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 2 of 5: Compute rAr_A
Previous stepNext step
ABAY=sin⁡Csin⁡(B+A/2),rA=sin⁡Bsin⁡Csin⁡2(B+A/2)\frac{AB}{AY}=\frac{\sin C}{\sin(B+A/2)},\qquad r_A=\frac{\sin B\sin C}{\sin^2(B+A/2)}
Detailed analysis

Applying the sine rule in ABYABY gives AB/AY=sin⁡C/sin⁡(B+A/2)AB/AY=\sin C/\sin(B+A/2). Multiplying with the previous ratio yields rA=sin⁡Bsin⁡C/sin⁡2(B+A/2)r_A=\sin B\sin C/\sin^2(B+A/2).