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

Latest revision as of 10:39, 30 May 2020

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 1

$[asy] draw((0,0)--(10,0)--(10,10)--(0,10)--cycle); draw((5,0)--(5,10),dotted); draw((0,5)--(10,5),dotted); [/asy]$ 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 $\tfrac18$ .

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 $\tfrac18$ . 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 $\tfrac18$ .-Trex4days

1997 JBMO (Problems • Resources)
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Difference between revisions of "1997 JBMO Problems/Problem 1"

Latest revision as of 10:39, 30 May 2020

Contents

Problem

Solution 1

Solution 2

See Also

@@ Line 17: / Line 17: @@
 == Solution 2 ==
 <asy>
-draw((0, 0--(10,0)--(10,10)--(0,10)--cycle);
+draw((0, 0)--(10,0)--(10,10)--(0,10)--cycle);
 draw((0, 2.5)--(10, 2.5),dotted);
 draw((0, 5)--(10,5),dotted);