MathLabs

Problem 2

In the triangle ABCABC, prove that there is a point DD on side ABAB such that CDCD is the geometric mean of ADAD and DBDB if and only if sin⁡Asin⁡B≤sin⁡2C2\sin A \sin B \le \sin^2 \frac{C}{2}.
Step 5 of 6: Expressing both areas through the circumradius
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[ABC]=AB⋅BC⋅CA4R,∠MAB=∠MCB=C2 ⇒ MA=2Rsin⁡C2,[MAB]=AB⋅MA24R[ABC]=\frac{AB\cdot BC\cdot CA}{4R},\quad \angle MAB=\angle MCB=\frac{C}{2}\ \Rightarrow\ MA=2R\sin\frac{C}{2},\quad [MAB]=\frac{AB\cdot MA^2}{4R}
Detailed analysis

Let RR be the circumradius of △ABC\triangle ABC (also the circumradius of △MAB\triangle MAB, since MM lies on the same circle). The standard formula gives [ABC]=AB⋅BC⋅CA4R[ABC]=\dfrac{AB\cdot BC\cdot CA}{4R}. Since MM is the midpoint of arc ABAB, it lies on the internal bisector from CC, and the inscribed angle theorem gives ∠MAB=∠MCB=C2\angle MAB=\angle MCB=\tfrac{C}{2}; the chord subtending this inscribed angle has length MA=2Rsin⁡C2MA=2R\sin\frac{C}{2}. As △MAB\triangle MAB is isosceles with MA=MBMA=MB, applying the same area formula to it gives [MAB]=AB⋅MA⋅MB4R=AB⋅MA24R[MAB]=\dfrac{AB\cdot MA\cdot MB}{4R}=\dfrac{AB\cdot MA^2}{4R}.