Problem 4

Inside triangle

P_1P_2P_3

a point

P

is given. Let

Q_1,Q_2,Q_3

be the intersections of

PP_1,PP_2,PP_3

with the opposite sides. Prove that among

\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}

there is one not larger than

2

and one not smaller than

2

Step 1 of 5: Convert each cevian to an area ratio

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In plain words

Triangles sharing a base have areas proportional to their altitudes, and points on one cevian give a linear distance ratio.

r_1=\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{PQ_1}{P_1Q_1},\quad r_2=\frac{PQ_2}{P_2Q_2},\quad r_3=\frac{PQ_3}{P_3Q_3}

Detailed analysis

Because $P,Q_1,P_1$ are collinear and triangles $PP_2P_3$ and $P_1P_2P_3$ share base $P_2P_3$ , their area ratio equals $PQ_1/P_1Q_1$ . Define $r_1$ by this ratio; the other two identities follow cyclically.