IBM Research | Ponder This | May 2007 solutions
Skip to main content

May 2007

<<April May June>>

Ponder This Solution:

The answer is 3*sqrt(2)/4 = 1.06066+.

Suppose the unit cube has vertices (x,y,z) where x,y,z are in {0,1}. Then an example of a maximum size embedded square has corners (0,0,.75), (1,.25,1), (1,1,.25) and (0,.75,0).

We sketch a proof that this is largest possible embedded square. Let (r,s,t) be the center of the square. Let (r+a,s+b,t+c) and (r-a,s-b,t-c) be one pair of opposite corners of the square and let (r+x,s+y,t+z) and (r-x,s-y,t-z) be the other pair of opposite corners. Since these corners must lie within the unit cube it is easy to see that |a|,|b|,|c|,|x|,|y|,|z| <= .5 and that we may assume that r=s=t=.5. Furthermore the diagonals of a square meet at right angles so a*x+b*y+c*z=0. Also the diagonals of a square have equal length so a*a+b*b+c*c=x*x+y*y+z*z=d*d/4 (d is the length of the diagonal). We want the maximum value of d so that we can find a,b,c,x,y and z satisfying the above. If we let d=1.5 then setting (a,b,c) = (.5,.5,-.25) and (x,y,z) = (.5,-.25,.5) satisfies these constraints and produces the above example. It is easy although a bit tedious to verify that we can't do better.

I came up with this problem and solution myself but it seems to be fairly well known as "Prince Rupert's Problem".

If you have any problems you think we might enjoy, please send them in. All replies should be sent to: