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  • 5 years ago

Given a unit square, show that if five points are placed anywhere inside or on this square, then two of them must be at most sqrt(2)/2 units apart. @MIT 6.00 Intro Co…

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  1. anonymous
    • 5 years ago
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    Not exactly sure what the problem is asking. But if you have a 1x1 square, and place five points, one at each corner, and one at the center. The ones at the corners adjacent to each other are at 1 unit of distance from each other, The ones at corners opposite of them are at sqrt(2) units of distance from each other, and every point is at sqrt(2)/2 units from distance from a fifth point at the center. |----1 unit----| X________X _ | | | | x | 1 unit | | | X________X - sqrt(2) from corner to corner sqrt(2)/2 from corners to center This is the farthest away you can place them from each other. If the problem asks for (a minimum of) two points to be at most at sqrt(2)/2 units, the statement holds true for every combination of points. You could move some closer to move one farther away, but that makes two points closer, thus proving the restriction right. XXXX_____ | | | | | | _________X X___X____X | | X | | | _________X

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