IMO 1974 · 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 \[ \sin A\sin B\leq \sin^2\frac C2. \]

Official text of the problem. Official problem paper (PDF)

Solution

The endpoints \(A,B\) cannot serve as \(D\): there \(AD\cdot DB=0\), whereas \(CD>0\).

For an interior point \(D\) of \(AB\), put \[ \alpha=\angle ACD,\qquad \beta=\angle DCB. \] Then \(\alpha,\beta>0\) and \(\alpha+\beta=C\). Applying the sine rule in triangles \(ACD\) and \(BCD\) gives \[ \frac{AD}{CD}=\frac{\sin\alpha}{\sin A}, \qquad \frac{DB}{CD}=\frac{\sin\beta}{\sin B}. \] Consequently, \[ CD^2=AD\cdot DB \quad\Longleftrightarrow\quad \sin\alpha\sin\beta=\sin A\sin B. \tag{1} \]

For necessity, suppose such a point \(D\) exists. Since \(\beta=C-\alpha\), \[ \sin A\sin B =\sin\alpha\sin(C-\alpha) =\sin^2\frac C2-\sin^2\left(\alpha-\frac C2\right) \leq\sin^2\frac C2. \]

For sufficiency, suppose \[ \sin A\sin B\leq\sin^2\frac C2. \] The function \[ f(t)=\sin t\sin(C-t) \] is continuous on \([0,C/2]\), with \[ f(0)=0,\qquad f(C/2)=\sin^2\frac C2. \] Because \(\sin A\sin B>0\), the intermediate value theorem supplies an \(\alpha\in(0,C/2]\) such that \[ f(\alpha)=\sin A\sin B. \]

Draw the ray from \(C\) inside \(\angle ACB\) making angle \(\alpha\) with \(CA\). Since \(0<\alpha<C\), it intersects \(AB\) at an interior point \(D\). For this point, \(\beta=C-\alpha\), so (1) yields \[ CD^2=AD\cdot DB, \] and hence \(CD=\sqrt{AD\cdot DB}\), as required.

In the equality case, \(\alpha=C/2\) works, so \(D\) is the intersection of the internal angle bisector from \(C\) with \(AB\). \(\square\)

The solution was checked against John Scholes' IMO solution archive (Kalva).