MathLabs

Problem 2

A non-isosceles triangle A1A2A3A_1A_2A_3 has sides a1a_1, a2a_2, a3a_3 with the side aia_i lying opposite to the vertex AiA_i. Let MiM_i be the midpoint of the side aia_i, and let TiT_i be the point where the inscribed circle of triangle A1A2A3A_1A_2A_3 touches the side aia_i. Denote by SiS_i the reflection of the point TiT_i in the interior angle bisector of the angle AiA_i. Prove that the lines M1S1M_1S_1, M2S2M_2S_2 and M3S3M_3S_3 are concurrent.
Step 7 of 8: Unequal radii rule out a translation, leaving a genuine homothety
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In plain words

Comparing a single numerical invariant (the circumradius) is enough to eliminate an entire case (translation) from a geometric classification.

R/2≠r ⇒ △S1S2S3 and △M1M2M3 are not congruent ⇒ the relation is a genuine homothetyR/2 \ne r \ \Rightarrow\ \triangle S_1S_2S_3 \text{ and } \triangle M_1M_2M_3 \text{ are not congruent} \ \Rightarrow\ \text{the relation is a genuine homothety}
Detailed analysis

A translation preserves size, so it would force the two triangles to be congruent, in particular to have equal circumradii. Step 6 shows the circumradii rr and R/2R/2 are unequal, so a translation is impossible; by Step 5 the only remaining option is a genuine homothety, with some ratio k≠1,−1k\ne 1,-1 (here k=r/(R/2)≠±1k=r/(R/2)\ne \pm1) and some center PP, sending △M1M2M3\triangle M_1M_2M_3 to △S1S2S3\triangle S_1S_2S_3 with Mi↦SiM_i\mapsto S_i.