MathLabs

Problem 3

Let ABCABC be a triangle. The bisector of angle AA meets segment BCBC at XX and the circumcircle at YY. Let rA=AX/AYr_A=AX/AY, and define rB,rCr_B,r_C similarly. Prove that rA/sin⁡2A+rB/sin⁡2B+rC/sin⁡2C≥3r_A/\sin^2A+r_B/\sin^2B+r_C/\sin^2C\ge3, with equality if and only if the triangle is equilateral.
Step 1 of 5: Apply the sine rule in AXBAXB
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AXAB=sin⁡Bsin⁡(B+A/2)\frac{AX}{AB}=\frac{\sin B}{\sin(B+A/2)}
Detailed analysis

Because ∠AXB=180∘−B−A/2\angle AXB=180^\circ-B-A/2, the sine rule gives AX/AB=sin⁡B/sin⁡(B+A/2)AX/AB=\sin B/\sin(B+A/2).