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 5: Conclusion: the inequality is exactly the existence condition
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sin⁡Asin⁡B∈[0,sin⁡2C2] ⟺ ∃x: f(x)=sin⁡Asin⁡B\sin A\sin B\in\Bigl[0,\sin^2\frac{C}{2}\Bigr]\ \Longleftrightarrow\ \exists x:\ f(x)=\sin A\sin B
Detailed analysis

Since sin⁡Asin⁡B>0\sin A\sin B>0 automatically (as A,B∈(0,π)A,B\in(0,\pi)), the condition sin⁡Asin⁡B≤sin⁡2C2\sin A\sin B\le\sin^2\frac{C}{2} says precisely that sin⁡Asin⁡B\sin A\sin B lies in the attainable range [0,sin⁡2C2]\bigl[0,\sin^2\tfrac{C}{2}\bigr] found in Step 4. So some split C=C1+C2C=C_1+C_2 achieves sin⁡C1sin⁡C2=sin⁡Asin⁡B\sin C_1\sin C_2=\sin A\sin B if and only if sin⁡Asin⁡B≤sin⁡2C2\sin A\sin B\le\sin^2\frac{C}{2}; by Step 3 this split determines a point DD on ABAB with CD2=AD⋅DBCD^2=AD\cdot DB, which completes the equivalence in both directions.