, and from here it's rather easy.