Laying it out
My neighbour invited me over to supper the other night. She covered the square table we were eating at with three 1m x 1m table cloths. That made me think. What is the largest square table that can be covered with three 1-meter squares?
Problem source: Dan’s problems.
September 26th, 2009 at 3:29 am
Imagine an inverted T such that each table cloth’s bottom left corner is at (0,0), (1,0), and (0.5,1).
Now place a square within these such that two of it’s sides touch this inverted T at (0.5,1) and (1.5,1).
Now consider just the section of the “table” that sits on the top square defined by (0.5,1), (1.5,1), (0.5,2), (1.5,2) {call this square R}.
A portion of that square table is a triangle at the bottom of R.
That triangle has one known side – they hypotenuse – and two equal legs.
2a^2 = 1^2
a^2 = 1/2
a = ?2/2
Now bisect that triangle with a vertical line right down the middle. Again we have a 45-45-90 right triangle, this time with two legs of 1/2 and they hypotenuse of ?2/2. Toss out the hypotenuse since we no longer need it. Consider the leg of 1/2. This forms 1/3 of the diagonal of this table. The rest of the diagonal is simply the height of the lower 2 squares.
So the full diagonal of the table is 1.5.
The largest table that can be covered (in a very ugly fashion) by the tablecloths has sides measuring (3?2)/4 meters. 1.125 square meters.
There is a lot of wasted tablecloth hanging off the edges. I believe there may be a better way found by overlapping the tablecloths but I can’t spot it.
September 28th, 2009 at 12:03 pm
Just continuing this very fun problem a bit…
Consider square “R” above. Its portion of the table is a triangle — a right triangle with hypotenuse of 1 . Let one angle be x. With the hypotenuse = 1, the two other legs are cos(x) and sin(x).
The height of this triangle is [sin(2x)]/2, giving an area for this triangle of [sin(2x)]/4.
It makes logical sense that this area will be maximized at x=45º but it’s always best to verify it. Graphically, this does peak at (45,25).
With this triangle being maximized, the rest of the square can only be drawn one way and that is as it is described in my earlier post.
September 28th, 2009 at 4:07 pm
Let the table measure t by t metres. Place tablecloth ABCD with AB along y=0 and AD along x=0. Place the other two tablecloths each with one corner at (t, t); EFGH with E at (t, t) and EH turned clockwise by some angle theta from the vertical, so that FG goes through Y = (0, t); IJKL with I at (t, t) and IJ turned anticlockwise by theta from the horizontal, so that KL goes through (t, 0).
Then for FG to pass through Y = (0, t),
cos theta = 1/t
(Consider triangle EFY, right-angled at F.)
Then for EFGH and IJKL to cover (1+e, 1+e) where e is small and positive, theta must be less than 45 degrees. The bigger theta is, the bigger t is, so we maximise theta subject to the other constraint, which is that EFGH must cover (0, 1+e) where e is small and positive. The largest value of theta that satisfies this constraint is the one where GH goes through D = (0, 1). Another way to put this is that the line through D parallel to GH goes through H.
cos² theta = 1/t²
sin² theta = 1 – 1/t²
sin theta = sqrt(1 – 1/t²)
= sqrt(t²-1)/t
tan theta = sin theta / cos theta
= sqrt(t²-1)
E = (t, t)
vector EH = (- sin theta, – cos theta)
=> H = (t – sin theta, t – cos theta).
The line through D parallel to GH is
y = 1 – (tan theta) x ___ [1]
For this line to go through H,
t – cos theta = 1 – (tan theta)(t – sin theta)
= 1 – (tan theta)t + (tan theta)(sin theta)
= 1 – (tan theta)t + ((sin theta)/(cos theta))(sin theta)
= 1 – (tan theta)t + (sin² theta)/(cos theta)
=> t – 1/t = 1 – sqrt(t²-1)t + (1 – 1/t²)/(1/t)
= 1 – sqrt(t²-1)t + t – 1/t
=> sqrt(t²-1)t – 1 = 0
=> (t²-1)t² – 1 = 0
=> t^4 – t² – 1 = 0
=> t² = phi and t = sqrt(phi) (as t must be greater than 1)
Thus the area of the largest square table that can be covered is phi ~~ 1.6180339887499, and the sides of that table are of length sqrt(phi) ~~ 1.27201964951407.