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 2 of 8: Two reflections show

S_1T_3=S_2T_3

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In plain words

Two independent reflections describe the same chord length in two different ways, giving a length equality between $S$ -points and $T$ -points for free.

S_1T_3 = T_1T_2 \quad\text{(reflection in }A_1I\text{)}, \qquad S_2T_3 = T_1T_2 \quad\text{(reflection in }A_2I\text{)}

Detailed analysis

Since reflection preserves chord lengths of $\omega$ , Step 1 turns the chord $T_1T_2$ into the chord $S_1T_3$ , so $S_1T_3=T_1T_2$ . Symmetrically, reflecting across the bisector of angle $A_2$ fixes $\omega$ , sends $T_2$ to $S_2$ , and swaps $T_1\leftrightarrow T_3$ , turning chord $T_1T_2$ into chord $T_3S_2$ , so $S_2T_3=T_1T_2$ as well. Hence $S_1T_3=S_2T_3$ .