1961 IMO Problems/Problem 4

Revision as of 01:23, 8 February 2009 by Brut3Forc3 (talk | contribs) ((Previous edit was to slightly reword problem) Added solution.)


In the interior of 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 opposing edges of triangle $P_1P_2P_3$. Prove that among the ratios $\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}$ there exists one not larger than $2$ and one not smaller than $2$.


Since triangles $P_1P_2P_3$ and $PP_2P_3$ share the base $P_2Q_2$, we have $\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{P_1Q_1}{PQ_1}$, where $[ABC]$ denotes the area of triangle $ABC$. Similarly, $\frac{[PP_1P_3]}{[P_1P_2P_3]}=\frac{P_2Q_2}{PQ_2}, \frac{[PP_1P_2]}{[P_1P_2P_3]}=\frac{P_3Q_3}{PQ_3}$. Adding all of these gives $\frac{[PP_1P_3]}{[P_1P_2P_3]}+\frac{[PP_1P_2]}{[P_1P_2P_3]}+\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{P_2Q_2}{PQ_2}+\frac{P_3Q_3}{PQ_3}+\frac{P_1Q_1}{PQ_1}$, or $\frac{P_2Q_2}{PQ_2}+\frac{P_3Q_3}{PQ_3}+\frac{P_1Q_1}{PQ_1}=\frac{[PP_1P_2]+[PP_2P_3]+[PP_3P_1]}{[P_1P_2P_3]}=1$ We see that we must have at least one of the three fractions less than or equal to $\frac{1}{3}$, and at least one greater than $\frac{1}{3}$. These correspond to ratios $\frac{PP_i}{PQ_i}$ being less than or equal to $2$, and greater than or equal to $2$, respectively, so we are done.

1961 IMO (Problems) • Resources
Preceded by
Problem 3
1 2 3 4 5 6 Followed by
Problem 5
All IMO Problems and Solutions
Invalid username
Login to AoPS