Difference between revisions of "1997 JBMO Problems/Problem 1"
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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 1 == |
| − | == 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>. | ||
| + | |||
| + | == Solution 2 == | ||
| + | <asy> | ||
| + | draw((0, 0)--(10,0)--(10,10)--(0,10)--cycle); | ||
| + | draw((0, 2.5)--(10, 2.5),dotted); | ||
| + | draw((0, 5)--(10,5),dotted); | ||
| + | draw((0,7.5)--(10,7.5),dotted); | ||
| + | </asy> | ||
| + | We proceed in a similar way to Solution 1. | ||
| + | The maximum possible area of a triangle from three points on a rectangle is <math>\tfrac18</math>. However, for this to happen, all three points must be on an edge, so they are not inside the square. Thus, the area is less than <math>\tfrac18</math>.-Trex4days | ||
| + | |||
| + | == 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]] |
Latest revision as of 10:39, 30 May 2020
Contents
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 1
![[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 .
Solution 2
![[asy] draw((0, 0)--(10,0)--(10,10)--(0,10)--cycle); draw((0, 2.5)--(10, 2.5),dotted); draw((0, 5)--(10,5),dotted); draw((0,7.5)--(10,7.5),dotted); [/asy]](http://latex.artofproblemsolving.com/1/e/8/1e83b8c4f47901a1a5a8235084c3e553658ef52f.png)
We proceed in a similar way to Solution 1.
The maximum possible area of a triangle from three points on a rectangle is . However, for this to happen, all three points must be on an edge, so they are not inside the square. Thus, the area is less than .-Trex4days
See Also
| 1997 JBMO (Problems • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All JBMO Problems and Solutions | ||
