IMO 1981 · Problem 1

\(P\) is a point inside a given triangle \(ABC\). \(D, E, F\) are the feet of the perpendiculars from \(P\) to the lines \(BC, CA, AB\) respectively. Find all \(P\) for which \[ \frac{BC}{PD}+\frac{CA}{PE}+\frac{AB}{PF} \] is least.

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

Answer

The unique such point is the incenter of \(ABC\). The minimum value is \[ \boxed{\frac{(AB+BC+CA)^2}{2[ABC]}}. \]

Solution

Write \[ a=BC,\qquad b=CA,\qquad c=AB,\qquad x=PD,\qquad y=PE,\qquad z=PF, \] and let \(\Delta=[ABC]\). Since \(P\) is strictly inside the triangle, \(x,y,z>0\), and the three triangles \(PBC,PCA,PAB\) partition \(ABC\). Thus \[ ax+by+cz=2\Delta. \] These area formulas remain valid when a perpendicular foot lies on an extension of a side.

By the Cauchy–Schwarz inequality, \[ \begin{aligned} \left(\frac ax+\frac by+\frac cz\right)(ax+by+cz) &\ge \left( \sqrt{\frac ax}\sqrt{ax} +\sqrt{\frac by}\sqrt{by} +\sqrt{\frac cz}\sqrt{cz} \right)^2\\ &=(a+b+c)^2. \end{aligned} \] Consequently, \[ \frac ax+\frac by+\frac cz \ge \frac{(a+b+c)^2}{2\Delta}. \] Equality holds precisely when \[ \frac{\sqrt{a/x}}{\sqrt{ax}} =\frac{\sqrt{b/y}}{\sqrt{by}} =\frac{\sqrt{c/z}}{\sqrt{cz}}, \] which is equivalent to \(x=y=z\).

The incenter has equal distances from the three side lines, so it attains this bound. To verify uniqueness, suppose an interior point \(P\) satisfies \(x=y=z\). Put \[ \alpha=\angle BAP,\qquad \beta=\angle PAC. \] Then \[ z=AP\sin\alpha,\qquad y=AP\sin\beta. \] Hence \(\sin\alpha=\sin\beta\). Because \(\alpha,\beta>0\) and \(\alpha+\beta=\angle BAC<\pi\), this forces \(\alpha=\beta\). Thus \(P\) lies on the internal bisector of angle \(A\). The same argument at \(B\) shows that \(P\) lies on its internal bisector as well. Their unique intersection is the incenter.

Therefore the incenter is the unique minimizing point.

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