IMO 1993 · Problem 4

For three points \(P,Q,R\) in the plane, we define \(m(PQR)\) as the minimum length of the three altitudes of \(\triangle PQR\). (If the points are collinear, we set \(m(PQR)=0\).)

Prove that for points \(A,B,C,X\) in the plane, \[ m(ABC)\leq m(ABX)+m(AXC)+m(XBC). \]

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

Solution

Write \([T]\) for the area of a triangle \(T\). If \(T\) is nondegenerate and its longest side has length \(L\), then \[ m(T)=\frac{2[T]}{L}. \]

We first prove that if a triangle \(T\) is contained in a triangle \(T'\), then \(m(T)\leq m(T')\).

To do so, consider any nondegenerate triangle \(PQR\) contained in a strip of width \(w\). Label its vertices so that \(P\) and \(R\) have the smallest and largest coordinates in the direction perpendicular to the strip. Let their separation in that direction be \(h\leq w\). The line through \(Q\) parallel to the strip meets \(PR\) at a point \(S\). Splitting the triangle along \(QS\) gives \[ 2[PQR]=h\cdot QS. \] Since \(S\) lies on \(PR\), write \(S=(1-t)P+tR\), where \(0\leq t\leq1\). The triangle inequality gives \[ QS\leq(1-t)QP+tQR\leq\max(QP,QR). \] Consequently, if \(L\) is the longest side of \(PQR\), \[ m(PQR)=\frac{h\cdot QS}{L}\leq h\leq w. \] This also holds for degenerate triangles, whose \(m\)-value is zero.

Now a nondegenerate triangle \(T'\) lies in the strip bounded by its longest side and the parallel line through its opposite vertex. This strip has width \(m(T')\), so every triangle \(T\subseteq T'\) satisfies \(m(T)\leq m(T')\). If \(T'\) is degenerate, so is \(T\), proving the claim in all cases.

We turn to the required inequality. If \(A,B,C\) are collinear, its left-hand side is zero, so it holds. Henceforth assume that \(ABC\) is nondegenerate, and put \[ L=\max(AB,BC,CA). \]

First suppose that \(X\) lies in the closed triangle \(ABC\). Every side of \(ABX,AXC,XBC\) has length at most \(L\). Indeed, writing \[ X=\alpha A+\beta B+\gamma C,\qquad \alpha,\beta,\gamma\geq0,\qquad \alpha+\beta+\gamma=1, \] we obtain \[ AX\leq\beta AB+\gamma AC\leq L, \] and similarly \(BX,CX\leq L\). Thus, using the area formula above, with degenerate subtriangles contributing zero, \[ \begin{aligned} m(ABX)+m(AXC)+m(XBC) &\geq \frac{2[ABX]+2[AXC]+2[XBC]}{L}\\ &=\frac{2[ABC]}{L}\\ &=m(ABC). \end{aligned} \]

It remains to consider \(X\) outside \(ABC\). If one of \(A,B,C\) lies in the closed triangle formed by the other three points, the containment claim proves the result immediately. For example, if \(C\in\triangle ABX\), then \[ \triangle ABC\subseteq\triangle ABX, \qquad m(ABC)\leq m(ABX), \] which is sufficient.

Otherwise, none of the four points lies in the triangle formed by the other three. Their convex hull is therefore a convex quadrilateral. Relabel \(A,B,C\), if necessary, so that its vertices occur in the order \(A,B,X,C\). Its diagonals \(AX\) and \(BC\) intersect at a point \(Y\).

Apply the already proved case to \(Y\in\triangle ABC\). Since \(B,Y,C\) are collinear, \[ m(ABC)\leq m(ABY)+m(AYC). \] Moreover, \(Y\in AX\), so \[ \triangle ABY\subseteq\triangle ABX, \qquad \triangle AYC\subseteq\triangle AXC. \] The containment claim now yields \[ m(ABC) \leq m(ABX)+m(AXC) \leq m(ABX)+m(AXC)+m(XBC). \] All configurations have been covered. \(\square\)

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