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 4 of 5: The range of sin⁡C1sin⁡C2\sin C_1\sin C_2 is exactly [0,sin⁡2C2]\bigl[0,\sin^2\frac{C}{2}\bigr]
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f(x)=sin⁡ ⁣(C2+x)sin⁡ ⁣(C2−x)=cos⁡(2x)−cos⁡C2,x∈[0,C2]f(x)=\sin\!\left(\tfrac{C}{2}+x\right)\sin\!\left(\tfrac{C}{2}-x\right)=\frac{\cos(2x)-\cos C}{2},\quad x\in\left[0,\tfrac{C}{2}\right]
Detailed analysis

Write C1=C2+xC_1=\tfrac{C}{2}+x, C2=C2−xC_2=\tfrac{C}{2}-x for x∈[0,C2]x\in[0,\tfrac{C}{2}] (so x=0x=0 splits CC evenly, x=C/2x=C/2 makes C2=0C_2=0). Then sin⁡C1sin⁡C2=f(x)=cos⁡(2x)−cos⁡C2\sin C_1\sin C_2=f(x)=\dfrac{\cos(2x)-\cos C}{2}, a piece of a cosine curve that decreases continuously from f(0)=1−cos⁡C2=sin⁡2C2f(0)=\dfrac{1-\cos C}{2}=\sin^2\frac{C}{2} down to f(C/2)=0f(C/2)=0. So as the split varies, sin⁡C1sin⁡C2\sin C_1\sin C_2 takes every value in [0,sin⁡2C2]\bigl[0,\sin^2\frac{C}{2}\bigr], and no value outside it.