MathLabs

Problem 6

Let ABCABC be an equilateral triangle. Let A1,B1,C1A_1,B_1,C_1 be interior points of ABCABC such that BA1=A1CBA_1=A_1C, CB1=B1ACB_1=B_1A, AC1=C1BAC_1=C_1B, and ∠BA1C+∠CB1A+∠AC1B=480∘\angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ. Let A2=BC1∩CB1A_2=BC_1\cap CB_1, B2=CA1∩AC1B_2=CA_1\cap AC_1, C2=AB1∩BA1C_2=AB_1\cap BA_1. Prove that if triangle A1B1C1A_1B_1C_1 is scalene, then the circumcircles of triangles AA1A2AA_1A_2, BB1B2BB_1B_2, and CC1C2CC_1C_2 all pass through two common points.
Step 2 of 6: Each interior point is a circumcenter
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In plain words

Chasing the angles at A2 formed by the two cevians BC1 and CB1 shows the angle it subtends at BC is exactly half of the isosceles angle at A1, and since A1 is already equidistant from B and C this identifies A1 as the circumcenter.

∠BA2C=60∘+β+γ=90∘−α=12∠BA1C\angle BA_2C=60^\circ+\beta+\gamma=90^\circ-\alpha=\tfrac12\angle BA_1C
Detailed analysis

Since ∠ABC=60∘\angle ABC=60^\circ, ∠CBC1=60∘−γ\angle CBC_1=60^\circ-\gamma and ∠B1CB=60∘−β\angle B_1CB=60^\circ-\beta, so in triangle BA2CBA_2C, ∠BA2C=180∘−∠CBC1−∠B1CB=60∘+β+γ\angle BA_2C=180^\circ-\angle CBC_1-\angle B_1CB=60^\circ+\beta+\gamma; using α+β+γ=30∘\alpha+\beta+\gamma=30^\circ this equals 90∘−α90^\circ-\alpha. Meanwhile ∠BA1C=180∘−2α=2(90∘−α)=2∠BA2C\angle BA_1C=180^\circ-2\alpha=2(90^\circ-\alpha)=2\angle BA_2C. Since A1A_1 is already equidistant from BB and CC (as BA1=A1CBA_1=A_1C) and lies inside △A2BC\triangle A_2BC on the correct side, this doubling relation is exactly the condition for A1A_1 to be the circumcenter of △A2BC\triangle A_2BC; the symmetric argument gives B1,C1B_1,C_1 as the circumcenters of △B2CA,△C2AB\triangle B_2CA,\triangle C_2AB.