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 6: Existence of

D

reduces to comparing two heights

Previous step Next step

M=\text{midpoint of arc }AB\text{ not containing }C,\quad d(L,AB)\le d(M,AB)\ \Longleftrightarrow\ L\cap w\ne\varnothing

Detailed analysis

Among all points of the circumcircle $w$ on the far side of $AB$ from $C$ , the one farthest from line $AB$ is $M$ , the midpoint of that arc (the point where the perpendicular bisector of $AB$ meets the circle). Hence the line $L$ meets $w$ exactly when its distance to $AB$ does not exceed the distance from $M$ to $AB$ , i.e. $CF\le MN$ where $N$ is the foot of the perpendicular from $M$ to $AB$ . By Step 2, a valid point $D$ on segment $AB$ exists precisely when $L$ meets $w$ , since then $E$ can be taken at a point of $L\cap w$ and $D=CE\cap AB$ .