Problem 2

A non-isosceles triangle

A_1A_2A_3

has sides

a_1

a_2

a_3

with the side

a_i

lying opposite to the vertex

A_i

. Let

M_i

be the midpoint of the side

a_i

, and let

T_i

be the point where the inscribed circle of triangle

A_1A_2A_3

touches the side

a_i

. Denote by

S_i

the reflection of the point

T_i

in the interior angle bisector of the angle

A_i

. Prove that the lines

M_1S_1

M_2S_2

and

M_3S_3

are concurrent.

Step 1 of 8: Reflection at vertex

A_1

fixes the incircle and swaps two touch points

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In plain words

The incircle is the natural mirror here: every reflection that fixes the incenter automatically fixes the incircle, so touch points can only be sent to other points on the same circle.

T_1 \xrightarrow{\ \text{reflect in } A_1I\ } S_1, \qquad T_2 \longleftrightarrow T_3

Detailed analysis

Let $I$ be the incenter and $\omega$ the incircle. Reflecting the plane across the interior bisector of angle $A_1$ fixes $I$ and hence fixes $\omega$ , and it swaps the two sides through $A_1$ , namely $A_1A_2$ and $A_1A_3$ . By definition this reflection sends $T_1$ to $S_1$ ; since it also swaps the sides carrying $T_2$ and $T_3$ , it swaps these two touch points as well.