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 6 of 8: Comparing circumradii: incircle vs. nine-point circle

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In plain words

The non-isosceles hypothesis is exactly what is needed here: it is the only place in the whole solution where it gets used, and it is essential — for an equilateral triangle the two circles would coincide and the argument would break down (the concurrency would trivially hold at the common center).

r = \text{circumradius of } S_1S_2S_3, \quad \tfrac{R}{2} = \text{circumradius of } M_1M_2M_3, \quad R>2r

Detailed analysis

All of $S_1,S_2,S_3$ lie on the incircle $\omega$ (Step 1), so $\omega$ , of radius $r$ , is the circumcircle of $\triangle S_1S_2S_3$ . The circumcircle of the medial triangle $M_1M_2M_3$ is the nine-point circle of $\triangle A_1A_2A_3$ , of radius $R/2$ where $R$ is the circumradius of $\triangle A_1A_2A_3$ . Euler's inequality gives $R\ge 2r$ with equality only for an equilateral triangle; since $\triangle A_1A_2A_3$ is explicitly non-isosceles (hence not equilateral), $R>2r$ strictly, so $R/2\ne r$ .