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 5 of 6: The three ratios multiply to one
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In plain words

Each ratio is built from a law-of-sines computation using the same three quantities 30 plus alpha, 30 plus beta, 30 plus gamma attached cyclically to the three vertices, so multiplying the three ratios together telescopes every factor away.

kakbkc=1k_ak_bk_c=1
Detailed analysis

The sine-rule formulas in Step 4 give ka=AXAY=AB1AC1sin⁡(30∘+β)sin⁡(30∘+γ)k_a=\dfrac{AX}{AY}=\dfrac{AB_1}{AC_1}\dfrac{\sin(30^\circ+\beta)}{\sin(30^\circ+\gamma)}. Cyclically, kb=BC1BA1sin⁡(30∘+γ)sin⁡(30∘+α)k_b=\dfrac{BC_1}{BA_1}\dfrac{\sin(30^\circ+\gamma)}{\sin(30^\circ+\alpha)} and kc=CA1CB1sin⁡(30∘+α)sin⁡(30∘+β)k_c=\dfrac{CA_1}{CB_1}\dfrac{\sin(30^\circ+\alpha)}{\sin(30^\circ+\beta)}. The hypotheses give BA1=CA1BA_1=CA_1, CB1=AB1CB_1=AB_1, and AC1=BC1AC_1=BC_1. Multiplying the three displayed expressions therefore cancels both all sine factors and all length factors, yielding kakbkc=1k_ak_bk_c=1.