Difference between revisions of "1961 IMO Problems/Problem 4"

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==Problem==
 
==Problem==
In the interior of [[triangle]] <math>ABC</math> a [[point]] <math>P</math> is given.  Let <math>Q_1,Q_2,Q_3</math> be the [[intersection]]s of <math>PP_1, PP_2,PP_3</math> with the opposing [[edge]]s of triangle <math>ABC</math>.  Prove that among the [[ratio]]s <math>\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}</math> there exists one not larger than <math>2</math> and one not smaller than <math>2</math>.
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In the interior of [[triangle]] <math>P_1P_2P_3</math> a [[point]] <math>P</math> is given.  Let <math>Q_1,Q_2,Q_3</math> be the [[intersection]]s of <math>PP_1, PP_2,PP_3</math> with the opposing [[edge]]s of triangle <math>P_1P_2P_3</math>.  Prove that among the [[ratio]]s <math>\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2},\frac{PP_3}{PQ_3}</math> there exists one not larger than <math>2</math> and one not smaller than <math>2</math>.
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==Solution==
 
==Solution==
 
{{solution}}
 
{{solution}}

Revision as of 01:07, 8 February 2009

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

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