Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Problem)
((Previous edit was to slightly reword problem) Added solution.)
Line 3: Line 3:
  
 
==Solution==
 
==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>
 +
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.
  
  
 
{{IMO box|year=1961|num-b=3|num-a=5}}
 
{{IMO box|year=1961|num-b=3|num-a=5}}

Revision as of 01:23, 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

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1961_IMO_Problems/Problem_4&oldid=29998"