Difference between revisions of "1997 JBMO Problems/Problem 1"
Rockmanex3 (talk | contribs) (Solution to Problem 1 -- Pigeonhole!) |
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== Problem == | == 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 <math>\frac{1}{8}</math>. | |
== Solution == | == Solution == | ||
| − | == See | + | <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. | ||
| + | |||
| + | <br> | ||
| + | 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>. | ||
| + | |||
| + | == See Also == | ||
{{JBMO box|year=1997|before=First Problem|num-a=2}} | {{JBMO box|year=1997|before=First Problem|num-a=2}} | ||
| − | [[Category:Intermediate | + | [[Category:Intermediate Geometry Problems]] |
Revision as of 23:20, 3 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 
.
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]](http://latex.artofproblemsolving.com/d/4/4/d446dd32115f03308d3342ede0186f379688fd1c.png)
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 
is 
, 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 .
See Also
| 1997 JBMO (Problems • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All JBMO Problems and Solutions | ||
