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1961 IMO Problems/Problem 4

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Problem

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 $2$ .

Solution

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 $\frac{[PP_1P_3]}{[P_1P_2P_3]}+\frac{[PP_1P_2]}{[P_1P_2P_3]}+\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{P_2Q_2}{PQ_2}+\frac{P_3Q_3}{P...$ , 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)
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5

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Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.

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