# contest problem in geometry

Suppose the inscribed circle of $\triangle A_1A_2A_3$ touches the sides $A_2A_3, A_3A_1, A_1A_2$ at $T_1,T_2,T_3$. From the midpoints $M_1,M_2,M_3$ of $A_2A_3,A_3A_1,A_1A_2$, draw lines perpendicular to $T_2T_3,T_3T_1,T_1T_2$. Prove that these perpendicular lines are concurrent.

I have a solution but is rather long and is not elegant. Since it is a contest type problem, I guess there is an elegant way of solving it, may someone please help, thanks.

A picture of the construct is added below:

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Shouldn't the lines be perpendicular to $A_2A_3,A_3A_1,A_1A_2?$ – Ross Millikan May 8 '13 at 16:01
@RossMillikan If the lines were perpendicular to $A_2A_3, A_3A_1, A_1A_2$, as seen in the diagram, then the problem would be trivial, as the perpendicular lines are concurrent at the circumcenter of $A_1A_2A_3$. I think it is more likely that the question is correct, but the diagram shown (which incidentally wasn't added by the OP) is wrong. – Ivan Loh May 9 '13 at 3:08

As shown in the above diagram, let $H_1, H_2, H_3$ be the foot of the perpendicular from $M_1, M_2, M_3$ to $T_2T_3, T_1T_3, T_1T_2$ respectively, and let $M_1H_1$ intersect $A_1A_2, A_1A_3$ at $P, Q$ respectively. We then want to show that $M_1H_1, M_2H_2, M_3H_3$ are concurrent.
Note that $M_1M_2 \parallel A_1A_2, M_1M_3 \parallel A_1A_3, M_2M_3 \parallel A_2A_3$.
$$\angle{H_1M_1M_3}=\angle{H_1QT_2}=90^{\circ}-\angle{H_1T_2Q}=90^{\circ}-\angle{T_2T_3A_1}=\angle{T_3PH_1}=\angle{H_1M_1M_2}$$
Thus $M_1H_1$ is the angle bisector of $\angle{M_2M_1M_3}$. Similarly, $M_2H_2, M_3H_3$ are the angle bisectors of $\angle{M_1M_2M_3}$ and $\angle{M_1M_3M_2}$ respectively, so $M_1H_1, M_2H_2, M_3H_3$ are concurrent at the incenter of triangle $M_1M_2M_3$.