A Problem by Ivan Pavlov

This problem was originally posted on Facebook.

Problem (Ivan Pavlov). Let \(ABC\) be a triangle with incenter \(I\). Let \(A_1,B_1,C_1\) be the reflections of \(I\) across the sides \(BC,CA,AB\), respectively, and let \(A_2,B_2,C_2\) be the reflections of \(A,B,C\) about \(I\), respectively. Prove that the lines \(A_1A_2\),\(B_1B_2\),\(C_1C_2\) are concurrent.

Proof. Let \(D,E,F\) be the intouch triangle of \(ABC\), and let \(I\) be its incenter. Since \(A_1,B_1,C_1\) are the reflections of \(I\) across \(BC,CA,AB\), respectively, we have

\[A_1B_1C_1=h_{I,2}(DEF),\]

while, since \(A_2,B_2,C_2\) are the reflections of \(A,B,C\) about \(I\), we have

\[A_2B_2C_2=h_{I,-1}(ABC).\]

Now apply the same homothety \(h_{I,\frac12}\) to both triangles. Then \(A_1B_1C_1\) is sent to \(DEF\), whereas \(A_2B_2C_2\) is sent to

\[h_{I,\frac12}\circ h_{I,-1}(ABC)=h_{I,-\frac12}(ABC).\]

By Theorem 3 of B.~Suceavă and P.~Yiu, The Feuerbach Point and Euler Lines, Forum Geometricorum 6 (2006), 191--197, every homothetic image \(h_{I,t}(ABC)\) is perspective with the intouch triangle \(DEF\); hence \(DEF\) and \(h_{I,-\frac12}(ABC)\) are perspective. Therefore their corresponding joins are concurrent. Since \(h_{I,\frac12}\) is an affine transformation, and affine transformations preserve incidence, the inverse images of those three concurrent joins, namely \(A_1A_2\),\(B_1B_2\),\(C_1C_2\) are concurrent as well.