Problem

In the interior of \Delta P_1 P_2 P_3 a point P is given. Let Q_1,Q_2,\text{ and }Q_3 respectively be the intersections of PP_1,PP_2,PP_3 with the opposing edges of \Delta P_1 P_2 P_3. Prove that among the ratios PP_1/PQ_1,PP_2/PQ_2,\text{ and }PP_3/PQ_3 there exists at least one not larger than 2 and at least one not smaller than 2.

Solution

To simplify, let i be a generic index from 0,1,2, and j=i+1, k=i-1, both modulo 3.

At first sight, we can employ van Obel's theorem:

{PP_i}/{PQ_i}={P_i Q_k}/{P_j Q_k} + {P_i Q_j}/{P_k Q_j}. Let's denote {P_i Q_k}/{P_j Q_k} as x_{ikj}. (Note that x_{ikj}=x_{jki}.)

For, summing for i=0,1,2 we'll get the ratios on the right and their inverses, so that

{PP_1}/{PQ_1}+{PP_2}/{PQ_2}+{PP_3}/{PQ_3}=x_{132}+1/x_{321}+x_{213}+1/x_{132}+x_{321}+1/x_{213}\ge 6

because for a positive x, x+1/x\ge 2. It follows that at least one of the ratios in question is not less than 2. It does not seem possible to derive from here that at least one of the ratios is not greater than 2. So, we'll have to try another approach.

van Obel's theorem is intimately related to the barycentric coordinates. The three coordinates are defined by the ratios

{PQ_1}/{P_1Q_1},{PQ_2}/{P_2Q_2}, \text{and} {PQ_3}/{P_3Q_3}.

The sum of the three coordinates is always 1 because, say,

{PQ_1}/{P_1Q_1} = Area(\Delta P P_2 P_3) / Area(\Delta P_1 P_2 P_3).

Thus at least one of the coordinates is not greater than 1/3 and at least one of the coordinates is not less than 1/3.

Assume {PQ_1}/{P_1Q_1} ≥ 1/3, then {P_1Q_1}/{PQ_1} = {P_1Q_1}/{PQ_1}=(PP_1 + PQ_1)/PQ_1 ≤ 3, implying {PP_1}/{PQ_1}≤2, as required.