MHB What is the maximum distance between two points in a square of side length 1?

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    2016
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In a square with a side length of 1, it can be shown that among any five points, at least one pair will be no further than √2/2 apart. This conclusion is derived from the pigeonhole principle, which indicates that the square can be divided into smaller sections where points must fall into the same section. The correct solutions to this problem were provided by users greg1313 and Fallen Angel. Their contributions highlight the mathematical reasoning behind the distance constraint. Understanding this concept is essential for solving similar geometric problems.
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Here is this week's POTW:

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Show that if $5$ points are all in, or on, a square of side length $1$, then some pair of them will be no further than $\dfrac{\sqrt{2}}{2}$ apart.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to greg1313 and Fallen Angel for their correct solutions. greg1313's solution follows:

Construct four quarter circles with radius $\dfrac{\sqrt2}{2}$, each with its centre on a vertex of the square. Notice that there is no area on the square that is not contained by, or on, a quarter circle. Hence one cannot place more that four points in or on the square without such a point being at most $\dfrac{\sqrt2}{2}$ from another point.
 

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