IMO 1961 · Problem 4

Consider triangle \(P_1P_2P_3\) and a point \(P\) within the triangle. Lines \(P_1P, P_2P, P_3P\) intersect the opposite sides in points \(Q_1, Q_2, Q_3\) respectively. Prove that, of the numbers \[ \frac{P_1P}{PQ_1},\ \frac{P_2P}{PQ_2},\ \frac{P_3P}{PQ_3} \] at least one is \(\leq 2\) and at least one is \(\geq 2\).

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

Solution

Let \([ABC]\) denote the area of triangle \(ABC\), and put \[ S=[P_1P_2P_3],\qquad S_1=[PP_2P_3],\quad S_2=[PP_3P_1],\quad S_3=[PP_1P_2]. \] Since \(P\) is interior to the triangle, these three smaller triangles have positive areas and partition the original triangle. Hence \[ S_1+S_2+S_3=S. \]

For each \(i\), the points \(P_i,P,Q_i\) lie in that order on a line. The triangle of area \(S_i\) and the original triangle share the side opposite \(P_i\) as a base. Their perpendicular heights to this base are proportional to \(PQ_i\) and \(P_iQ_i\), respectively: both heights equal the corresponding segment length multiplied by the sine of the angle that \(P_iQ_i\) makes with the base. Thus \[ \frac{S_i}{S}=\frac{PQ_i}{P_iQ_i}. \] Using \(P_iQ_i=P_iP+PQ_i\), we obtain \[ \frac{P_iP}{PQ_i} =\frac{P_iQ_i}{PQ_i}-1 =\frac{S}{S_i}-1. \]

Because the three positive numbers \(S_1,S_2,S_3\) sum to \(S\), at least one is at least \(S/3\), and at least one is at most \(S/3\). For the first of these, the displayed identity gives \[ \frac{P_iP}{PQ_i}\leq 3-1=2; \] for the second, it gives \[ \frac{P_iP}{PQ_i}\geq 3-1=2. \] These inequalities include the equality cases and prove the assertion. \(\square\)

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