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

((Previous edit was to slightly reword problem) Added solution.)
(→‎Solution: Fixed errors which have gone unnoticed for >6 years)
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==Solution==
 
==Solution==
−
Since triangles <math>P_1P_2P_3</math> and <math>PP_2P_3</math> share the base <math>P_2Q_2</math>, we have <math>\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{P_1Q_1}{PQ_1}</math>, where <math>[ABC]</math> denotes the area of triangle <math>ABC</math>. Similarly, <math>\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}</math>. Adding all of these gives <math>\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}</math>, or <math>\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</math>
+
Let <math>[ABC]</math> denote the area of triangle <math>ABC</math>.
−
We see that we must have at least one of the three fractions less than or equal to <math>\frac{1}{3}</math>, and at least one greater than <math>\frac{1}{3}</math>. These correspond to ratios <math>\frac{PP_i}{PQ_i}</math> being less than or equal to <math>2</math>, and greater than or equal to <math>2</math>, respectively, so we are done.
+
 
 +
Since triangles <math>P_1P_2P_3</math> and <math>PP_2P_3</math> share the base <math>P_2P_3</math>, we have <math>\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{PQ_1}{P_1Q_1}</math>.
 +
 
 +
Similarly, <math>\frac{[PP_1P_3]}{[P_1P_2P_3]}=\frac{PQ_2}{P_2Q_2}, \frac{[PP_1P_2]}{[P_1P_2P_3]}=\frac{PQ_3}{P_3Q_3}</math>.
 +
 
 +
Adding all of these gives <math>\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{PQ_2}{P_2Q_2}+\frac{PQ_3}{P_3Q_3}+\frac{PQ_1}{P_1Q_1}=1</math>.
 +
 
 +
We see that we must have at least one of the three fractions not greater than <math>\frac{1}{3}</math>, and at least one not less than <math>\frac{1}{3}</math>. These correspond to ratios <math>\frac{PP_i}{PQ_i}</math> being less than or equal to <math>2</math>, and greater than or equal to <math>2</math>, respectively, so we are done.
  
  
 
{{IMO box|year=1961|num-b=3|num-a=5}}
 
{{IMO box|year=1961|num-b=3|num-a=5}}

Revision as of 05:42, 18 December 2015

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

Let $[ABC]$ denote the area of triangle $ABC$.

Since triangles $P_1P_2P_3$ and $PP_2P_3$ share the base $P_2P_3$, we have $\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{PQ_1}{P_1Q_1}$.

Similarly, $\frac{[PP_1P_3]}{[P_1P_2P_3]}=\frac{PQ_2}{P_2Q_2}, \frac{[PP_1P_2]}{[P_1P_2P_3]}=\frac{PQ_3}{P_3Q_3}$.

Adding all of these gives $\frac{[PP_1P_3]}{[P_1P_2P_3]}+\frac{[PP_1P_2]}{[P_1P_2P_3]}+\frac{[PP_2P_3]}{[P_1P_2P_3]}=\frac{PQ_2}{P_2Q_2}+\frac{PQ_3}{P_3Q_3}+\frac{PQ_1}{P_1Q_1}=1$.

We see that we must have at least one of the three fractions not greater than $\frac{1}{3}$, and at least one not less 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
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