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 1 of 6: Translate the 480 degree condition into a clean angle sum
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In plain words

Since A1, B1, C1 sit at the apex of isosceles triangles built on the sides, each of the three given angles equals 180 degrees minus twice a base angle, so the single 480 degree condition collapses into a sum of just three small angles.

α+β+γ=30∘,α=∠A1CB=∠CBA1, β=∠ACB1=∠B1AC, γ=∠C1AB=∠C1BA\alpha+\beta+\gamma=30^\circ,\quad \alpha=\angle A_1CB=\angle CBA_1,\ \beta=\angle ACB_1=\angle B_1AC,\ \gamma=\angle C_1AB=\angle C_1BA
Detailed analysis

Since BA1=A1CBA_1=A_1C, triangle BA1CBA_1C is isosceles, so ∠BA1C=180∘−2α\angle BA_1C=180^\circ-2\alpha where α=∠A1CB=∠CBA1\alpha=\angle A_1CB=\angle CBA_1; similarly ∠CB1A=180∘−2β\angle CB_1A=180^\circ-2\beta and ∠AC1B=180∘−2γ\angle AC_1B=180^\circ-2\gamma for β,γ\beta,\gamma defined analogously. The hypothesis ∠BA1C+∠CB1A+∠AC1B=480∘\angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ becomes 540∘−2(α+β+γ)=480∘540^\circ-2(\alpha+\beta+\gamma)=480^\circ, i.e. α+β+γ=30∘\alpha+\beta+\gamma=30^\circ.