Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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: Triangle

share|improve this question
1  
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

1 Answer 1

Diagram

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 $M_1M_2M_3$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.