Difference between revisions of "1997 JBMO Problems/Problem 1"

(Solution to Problem 1 -- Pigeonhole!)
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== Problem ==
 
== Problem ==
  
''(Bulgaria)'' Show that given any 9 points inside a square of side length 1 we can always find 3 that form a triangle with area less than <math>\frac{1}{8}</math>
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Show that given any 9 points inside a square of side length 1 we can always find 3 that form a triangle with area less than <math>\frac{1}{8}</math>.
  
 
== Solution ==
 
== Solution ==
  
== See also ==
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<asy>
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draw((0,0)--(10,0)--(10,10)--(0,10)--cycle);
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draw((5,0)--(5,10),dotted);
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draw((0,5)--(10,5),dotted);
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</asy>
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Divide up the square into four congruent squares, as seen in the diagram.  By the [[Pigeonhole Principle]], at least one square has at least three points.
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<br>
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The maximum possible area of a triangle from three points on the square with side length <math>\tfrac12</math> is <math>\tfrac18</math>, and this is achieved when two of the points are on two adjacent corners and the third one is on the opposite side.  However, since only one corner is truly ''inside'' the square, only one of the points can be on the corner, so the area of the triangle would be less than <math>\tfrac18</math>.
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== See Also ==
  
 
{{JBMO box|year=1997|before=First Problem|num-a=2}}
 
{{JBMO box|year=1997|before=First Problem|num-a=2}}
  
[[Category:Intermediate Combinatorics Problems]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 00:20, 4 August 2018

Problem

Show that given any 9 points inside a square of side length 1 we can always find 3 that form a triangle with area less than $\frac{1}{8}$.

Solution

[asy] draw((0,0)--(10,0)--(10,10)--(0,10)--cycle); draw((5,0)--(5,10),dotted); draw((0,5)--(10,5),dotted); [/asy] Divide up the square into four congruent squares, as seen in the diagram. By the Pigeonhole Principle, at least one square has at least three points.


The maximum possible area of a triangle from three points on the square with side length $\tfrac12$ is $\tfrac18$, and this is achieved when two of the points are on two adjacent corners and the third one is on the opposite side. However, since only one corner is truly inside the square, only one of the points can be on the corner, so the area of the triangle would be less than $\tfrac18$.

See Also

1997 JBMO (ProblemsResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5
All JBMO Problems and Solutions
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