Trigonometry
Trigonometry
Allen Star Batch
Grade 11
Question:
Let $\triangle ABC$ be inscribed in a circle having radius unity. The three internal bisectors of the angles $A, B$ and $C$ are extended to intersect the circumcircle of $\triangle ABC$ at $A_1, B_1$ and $C_1$ respectively. Find $$\frac{AA_1\cos\frac{A}{2} + BB_1\cos\frac{B}{2} + CC_1\cos\frac{C}{2}}{\sin A + \sin B + \sin C}$$
Step-by-Step Solution
Key Concept: Use sine rule and product formulas to relate the angle bisector lengths and express the sum as a constant.
From $\frac{c}{\sin C} = \frac{AA_1}{\sin(B + \frac{A}{2})}$, we get $AA_1\cos\frac{A}{2} = \sin B + \sin C$ (with $R = 1$). Using product-to-sum formulas: $AA_1\cos\frac{A}{2} + BB_1\cos\frac{B}{2} + CC_1\cos\frac{C}{2} = 2$.
Correct Answer: 2