Given 5 points in the interior of a square of side length one. Prove at least two two of the points are less than
sq. rt. of 2/2
units apart. Thx
2007-09-19 08:51:17 · 7 answers · asked by Meow* 2 in Science & Mathematics ➔ Mathematics
The diagonal of the square has to be the square root of 2.
This results from the fact that this diagonal is the hypotenuse of a triangle with two catheti of 1. According to the Pythagoras theroem this hypotenuse has to be sqrt of 2 then: Pytagoras theorem: a^2 + b^2 = c^2 with c for the hypotenuse. In this case we have 1^2 + 1^2= c^2 with a and b = 1 ---> we have 2=c^2 ---> c=sqrt of 2= 1,4...
The maximum distance in this square - the hypotenuse - is therefore sqrt 2. As all points are in the interior of the square (not at the edqe) every distance between two points has to be below sqrt 2, because all other distances have to be shorter!
2007-09-19 09:09:20 · answer #1 · answered by hollunder 5 · 0⤊ 0⤋
The diagonals are each sqrt(2) in length
If you put 4 points just inside each of the 4 corners, then the distance between two adjacent points will be slightly less than 1 and the ditance between diagonally opposite sides will be just under sqrt(2). But when you now place the fifth point, you can't find a place to put it so that it its distance to any of the 4 points is =/>sqrt(2).
If you put in at the intersection of the diagonals or anywhere on either diagonal it will automatically be within a distance less than sqrt(2)/2 from one of the other points.
If you put the point in any ofthe 4 triangles formed by the diagonals and the sides of the squares, the point must be closer to one the two points located near the corners than Sqrt(2)/2.
Draw the unit square and the diagonals which = sqrt(2) and half diagonal = sqrt(2)/2 and you will quickly see this is true
2007-09-19 16:21:55 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
Divide the square into 4 squares of side length 1/2. Now since you have 5 points and 4 smaller squares, any placement of the 5 point must put at least 2 in the same square, call these points x and y. Since x and y lie in the same square of side length 1/2, they can be at most sqrt(2)/2 apart (since sqrt(2)/2 is the length of the diagonal of the smaller squares).
2007-09-19 16:06:24 · answer #3 · answered by Phineas Bogg 6 · 0⤊ 0⤋
The length of the diagonal of the unit square equals sqrt 2
Now as mentioned the points are in the interior of the square and do not lie on the side. And we also know the longest line that can be drawn within the square is its diagonol.
So anything inside the square will have the distance of less than sqrt 2 between them
2007-09-19 16:00:19 · answer #4 · answered by Varshita 3 · 1⤊ 0⤋
The points must be placed on the interior, so imagin you have a slightly smaller square(side<1) If you place 4 of the points at each corner(which is the furthest you can place them from each other) the fifth point would have to be in the center, to be as far as possible from all the other points. and since the side of the square is less than 1, the distance to each of these points would be less than root(2)/2
2007-09-19 15:57:56 · answer #5 · answered by unonimy 2 · 1⤊ 0⤋
The distance between ANY two points is bounded!
Draw the square (0,0), (0,1),(1,0) and (1.1)
The points that will be farthest apart from each other will be located at the diagonals, say (0,0) and (1,1).
What's the distance between them?
By the pythagorean theorem, it's sqrt(2). This means NO two points are more than sqrt(2) apart.
To find two points less than sqrt(2)/2 apart, I suggest you find the intersectin of the two diagonals, it's (1/2, 1/2), and find it's distance to the corner (1,1).
2007-09-19 16:01:31 · answer #6 · answered by pbb1001 5 · 0⤊ 0⤋
can the square be plotted in negative cooridinates?
2007-09-19 15:57:50 · answer #7 · answered by Kyle C 1 · 0⤊ 0⤋