Consider a rectangle of length 4 and width 1 Show that given

Consider a rectangle of length 4 and width 1. Show that given any 5 points in this rectangle, there must be at least one pair that are of distance at most squareroot 2 from each other. Show that given any 9 points in this rectangle, there must be at least one pair that are of distance at most ^ from each other. How many ways can two adjacent squares be selected form an 8-by-8 chess board if we allow the squares to align vertically, or align horizontally, or align northwest-southeast, but not northeast-southwest.

Solution

If 5 points are there in the recatngle we need to ensure that the points are placed as farther apart from each other as possible. And inspite of placing them as farther apart as possible if still the maximum separation between any 2 points is at most root 2 then we can say that there will be at least 1 pair with at most root 2 separation distance.

To place 5 points on a rectangle as farther apart as possible: Start with a point on bottom left vertice, next point on top line, 3rd point on the mid point of the base fourth point draw it in a similar fashion as you drew 2nd point with respect to the first point and the fifth point will be on the bottom right vertice; Attaching the figure as below, notice that within hte rectangle 2 mountain like shapes are beign formed by joining these lines;

Here are 2 rectanlges with length = 4 and breadth =1
In figure 1; AC=2 and BD=2 so and Angle (CAB=45) so AB cos45= AC/2=1 so AB= 1/cos45= root 2; Similarly BC=CD=DE= root 2; So in this rectangle there will be at least 1 pair taht are at most root 2 distance apart.


Draw the next figure similarly as you drew the figure 1, try and accomodate 5 points on the base at equal itnervals and 4 points on the top line of the rectangle at equidistat intervals;
In figure 2: PR=RT=TY=VX=1 and QS=SU=UW=1 ; In triangle PQR the height is = height of rectangle = 1 and let us drop a perpendicular from Q on PR and call it point Z; so QZ=height of rectangle = 1 and PZ= PR/2= 1/2;
Since PQZ will be a right angled traignle we can say taht PQ^2= PZ^2 + QZ^2= (0.5)^2 + (1)^2 = 5/4 thus PZ=root (5/4) = root(5) / 2
So in this rectangle there will be at least 1 pair that are at most root(5) / 2 distance apart.

For chess problem:
Horizontal alignment: In first row you can select 7 such pairs, A1-A2, A2-A3 ..and till A7-A8; Thus 7 pairs of for each row, Total pairs =7*8=56 pairs for horizontal alignment;
Vertical alignment: In first column you can select 7 such pairs; A1-B1, B1-C1 and ... till G1-H1; and similarly 7 pairs for each of the 8 columns thus 7*8=56 pairs for vertical alignment;

For northwest -southeast alignment:
Start with the northwest-southeast diagonal and we get 7 such pairs on the diagonal from A1-B2, B2-C3 and so on till G7-H8; thus 7 pairs
Now go on either side of the main diagonal to smaller crossed lines and for either side you will get 6 pairs; from A2-B3, B3-C4 and so on till F7-G8 for the first diagonal to B1-C2, C2-D3 and so on till G6-H7 so 6 more
Thus for nothwest southeast alignment you get = 7+6+6+5+5+4+4+3+3+2+2+1+1= 49

So total alignments = 56+56+ 49= 161