Square Division

Show how to divide a unit square into two rectangles so that the smaller rectangle can be placed on the larger with every vertex of the smaller on exactly one of the edges of the larger.

Source: 1994 Dutch Mathematical Olympiad; as reported in Crux Mathematicorum, Sept 1998, Vol. 24, No. 5, p. 264.

This problem was sent recently as a Macalaster Problem of the Week (#887). The solution was sent with the following week's puzzle, but no derivation was included, so I thought it was worth seeing if some of my readers could derive the solution.

Solutions were received from many people: Larry Baum, Philippe Fondanaiche, Kobskii Ha, Denis Borris, Mit Ratilal Shah. A nice trigonometric solution was provided by Mit: 
If the square of side a is cut into two parts at length x, then we have
to determine x.

If a-x denotes the smaller side of the rectangle enclosed by the other
rectangle, then we get the following two equations :

x = (a-x)cosA + asinA
a = (a-x)sinA + acosA

Solving, we find that sinA = 1/2

And therefore x = (sqroot(3) - 1)*a

[To do this geometrically, construct an adjacent identical square.
With one of its outside vertices as center, construct an arc of length 2a.
This will intersect the original square to give a-x & x. - KD]
Larry Baum sent a general solution for any orignally kx1 rectangle: 
Start with a rectangle of size kx1 and cut it into two rectangles ax1
and (k-a)x1 where a < k/2.

To positon the smaller one on the larger, we need b and c such that:

I)   b^2 + c^2 = a^2
II)  b/a = (1-c)/1
III) (k-a-b)^2 + (1-c)^2 = 1

I & II => b=2a^2/(1+a^2), c=a(1-a^2)/(1+a^2)

Substituting into III: 
[(k-1) + (1-a)^2/(1+a^2)]^2 + [(1-a^2)/(1+a^2)]^2 = 1
[(k-1)(1+a^2) + (1-a)^2]^2 + (1-a^2)^2 = (1+a^2)^2
[ka^2 - 2a + k]^2 = 4a^2

Since b > 0, the expression inside the brackets is positive, so taking
positive square roots:
ka^2 - 2a + k = 2a
ka^2 - 4a + k = 0

a = 2 +- sqrt(4-k^2)
a < k/2 => a=2 - sqrt(4-k^2)  -- general solution (Note k cannot be > 2)
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