Problem 2

In the triangle

ABC

, prove that there is a point

D

on side

AB

such that

CD

is the geometric mean of

AD

and

DB

if and only if

\sin A \sin B \le \sin^2 \frac{C}{2}

Step 3 of 5: Reformulating as a splitting problem

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\sin A\sin B\le\sin^2\frac{C}{2}\ \Longleftrightarrow\ \exists\,C_1+C_2=C:\ \sin C_1\sin C_2=\sin A\sin B

Detailed analysis

So the original question — does a point $D$ with $CD^2=AD\cdot DB$ exist on $AB$ ? — is equivalent to: can $\angle C$ be split as $C_1+C_2$ (with $C_1,C_2>0$ ) so that $\sin C_1\sin C_2=\sin A\sin B$ ? We now determine exactly which values $\sin C_1\sin C_2$ can take as the split $C_1+C_2=C$ varies.