IMO 1974 SL 10

Problem

II 4 (FIN 3)IMO2 Let \triangleABC be a triangle. Prove that there exists a point D on the side AB such that CD is the geometric mean of AD and BD if and only if \sqrt sin A sin B \leqsin C 2 .

Solution

If we set \angleACD = \gamma1 and \angleBCD = \gamma2 for a point D on the segment AB, then by the sine theorem, f(D) = CD2 AD \cdot BD = CD AD \cdot CD BD = sin \alpha sin \beta sin \gamma1 sin \gamma2 . The denominator of the last fraction is sin \gamma1 sin \gamma2 = 1 2 (cos(\gamma1 −\gamma2) −cos(\gamma1 + \gamma2)) = 1 2 (cos(\gamma1 −\gamma2) −cos \gamma) \leq1 −cos \gamma = sin2 \gamma 2 , from which we deduce that the set of values of f(D) is the interval  sin \alpha sin \beta sin2 \gamma , +\infty

. Hence f(D) = 1 (equivalently, CD2 = AD \cdot BD) is possible if and only if sin \alpha sin \beta \leqsin2 \gamma 2, i.e.,

sin \alpha sin \beta \leqsin \gamma 2 .

Second solution. Let E be the second point of intersection of the line CD with the circumcircle k of ABC. Since AD \cdot BD = CD \cdot ED (power of D with respect to k), CD2 = AD \cdot BD ie equivalent to ED \geqCD. Clearly the ratio ED CD (D \inAB) takes a minimal value when E is the midpoint of the arc AB not containing C. (This follows from ED : CD = E′D : C′D when C′ and E′ are respectively projections from C and E onto AB.) On the other hand, it is directly shown that in this case ED CD = sin2 \gamma sin \alpha sin \beta , and the assertion follows.